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. Calculate the least-squares line. Put the equation in the form ŷ = a + bx. e. Find the correlation coefficient. What does it imply about the significance of the relationship? f. Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas? g. Use the two points in Part F to plot the least-squares line on your graph from Part B. h. Does it appear that a line is the best way to fit the data? Why or why not? i. Are there any outliers? j. Use the least-squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not? k. Delete Hawaii and substitute Alaska for it. Alaska is a state with an area of 656,424 square miles. l. Calculate the new least-squares line. m. Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case? n. Do you think that, in general, newer states are larger than the original states? SOLUTIONS 1 dependent variable: fee amount independent variable: time This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 | Linear Regression and Correlation 751 3 Figure 12.25 5 Figure 12.26 7 y = 6x + 8, 4y = 8, and y + 7 = 3x are all linear equations. 9 The number of flu cases depends on the year. Therefore, year becomes the independent variable and the number of flu cases is the dependent variable. 11 The y-intercept is 50 (a = 50). At the start of the cleaning, the company charges a one-time fee of $50 (this is when x = 0). The slope is 100 (b = 100). For each session, the company charges $100 for each hour they clean. 13 12,000 lb of soil 15 The slope is –1.5 (b = –1.5). This means the stock is losing value at a rate of $1.50 per hour. The y-intercept is $15 (a = 15). This means the price of stock before the trading day was $15. 752 17 Chapter 12 | Linear Regression and |
Correlation Group x (no. of hours spent studying) y (final exam grades) Median x value Median y value Table 12.38 19 ŷ = 2.23 + 1.99x 45 50 51 65 72 80 90 96 2 4.5 7 50 68.5 90 21 The slope is 1.99 (b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another $1.99 million in pay each year. 23 It means that there is no correlation between the data sets. 25 Yes. There are enough data points and the value of r is strong enough to show there is a strong negative correlation between the data sets. 27 Ha: ρ ≠ 0 29 $250,120 31 1326 acres 33 1125 hours, or when x = 1125 35 Check student solution. 36 a. When x = 1985, ŷ = 25,52. b. When x = 1990, ŷ = 34,275. c. When x = 1970, ŷ = –725. Why doesn’t this answer make sense? The range of x values was 1981 to 2002; the year 1970 is not in this range. The regression equation does not apply, because predicting for the year 1970 is extrapolation, which requires a different process. Also, a negative number does not make sense in this context, when we are predicting flu cases diagnosed. 38 Also, the correlation r = 0.4526. If r is compared with the value in the 95 Percent Critical Values of the Sample Correlation Coefficient Table, because r > 0.423, r is significant, and you would think that the line could be used for prediction. But, the scatter plot indicates otherwise. 39 Check student’ solution. 40 ŷ = 3,448,225 + 1750x 42 There was an increase in flu cases diagnosed until 1993. From 1993 through 2002, the number of flu cases diagnosed declined each year. It is not appropriate to use a linear regression line to fit to the data. 44 Because there is no linear association between year and number of flu cases diagnosed, it is not appropriate to calculate a linear correlation coefficient. When there is a linear association and it is appropriate to calculate a correlation, we cannot say that one variable causes the other variable. 46 We don’t know if the pre-1981 data were collected from a single year. So, we don’t have an accurate x value for this figure. Regression equation |
: ŷ (number of flu cases) = –3,448,225 + 1749.777 (year). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 | Linear Regression and Correlation 753 Coefficients Intercept –3,448,225 x Variable 1 1,749.777 Table 12.39 47 • a = –3,488,225 • b = 1,750 • correlation = 0.4526 • n = 22 48 No, he is not correct. An outlier is only an influential point if it significantly impacts the slope of the least-squares regression line and the correlation coefficient, r. If omission of this data point from the calculation of the regression line does not show much impact on the slope or r-value, then the outlier is not considered an influential point. For different reasons, it still may be determined that the data point must be omitted from the data set. 49 Yes. There appears to be an outlier at (6, 58). 51 The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate as a predictor for the data. 53 s = 1.75 55 a. b. c. d. e. independent variable: age; dependent variable: fatalities independent variable: number of family members; dependent variable: grocery bill independent variable: age of applicant; dependent variable: insurance premium independent variable: power consumption; dependent variable: utility independent variable: higher education (years); dependent variable: crime rates 58 It means that 72 percent of the variation in the dependent variable (y) can be explained by the variation in the independent variable (x). 754 60 Chapter 12 | Linear Regression and Correlation x (SAT math scores) y (GPAs) 261 262 327 350 363 364 373 544 587 624 741 Table 12.40 50 60 62 65 70 67 71 86 87 90 98 We must remember to check the order of the y values within each group as well. We notice that the y values in the second group are not in order from the least value to the greatest value; these values thus must be reordered, meaning the median y value for that group is 70. Group x (SAT math scores) y (GPAs) Median x value Median y value 1 2 3 261 262 327 350 363 364 373 544 587 624 7 |
41 Table 12.41 50 60 62 65 67 70 71 86 87 90 98 294.5 364 61 70 605.5 88.5 The ordered pairs are (294.5, 61), (364, 70), and (605.5, 88.5). The slope can be calculated using the formula m = y3 − y1 x3 − x1. Substituting the median x and y values, from the first and third groups gives m = 88.5 − 61 605.5 − 294.5, which simplifies to m ≈ 0.09. The y-intercept may be found using the formula b = ∑ y − m∑ x 3. The sum of the median x values is 1264, and the sum of the median y values is 219.5. Substituting these sums and the slope into the formula gives b = 219.5 − 0.09(1264) 3, which simplifies to b ≈ 35.25. The line of best fit is represented as y = mx + b. Thus, the equation can be written as y = 0.09x + 35.25. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 | Linear Regression and Correlation 755 61 b. Check student solution. c. ŷ = 35.5818045 – 0.19182491x d. r = –0.57874 For four degrees of freedom and alpha = 0.05, the LinRegTTest gives a p value of 0.2288, so we do not reject the null hypothesis; there is not a significant linear relationship between deaths and age. Using the table of critical values for the correlation coefficient, with four degrees of freedom, the critical value is 0.811. The correlation coefficient r = –0.57874 is not less than –0.811, so we do not reject the null hypothesis. f. There is not a linear relationship between the two variables, as evidenced by a p value greater than 0.05. 63 a. We wonder if the better discounts appear earlier in the book, so we select page as x and discount as y. b. Check student solution. c. ŷ = 17.21757 – 0.01412x d. r = – 0.2752 For seven degrees of freedom and alpha = 0.05, LinRegTTest gives |
a p value = 0.4736, so we do not reject; there is a not a significant linear relationship between page and discount. Using the table of critical values for the correlation coefficient, with seven gives degrees of freedom, the critical value is 0.666. The correlation coefficient xi = –0.2752 is not less than 0.666, so we do not reject. f. There is not a significant linear correlation so it appears there is no relationship between the page and the amount of the discount. As the page number increases by one page, the discount decreases by $0.01412. 65 a. Year is the independent or x variable; the number of letters is the dependent or y variable. b. Check student’s solution. c. No. d. ŷ = 47.03 – 0.0216x e. –0.4280. The r value indicates that there is not a significant correlation between the year the state entered the Union and the number of letters in the name. g. No. The relationship does not appear to be linear; the correlation is not significant. 66 Using LinRegTTest, the output for the original least-squares regression line is y = 26.14 + 0.7539x and r = 0.6657. The output for the new least-squares regression line, after omitting the outlier of (56, 95), is ŷ = 6.36 + 1.0045x and r = 0.9757. The slope of the new line is quite a bit different from the slope of the original least-squares regression line, but the larger change is shown in the r-values, such that the new line has an r-value that has increased to a value that is almost equal to one. Thus, it may be stated that the outlier (56, 95) is also an influential point. 68 a. and b. Check student solution. c. The slope of the regression line is –0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32 percent new sparrow hawks, which doesn’t make sense, because if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the |
percentage of new birds in the colony decreases by 30.3 percent. d. If we examine r2, we see that only 57.52 percent of the variation in the percentage of new birds is explained by the model and the correlation coefficient, r = –.7584 only indicates a somewhat strong correlation between returning and new percentages. e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66 percent, our observed new percentage, 6 percent, is almost 6 percent less than the predicted new value of 11.98 percent. If we remove this data pair, we see only an adjusted slope of –0.2789 and an adjusted intercept of 30.9816. In other words, although these data generate the largest residual, it is not an outlier, nor is the data pair an influential point. f. If there are 70 percent returning birds, we would expect to see y =– 0.2789(70) + 30.9816 = 0.114 or 11.4 percent new birds in the colony. 756 70 a. Check student solution. b. Check student solution. Chapter 12 | Linear Regression and Correlation c. We have a slope of –1.4946 with a y-intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate decreases by 1.5 beats per minute. If the student is not swimming at all, the y-intercept indicates that his heart rate will be 193.88 beats per minute. Although the slope has meaning (the longer it takes to swim 2000 m, the less effort the heart puts out), the y-intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low. d. Because only 1.5 percent of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship. e. Point (34.72, 124) generates the largest residual: –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes –2.953, with the y-intercept changing to 247.1616. Although the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the |
y-intercept becomes more meaningful. 72 If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = –0.0009x + 160, with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set. 73 c. No. The y-intercept would occur at year 0, which doesn’t exist. 74 a. Check student's solution. b. Yes. c. No, the y-intercept would occur at year 0, which doesn’t exist. d. ŷ = −266.8863 + 0.1656x. e. 0.9448, yes. f. 62.8233, 62.3265. g. Yes. h. No, (1987, 62.7). i. 72.5937, no. j. Slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. 76 a. Size (ounces) Cost ($) Cost per ounce 16 32 64 200 Table 12.42 3.99 4.99 5.99 10.99 24.94 15.59 9.36 5.50 b. Check student solution. c. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. d. ŷ = 20.2368 – 0.0819x e. r = –.8086 f. 40-oz: 16.96 cents/oz This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 | Linear Regression and Correlation 757 g. 90-oz: 12.87 cents/oz h. The relationship is not linear; the least-squares line is not appropriate. i. There are no outliers. j. No. You would be extrapolating. The 300-oz size is outside the range of x. k. X = –0.08194. For each additional ounce in size, the cost per ounce decreases by 0.082 cents. 78 a. Size is x, the independent variable, and price is y, the dependent variable. b. Check student solution. |
c. The relationship does not appear to be linear. d. ŷ = –745.252 + 54.75569x. e. r =.8944 and yes, it is significant. f. 32-inch: $1006.93, 50-inch: $1992.53. g. No, the relationship does not appear to be linear. However, r is significant. h. No, the 60-inch TV. i. For each additional inch, the price increases by $54.76. 758 80 Chapter 12 | Linear Regression and Correlation a. Rank is the independent variable and area is the dependent variable. b. Check student solution. c. There appears to be a linear relationship, with one outlier. d. ŷ (area) = 24177.06 + 1010.478x e. r =.50047. r is not significant, so there is no relationship between the variables. f. Alabama: 46,407.576 square miles, Colorado: 62,575.224 square miles. g. The Alabama estimate is closer than the Colorado estimate. h. If the outlier is removed, there is a linear relationship. i. There is one outlier (Hawaii). j. rank 51: 75,711.4 square miles, no. k. Alabama Colorado Hawaii Iowa Maryland Missouri New Jersey Ohio 7 8 6 4 8 8 9 4 1819 1876 1959 1846 1788 1821 1787 1803 South Carolina 13 1788 Utah Wisconsin 4 9 1896 1848 Table 12.43 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 l. ŷ = –87065.3 + 7828.532x. m. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. n. Yes, with the exception of Hawaii. 73 c. No. The y-intercept would occur at year 0, which doesn’t exist. 74 a. Check student's solution. b. Yes. c. No, the y-intercept would occur at year 0, which doesn’t exist. d. ŷ = −266.8863 + 0.1656x. e. 0.9448, yes. f. 62.8233, 62.3 |
265. g. Yes. h. No, (1987, 62.7). i. 72.5937, no. j. Slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 12 | Linear Regression and Correlation 759 76 a. Size (ounces) Cost ($) Cost per ounce 16 32 64 200 Table 12.44 3.99 4.99 5.99 10.99 24.94 15.59 9.36 5.50 b. Check student solution. c. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. d. ŷ = 20.2368 – 0.0819x e. r = –.8086 f. 40-oz: 16.96 cents/oz g. 90-oz: 12.87 cents/oz h. The relationship is not linear; the least-squares line is not appropriate. i. There are no outliers. j. No. You would be extrapolating. The 300-oz size is outside the range of x. k. X = –0.08194. For each additional ounce in size, the cost per ounce decreases by 0.082 cents. 78 a. Size is x, the independent variable, and price is y, the dependent variable. b. Check student solution. c. The relationship does not appear to be linear. d. ŷ = –745.252 + 54.75569x. e. r =.8944 and yes, it is significant. f. 32-inch: $1006.93, 50-inch: $1992.53. g. No, the relationship does not appear to be linear. However, r is significant. h. No, the 60-inch TV. i. For each additional inch, the price increases by $54.76. 80 a. Rank is the independent variable and area is the dependent variable. b. Check student solution. c. There appears to be a linear relationship, with one outlier. d. ŷ (area) = 24177.06 + 1010.478x e. r =.50047. r is not significant, so there is no relationship between the variables. f. Alabama |
: 46,407.576 square miles, Colorado: 62,575.224 square miles. g. The Alabama estimate is closer than the Colorado estimate. h. If the outlier is removed, there is a linear relationship. i. There is one outlier (Hawaii). j. rank 51: 75,711.4 square miles, no. 760 k. Chapter 12 | Linear Regression and Correlation Alabama Colorado Hawaii Iowa Maryland Missouri New Jersey Ohio 7 8 6 4 8 8 9 4 1819 1876 1959 1846 1788 1821 1787 1803 South Carolina 13 1788 Utah Wisconsin 4 9 1896 1848 Table 12.45 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 l. ŷ = –87065.3 + 7828.532x. m. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. n. Yes, with the exception of Hawaii. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 761 13 | F DISTRIBUTION AND ONE-WAY ANOVA Figure 13.1 One-way ANOVA is used to measure information from several groups. Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: Interpret the F probability distribution as the number of groups and the sample size change • • Discuss two uses for the F distribution: one-way ANOVA and the test of two variances • Conduct and interpret one-way ANOVA • Conduct and interpret hypothesis tests of two variances Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies among several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to his or her 762 Chapter 13 | F Distribution and One-way Anova upbringing. A consumer looking for a new car might compare the average gas mileage of several models. For hypothesis tests comparing averages across more than two groups, statisticians have developed a method called analysis of variance (abbreviated ANOVA). In this chapter, you will study the |
simplest form of ANOVA called single factor or oneway ANOVA. You will also study the F distribution, used for one-way ANOVA, and the test of two variances. This is a very brief overview of one-way ANOVA. You will study this topic in much greater detail in future statistics courses. One-way ANOVA, as it is presented here, relies heavily on a calculator or computer. 13.1 | One-Way ANOVA The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test uses variances to help determine if the means are equal or not. To perform a one-way ANOVA test, there are five basic assumptions to be fulfilled: • Each population from which a sample is taken is assumed to be normal. • All samples are randomly selected and independent. • The populations are assumed to have equal standard deviations (or variances). • The factor is a categorical variable. • The response is a numerical variable. The Null and Alternative Hypotheses The null hypothesis is that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups H0: μ1 = μ2 = μ3 =... = μk Ha: At least two of the group means μ1, μ2, μ3,..., μk are not equal. That is, μi ≠ μj for some i ≠ j. The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (red box plots), H0: μ1 = μ2 = μ3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations. If the null hypothesis is false, then the variance of the combined data is larger, which is caused by the different means as shown in the second graph (green box plots). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 763 Figure 13.2 (a) We fail to reject H0 as it may be true. All the means are about the same; the differences may be due to random variation. ( |
b) We reject H0 as all the means are not the same; the differences are too large to be due to random variation. 13.2 | The F Distribution and the F Ratio The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom: one for the numerator and one for the denominator. For example, if F follows an F distribution and the number of degrees of freedom for the numerator is 4, and the number of degrees of freedom for the denominator is 10, then F ~ F4,10. NOTE The F distribution is derived from the Student’s t-distribution. The values of the F distribution are squares of the corresponding values of the t-distribution. One-way ANOVA expands the t-test for comparing more than two groups. The scope of that derivation is beyond the level of this course. It is preferable to use ANOVA when there are more than two groups instead of performing pairwise t-tests because performing multiple tests introduces the likelihood of making a Type 1 error. To calculate the F ratio, two estimates of the variance are made. 1. Variance between samples: an estimate of σ2 that is the variance of the sample means multiplied by n, when the sample sizes are the same. If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation. 2. Variance within samples: an estimate of σ2 that is the average of the sample variances, also known as a pooled variance. When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation. • SSbetween = the sum of squares that represents the variation among the different samples • SSwithin = the sum of squares that represents the variation within samples that is due to chance To find a sum of squares mean, add together squared quantities which, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Descriptive Statistics. MS means mean square. MSbetween is the variance between groups, and MSwithin is the variance within groups. 764 Chapter 13 | F Distribution and One-way Anova Calculation of Sum of Squares and Mean Square • k |
= the number of different groups • nj = the size of the jth group • sj = the sum of the values in the jth group • n = total number of all the values combined (total sample size: ∑nj) • x = one value: ∑x = ∑sj • Sum of squares of all values from every group combined: ∑x2 • Between group variability: SStotal = ∑x2 – ⎛ ⎝∑ x2⎞ n ⎠ • Total sum of squares: ∑x2 – 2 ⎠ ⎛ ⎝∑ x⎞ n • Explained variation: SS(between) = ∑ (s j)2 n j ⎡ ⎢ ⎣ sum of 2 ⎞ ⎠ ⎛ ⎝∑ s j n ⎤ ⎥ − ⎦ squares representing variation among the different samples • Unexplained variation: sum of squares representing variation within samples due to chance SSwithin = SStotal – SSbetween • dfs for different groups (dfs for the numerator): df = k – 1 • Equation for errors within samples (dfs for the denominator): dfwithin = n – k • Mean square (variance estimate) explained by the different groups: MSbetween = SSbetween d fbetween • Mean square (variance estimate) that is due to chance (unexplained): MSwithin = SSwithin d fwithin MSbetween and MSwithin can be written as follows: • MSbetween = SSbetween d fbetween = SSbetween k − 1 • MSwithin = SSwithin d fwithin = SSwithin n − k The one-way ANOVA test depends on the fact that MSbetween can be influenced by population differences among means of the several groups. Since MSwithin compares values of each group to its own group mean, the fact that group means might be different does not affect MSwithin. The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MSbetween and MSwithin should both estimate the same value. NOTE The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution because it is assumed that the populations are normal and that they |
have equal variances. F Ratio or F Statistic F = MSbetween MSwithin If MSbetween and MSwithin estimate the same value, following the belief that H0 is true, then the F ratio should be approximately equal to 1. Mostly, just sampling errors would contribute to variations away from 1. As it turns out, MSbetween This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 765 consists of the population variance plus a variance produced from the differences between the samples. MSwithin is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MSbetween will generally be larger than MSwithin. Then the F ratio will be larger than 1. However, if the population effect is small, it is not unlikely that MSwithin will be larger in a given sample. The previous calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F ratio can be written as follows: F Ratio formula when the groups are the same size F = 2 n ⋅ s x¯ s2 pooled where • n = the sample size • dfnumerator = k – 1 • dfdenominator = n – k • s2 pooled = the mean of the sample variances (pooled variance) • s x¯ 2 = the variance of the sample means Data is typically put into a table for easy viewing. One-way ANOVA results are often displayed in this manner by computer software. Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F SS(Factor) SS(Error) SS(Total MS(Factor) = SS(Factor)/(k – 1) F = MS(Factor)/MS(Error) MS(Error) = SS(Error)/(n – k) Factor (Between) Error (Within) Total Table 13.1 Example 13.1 Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in Table 13.2. Plan 1: n1 = 4 Plan 2: n2 = 3 Plan 3: n3 = 3 5 4.5 4 3 Table 13.2 3.5 7 4.5 8 4 3.5 s1 = 16.5, s |
2 = 15, s3 = 15.5 Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test. 766 Chapter 13 | F Distribution and One-way Anova SS(between) = ∑ (s j) ⎞ ⎛ ⎝∑ s j ⎠ n = 2 s1 4 2 s2 3 + + 2 s3 3 (s1 + s2 + s3)2 10 − where n1 = 4, n2 = 3, n3 = 3, and n = n1 + n2 + n3 = 10 = (16.5)2 4 + (15)2 3 + (15.5)2 3 − (16.5 + 15 + 15.5)2 10 SS(between) = 2.2458 ⎝∑ x⎞ n S(total) = ∑ x2 − ⎛ ⎠ 2 = ⎝52 + 4.52 + 42 + 32 + 3.52 + 72 + 4.52 + 82 + 42 + 3.52⎞ ⎛ ⎠ − (5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5)2 10 = 244 − 472 10 = 244 − 220.9 SS(total) = 23.1 SS(within) = SS(total) − SS(between) = 23.1 − 2.2458 SS(within) = 20.8542 One-way ANOVA Table: The formulas for SS(Total), SS(Factor) = SS(Between), and SS(Error) = SS(Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA[L1, L2, L3] where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3, respectively). Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) SS(Factor) = SS(Between) = 2.2458 k – 1 = 3 groups – 1 = 2 SS(Error) = SS(Within) = 20.8542 SS(Total) = 2.2458 + 20.8542 = 23.1 n |
– k = 10 total data – 3 groups = 7 n – 1 = 10 total data – 1 = 9 Error (Within) Total Table 13.3 F = MS(Factor)/MS(Error) = 1.1229/2.9792 = 0.3769 MS(Factor) = SS(Factor)/(k – 1) = 2.2458/2 = 1.1229 MS(Error) = SS(Error)/(n – k) = 20.8542/7 = 2.9792 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 767 13.1 As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments: • Bare soil • A commercial ground cover • Black plastic • Straw • Compost All plants grew under the same conditions and were the same variety. Students recorded the weight in grams of tomatoes produced by each of the n = 15 plants, as seen in Table 13.4. Bare: n1 = 3 Ground Cover: n2 = 3 Plastic: n3 = 3 Straw: n4 = 3 Compost: n5 = 3 2,625 2,997 4,915 Table 13.4 5,348 5,682 5,482 6,583 8,560 3,830 7,285 6,897 9,230 6,277 7,818 8,677 Create the one-way ANOVA table. The one-way ANOVA hypothesis test is always right-tailed because larger F values are way out in the right tail of the F distribution curve and tend to make us reject H0. Notation The notation for the F distribution is F ~ Fdf(num),df(denom), where df(num) = dfbetween and df(denom) = dfwithin. The mean for the F distribution is μ = d f (denom) d f (denom) – 2. 13.3 | Facts About the F Distribution The following are facts about the F distribution: • The curve is not symmetrical but skewed to the right. • There is a different curve for each set of dfs. • The F statistic is greater than or equal to zero. • As the degrees of freedom for the numerator and for the |
denominator get larger, the curve approximates the normal. • Other uses for the F distribution include comparing two variances and two-way analysis of variance. Two-way analysis is beyond the scope of this chapter. 768 Chapter 13 | F Distribution and One-way Anova Figure 13.3 Example 13.2 Let’s return to the slicing tomato exercise in Try It. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5 percent, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest. Solution 13.2 The null and alternative hypotheses are as follows: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: μi ≠ μj for some i ≠ j The one-way ANOVA results are shown in Table 13.4 Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) 36,648,561 5 – 1 = 4 Error (Within) 20,446,726 15 – 5 = 10 Total 57,095,287 15 – 1 = 14 36,648,561 4 20,446,726 10 = 9,162,140 9,162,140 2,044,672.6 = 4.4810 = 2,044,672.6 Table 13.5 Distribution for the test: F4,10 df(num) = 5 – 1 = 4 df(denom) = 15 – 5 = 10 Test statistic: F = 4.4810 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 769 Figure 13.4 Probability statement: p-value = P(F > 4.481) = 0.0248 Compare α and the p-value: α = 0.05, p-value = 0.0248 Make a decision: Since α > p-value, we reject H0. Conclusion: At the 5 percent significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely |
to be due to chance alone. We may conclude that at least some of the mulches led to different mean yields. To find these results on the calculator: Press STAT. Press 1:EDIT. Put the data into the lists L1,L2,L3,L4,L5. Press STAT, arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter (L1,L2,L3,L4,L5). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p-value of the test. The calculator displays: F = 4.4810 p = 0.0248 (p-value) Factor df = 4 SS = 36648560.9 MS = 9162140.23 Error df = 10 SS = 20446726 MS = 2044672.6 770 Chapter 13 | F Distribution and One-way Anova 13.2 MRSA, or Staphylococcus aureus, can cause serious bacterial infections in hospital patients. Table 13.6 shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in Figure 13.5. Conc = 0.6 Conc = 0.8 Conc = 1.0 Conc = 1.2 Conc = 1.4 9 66 98 Table 13.6 16 93 82 22 147 120 30 199 148 27 168 132 Plot of the data for the different concentrations: Figure 13.5 Test whether the mean numbers of colonies are the same or are different. Construct the ANOVA table by hand or by using a TI-83, 83+, or 84+ calculator, find the p-value, and state your conclusion. Use a 5 percent significance level. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 771 Example 13.3 Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.7. Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 1.85 2.83 1.69 3.33 2.63 1.77 3.25 1.86 2.21 2.63 3.78 4.00 2.55 2.45 3.79 3.45 3.08 |
2.26 3.18 Table 13.7 Mean Grades for Four Sororities Using a significance level of 1 percent, is there a difference in mean grades among the sororities? Solution 13.3 Let μ1, μ2, μ3, μ4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five. NOTE This is an example of a balanced design, because each factor (i.e., sorority) has the same number of observations. H0: μ1 = μ2 = μ3 = μ4 Ha: Not all of the means μ1, μ2, μ3, μ4 are equal. Distribution for the test: F3,16 where k = 4 groups and n = 20 samples in total. df(num)= k – 1 = 4 – 1 = 3 df(denom) = n – k = 20 – 4 = 16 Calculate the test statistic: F = 2.23 Graph 772 Chapter 13 | F Distribution and One-way Anova Figure 13.6 Probability statement: p-value = P(F > 2.23) = 0.1241 Compare α and the p-value: α = 0.01 p-value = 0.1241 α < p-value Make a decision: Since α < p-value, we cannot reject H0. Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities. Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and enter (L1,L2,L3,L4). The calculator displays the F statistic, the p-value, and the values for the one-way ANOVA table: F = 2.2303 p = 0.1241 (p-value) Factor df = 3 SS = 2.88732 MS = 0.96244 Error df = 16 SS = 6.9044 MS = 0.431525 13.3 Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.8. This OpenStax book is available for free |
at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 773 Basketball Baseball Hockey Lacrosse 3.6 2.9 2.5 3.3 3.8 2.1 2.6 3.9 3.1 3.4 4.0 2.0 2.6 3.2 3.2 2.0 3.6 3.9 2.7 2.5 Table 13.8 GPAs for four sports teams Use a significance level of 5 percent and determine if there is a difference in GPA among the teams. 774 Chapter 13 | F Distribution and One-way Anova Example 13.4 A fourth-grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother’s garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data in inches in Table 13.9. Tommy’s Plants Tara’s Plants Nick’s Plants 24 21 23 30 23 Table 13.9 25 31 23 20 28 23 27 22 30 20 Does it appear that the three soils in which the bean plants were grown produce the same mean height? Test at a 3 percent level of significance. Solution 13.4 This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = 2. n ⋅ s x¯ s2 pooled First, calculate the sample mean and sample variance of each group. Tommy's Plants Tara's Plants Nick's Plants 24.2 11.7 25.4 18.3 24.4 16.3 Sample Mean Sample Variance Table 13.10 Next, calculate the variance of the three group means by calculating the variance of 24.2, 25.4, and 24.4. Variance of the group means = 0.413 = s x¯ 2, 2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew). then MSbetween = ns x |
¯ Calculate the mean of the three sample variances (11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s2 then MSwithin = s2 pooled = 15.433. pooled, The F statistic (or F ratio) is F = MSbetween MSwithin = 2 ns x¯ s2 pooled = (5)(0.413) 15.433 = 0.134. The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2. The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 775 The distribution for the test is F2,12 and the F statistic is F = 0.134. The p-value is P(F > 0.134) = 0.8759. Decision: Since α = 0.03 and the p-value = 0.8759, we do not reject H0. Why? Conclusion: With a 3 percent level of significance from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different. To calculate the p-value: •Press 2nd DISTR, •Arrow down to Fcdf and press ENTER, •Enter 0.134, E99, 2, 12, and •Press ENTER. The p-value is 0.8759. 13.4 Another fourth grader also grew bean plants, but in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.4. From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1 percent level of significance. Use one of the solution sheets in Appendix E. 13.4 | Test of Two Variances Another use of the F |
distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. For a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of check-out times for two checkers. To perform a F test of two variances, it is important that the following are true: • The populations from which the two samples are drawn are normally distributed. • The two populations are independent of each other. Unlike most other tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, the test can give higher p-values than it should, or lower ones, in ways that are unpredictable. Many texts suggest that students not use this test at all, but in the interest of completeness we include it here. Suppose we sample randomly from two independent normal populations. Let σ1 2 and σ2 2 be the population variances and 776 Chapter 13 | F Distribution and One-way Anova 2 be the sample variances. Let the sample sizes be n1 and n2. Since we are interested in comparing the two sample 2 and s2 s1 variances, we use the F ratio F = ⎡ ⎢(s1)2 (σ1)2 ⎣ ⎡ ⎢(s2)2 (σ2) has the distribution F ~ F(n1 – 1, n2 – 1), where n1 – 1 are the degrees of freedom for the numerator and n2 – 1 are the degrees of freedom for the denominator. If the null hypothesis is σ1 2 = σ2 2, then the F ratio becomes F = ⎡ ⎢(s1)2 (σ1)2 ⎣ ⎡ ⎢(s2)2 (σ2)s1)2 (s2)2. NOTE The F ratio could also be (s2)2 (s1)2. It depends on Ha and on which sample variance is larger. If the two populations have equal variances, then s1 2 and s2 2 are close in value and F = (s1)2 (s2)2 is close to 1. But if the two population variances are very different |
, s1 variance causes the ratio (s1)2 (s2)2 to be greater than 1. If s1 2 and s2 2 are far apart, then F = (s1)2 (s2)2 is a large number. 2 and s2 2 tend to be very different, too. Choosing s1 2 as the larger sample Therefore, if F is close to 1, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than 1, then the evidence is against the null hypothesis. A test of two variances may be left-tailed, right-tailed, or two-tailed. Example 13.5 Two college instructors are interested in whethe there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor’s grades have a variance of 52.3. The second instructor’s grades have a variance of 89.9. Test the claim that the first instructor’s variance is smaller. In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors. The level of significance is 10 percent. Solution 13.5 Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively. n1 = n2 = 30. 2 = σ2 H0: σ1 2 and Ha: σ1 2. 2 < σ2 Calculate the test statistic: By the null hypothesis (σ1 2 = σ2 2 ), the F statistic is This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 777 F = ⎡ ⎢(s1)2 (σ1)2 ⎣ ⎡ ⎢(s2)2 (σ2)s1)2 (s2)2 = 52.3 89.9 = 0.5818. Distribution for the test: F29,29 where n1 – 1 = 29 and n2 – 1 = 29. Graph: This test is left-tailed. Draw the graph, labeling and shading appropriately. Figure 13.7 Probability statement: p-value = P(F < 0.5818) = 0.0753. Compare α and the p-value: α = 0.10; α > p-value. |
Make a decision: Since α > p-value, reject H0. Conclusion: With a 10 percent level of significance from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller. Press STAT and arrow over to TESTS. Arrow down to D:2-SampFTest. Press ENTER. Arrow to Stats (89.9), and 30. Press ENTER after and press ENTER. For Sx1, n1, Sx2, and n2, enter each. Arrow to σ1: and < σ2. Press ENTER. Arrow down to Calculate and press ENTER. F = 0.5818 and p-value = 0.0753. Do the procedure again and try Draw instead of Calculate. (52.3), 30, 778 Chapter 13 | F Distribution and One-way Anova 13.5 The New York Choral Society divides male singers into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, and Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different? Tenor1 Bass2 Tenor1 Bass2 Tenor1 Bass2 69 72 71 66 76 74 71 66 68 72 75 67 75 74 72 72 74 72 67 70 65 72 70 68 64 73 66 72 74 70 66 68 75 68 70 72 Table 13.11 68 67 64 67 70 70 69 72 71 74 75 13.5 | Lab: One-Way ANOVA This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 779 13.1 One-Way ANOVA Student Learning Outcome • The student will conduct a simple one-way ANOVA test involving three variables. Collect the Data 1. Record the price per pound of eight fruits, eight vegetables, and eight breads in your local supermarket. Fruits Vegetables Breads Table 13.12 2. Explain how you could try to collect the data randomly. Analyze the Data and Conduct a Hypothesis Test 1. State the null hypothesis and the alternative hypothesis. 2. Comp |
ute the following: a. Fruit i. ii. x¯ = ______ s x = ______ iii. n = ______ b. Vegetables i. ii. x¯ = ______ s x = ______ iii. n = ______ c. Bread i. ii. x¯ = ______ s x = ______ iii. n = ______ 3. Find the following: a. df(num) = ______ 780 Chapter 13 | F Distribution and One-way Anova b. df(denom) = ______ 4. State the approximate distribution for the test. 5. Test statistic: F = ______ 6. Sketch a graph of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p-value. 7. p-value = ______ 8. Test at α = 0.05. State your decision and conclusion. 9. a. Decision: why did you make this decision? b. Conclusion (write a complete sentence): c. Based on the results of your study, is there a need to investigate any of the food groups’s prices? Why or why not? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 781 KEY TERMS analysis of variance also referred to as ANOVA; a method of testing whether the means of three or more populations are equal The method is applicable if • all populations of interest are normally distributed, • • the populations have equal standard deviations, and samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F ratio. one-way ANOVA applicable if a method of testing whether the means of three or more populations are equal; the method is • all populations of interest are normally distributed, • • • the populations have equal standard deviations, samples (not necessarily of the same size) are randomly and independently selected from each population, and there is one independent variable and one dependent variable. The test statistic for analysis of variance is the F ratio variance mean of the squared deviations from the mean; the square of the standard deviation For a set of data, a deviation can be represented as x – x¯ where x is a value of the data and x¯ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and 1. CHAPTER REVIEW 13.1 One |
-Way ANOVA Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom. Assumptions: • Each population from which a sample is taken is assumed to be normal. • All samples are randomly selected and independent. • The populations are assumed to have equal standard deviations (or variances). 13.2 The F Distribution and the F Ratio Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table. 13.3 Facts About the F Distribution The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p-value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p-value) 782 Chapter 13 | F Distribution and One-way Anova to the right of the statistic under the F curve. When the data have unequal group sizes (unbalanced data), then techniques from The F Distribution and the F Ratio need to be used for hand calculations. In the case of balanced data, where the groups are the same size, simplified calculations based on group means and variances may be used. In practice, software is usually employed in the analysis. As in any analysis, graphs of various sorts should |
be used in conjunction with numerical techniques. Always look at your data! 13.4 Test of Two Variances The F test for the equality of two variances rests heavily on the assumption of normal distributions. The test is unreliable if this assumption is not met. If both distributions are normal, then the ratio of the two sample variances is distributed as an F statistic, with numerator and denominator degrees of freedom that are one less than the samples sizes of the corresponding two groups. A test of two variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with two different degrees of freedom. Assumptions: • The populations from which the two samples are drawn are normally distributed. • The two populations are independent of each other. • k = the number of groups • nj = the size of the jth group • sj = the sum of the values in the jth group • n = the total number of all values (observations) combined • x = one value (one observation) from the data • • 2 = the variance of the sample means s x¯ = the mean of the sample variances (pooled s2 pooled variance) 13.4 Test of Two Variances F has the distribution F ~ F(n1 – 1, n2 – 1) F = 2 s1 2 σ1 2 s2 2 σ2 If σ1 = σ2, then F = s1 s2 2 2 FORMULA REVIEW 13.2 The F Distribution and the F Ratio SSbetween = ∑ (s j) ⎞ ⎠ ⎛ ⎝∑ s j n SStotal = ∑ x2 − 2 ⎞ ⎠ ⎛ ⎝∑ x n SSwithin = SStotal − SSbetween dfbetween = df(num) = k – 1 dfwithin = df(denom) = n – k MSbetween = SSbetween d fbetween MSwithin = SSwithin d fwithin F = MSbetween MSwithin F ratio when the groups are the same size: F = 2 ns x¯ s2 pooled Mean of the F distribution: µ = d f (num) d f (denom) − 1 where This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 783 PR |
ACTICE 13.1 One-Way ANOVA Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled to perform a one-way ANOVA test. What are they? 1. Write one assumption. 2. Write another assumption. 3. Write a third assumption. 4. Write a fourth assumption. 5. Write the final assumption. 6. State the null hypothesis for a one-way ANOVA test if there are four groups. 7. State the alternative hypothesis for a one-way ANOVA test if there are three groups. 8. When do you use an ANOVA test? 13.2 The F Distribution and the F Ratio Use the following information to answer the next seven exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in Table 13.13 are the weights for the different groups. Group 1 Group 2 Group 3 216 198 240 187 176 202 213 284 228 210 Table 13.13 170 165 182 197 201 9. What is the sum of squares factor? 10. What is the sum of squares error? 11. What is the df for the numerator? 12. What is the df for the denominator? 13. What is the mean square factor? 14. What is the mean square error? 15. What is the F statistic? Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in Table 13.14 are the goals per game for the different teams. Team 1 Team 2 Team 3 Team Table 13.14 784 Chapter 13 | F Distribution and One-way Anova Team 1 Team 2 Team 3 Team 4 3 2 4 4 0 0 3 2 Table 13.14 16. What is SSbetween? 17. What is the df for the numerator? 18. What is MSbetween? 19. What is SSwithin? 20. What is the df for the denominator? 21. What is MSwithin? 22. What is the F statistic? 23. Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis? 13.3 Facts About the F Distribution 24. An F statistic can have what values? 25. What happens to the curves as the degrees of freedom for the numerator and the denominator get larger? Use the following information to answer the next seven exercises. Four basketball teams took a random sample of players regarding how high each |
player can jump (in inches). The results are shown in Table 13.15. Team 1 Team 2 Team 3 Team 4 Team 5 36 42 51 32 35 38 48 50 39 38 44 46 41 39 40 Table 13.15 26. What is the df(num)? 27. What is the df(denom)? 28. What are the sum of squares and mean squares factors? 29. What are the sum of squares and mean squares errors? 30. What is the F statistic? 31. What is the p-value? 32. At the 5 percent significance level, is there a difference in the mean jump heights among the teams? Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Table 13.16. Group A Group B Group C 101 108 151 149 101 109 Table 13.16 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 785 Group A Group B Group C 98 107 111 160 112 126 198 186 160 Table 13.16 33. What is the df(num)? 34. What is the df(denom)? 35. What are the SSbetween and MSbetween? 36. What are the SSwithin and MSwithin? 37. What is the F Statistic? 38. What is the p-value? 39. At the 10 percent significance level, are the scores among the different groups different? Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = s2 = ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ Table 13.17 Enter the data into your calculator or computer. 40. p |
-value = ______ State the decisions and conclusions (in complete sentences) for the following preconceived levels of α. 41. α = 0.05 a. Decision: ____________________________ b. Conclusion: ____________________________ 42. α = 0.01 a. Decision: ____________________________ b. Conclusion: ____________________________ 13.4 Test of Two Variances Use the following information to answer the next two exercises. There are two assumptions that must be true to perform an F test of two variances. 43. Name one assumption that must be true. 786 Chapter 13 | F Distribution and One-way Anova 44. What is the other assumption that must be true? Use the following information to answer the next seven exercises. Two coworkers commute from the same building. They are interested in whether there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times. Test the claim at the 10 percent level. Assume that commute times are normally distributed. 45. State the null and alternative hypotheses. 46. What is s1 in this problem? 47. What is s2 in this problem? 48. What is n? 49. What is the F statistic? 50. What is the p-value? 51. Is the claim accurate? Use the following information to answer the next four exercises. Two students are interested in whether there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are more consistent. 52. State the null and alternative hypotheses. 53. What is the F statistic? 54. What is the p-value? 55. At the 5 percent significance level, do we reject the null hypothesis? Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records his or her speeds going up 35 hills. The first cyclist has a variance of 23.8, and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different. Ass |
ume that speeds are normally distributed. 56. State the null and alternative hypotheses. 57. What is the F statistic? 58. At the 5 percent significance level, what can we say about the cyclists’ variances? HOMEWORK This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 787 13.1 One-Way ANOVA 59. Three different traffic routes are tested for mean driving time. The entries in the Table 13.18 are the driving times in minutes on the three different routes. Route 1 Route 2 Route 3 30 32 27 35 27 29 28 36 16 41 22 31 Table 13.18 State SSbetween, SSwithin, and the F statistic. 60. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = s2 = ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ Table 13.19 State the hypotheses. H0: ____________ Ha: ____________ 13.2 The F Distribution and the F Ratio Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 Table 13.20 788 Chapter 13 | F Distribution and One-way Anova Northeast South West Central East ________ ________ ________ ________ ________ ________ ________ ________ ________ ________ |
x¯ = s2 = Table 13.20 H0: µ1 = µ2 = µ3 = µ4 = µ5 Hα: At least any two of the group means µ1, µ2, …, µ5 are not equal. 61. degrees of freedom – numerator: df(num) = _________ 62. degrees of freedom – denominator: df(denom) = ________ 63. F statistic = ________ 13.3 Facts About the F Distribution DIRECTIONS Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in Appendix E. 64. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10 percent, test the hypothesis that the three formulas produce the same mean weight gain. Linda’s Rats (g) Tuan’s Rats (g) Javier’s Rats (g) 43.5 39.4 41.3 46.0 38.2 Table 13.21 47.0 40.5 38.9 46.3 44.2 51.2 40.9 37.9 45.0 48.6 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 789 65. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in Table 13.22. Using a 5 percent significance level, test the hypothesis that the three mean commuting mileages are the same. Working-Class Professional (middle incomes) Professional (wealthy) 17.8 26.7 49.4 9.4 65.4 47.1 19.5 51.2 Table 13.22 16.5 17.4 22.0 7.4 9.4 2.1 6.4 13.9 8.5 6.3 4.6 12.6 11.0 28.6 15.4 |
9.3 Use the following information to answer the next two exercises. Table 13.23 lists the number of pages in four different types of magazines. Home Decorating News Health Computer 172 286 163 205 197 Table 13.23 87 94 123 106 101 82 153 87 103 96 104 136 98 207 146 66. Using a significance level of 5 percent, test the hypothesis that the four magazine types have the same mean length. 67. Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same? 790 Chapter 13 | F Distribution and One-way Anova 68. A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that Table 13.24 shows the results of a study. CNN FOX Local 72 37 56 60 51 45 12 18 38 23 35 15 43 68 50 31 22 Table 13.24 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 69. Are the means for the final exams the same for all statistics class delivery types? Table 13.25 shows the scores on final exams from several randomly selected classes that used the different delivery types. Online Hybrid Face-to-Face 83 73 84 81 72 84 77 80 81 Table 13.25 80 78 84 81 86 79 82 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 70. Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics, and Asians? Suppose that Table 13.26 shows the results of a study. White Black Hispanic Asian 4 1 5 2 6 8 2 4 6 Table 13.26 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 791 71. Are the mean numbers of daily visitors to a ski resort the same for |
the three types of snow conditions? Suppose that Table 13.27 shows the results of a study. Powder Machine Made Hard Packed 1,210 1,080 1,537 941 Table 13.27 2,107 1,149 862 1,870 1,528 1,382 2,846 1,638 2,019 1,178 2,233 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 792 Chapter 13 | F Distribution and One-way Anova 72. Sanjay made identical paper airplanes out of three different weights of paper: light, medium, and heavy. He made four airplanes from each of the weights and launched them himself across the room. Here are the distances (in meters) that his planes flew. Paper Type/Trial Trial 1 Trial 2 Trial 3 Trial 4 Heavy Medium 5.1 meters 3.1 meters 4.7 meters 5.3 meters 4 meters 3.5 meters 4.5 meters 6.1 meters Light 3.1 meters 3.3 meters 2.1 meters 1.9 meters Table 13.28 Figure 13.8 a. Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? b. Why is this a balanced design? c. Calculate the sample mean and sample standard deviation for each group. d. Does the weight of the paper have an effect on how far the plane will travel? Use a 1 percent level of significance. Complete the test using the method shown in the bean plant example in Example 13.4. ◦ Variance of the group means __________ ◦ MSbetween= ___________ ◦ Mean of the three sample variances ___________ ◦ MSwithin = _____________ ◦ F statistic = ____________ ◦ df(num) = __________, df(denom) = ___________ ◦ Number of groups _______ ◦ Number of observations _______ ◦ p-value = __________ (P(F > _______) = __________) ◦ Graph the p-value. ◦ Decision: _______________________ ◦ Conclusion: _______________________________________________________________ This OpenStax book is available for free at http://cnx.org/content/col30309/ |
1.8 Chapter 13 | F Distribution and One-way Anova 793 73. DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective but persisted in the environment and over time proved to be harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds. An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). The third group was a control line of nonselected or typical fruit flies (NS). Here are the data: RS SS NS RS SS NS 12.8 38.4 35.4 22.4 23.1 22.6 21.6 32.9 27.4 27.5 29.4 40.4 14.8 48.5 19.3 20.3 16 34.4 23.1 20.9 41.8 38.7 20.1 30.4 34.6 11.6 20.3 26.4 23.3 14.9 19.7 22.3 37.6 23.7 22.9 51.8 22.6 30.2 36.9 26.1 22.5 33.8 29.6 33.4 37.3 29.5 15.1 37.9 416.4 26.7 228.2 38.6 31 29.5 20.3 39 23.4 44.4 16.9 42.4 29.3 12.8 33.7 23.2 16.1 36.6 914.9 14.6 29.2 23.6 10.8 47.4 27.3 12.2 41.7 Table 13.29 The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1 percent level of significance, are the mean rates of egg selection for the three strains of fruit fly different? If so, in what way? Specifically, the researchers were interested in whether the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other. Here is a chart of the three |
groups: Figure 13.9 794 Chapter 13 | F Distribution and One-way Anova 74. The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms. Traditionally, we are taught that the normal human body temperature is 98.6 °F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate 95 percent confidence intervals for the mean body temperature in each group and comment about the confidence intervals. FL FH ML MH FL FH ML MH 96.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6 96.7 97.7 96.7 97 98.7 98.6 98.1 98.6 97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7 97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8 97.4 97.6 97.7 97.8 98 98 98 98 97.3 97.4 98.7 98.7 98.2 98.8 97.4 97.5 98.8 98.8 98.2 98.8 97.4 97.6 98.8 98.8 98.3 98.9 97.4 97.7 98.8 98.8 98.4 97.8 98.1 97.5 97.8 98.8 98.9 98.4 97.9 98.3 97.6 97.9 99.2 97.9 98.3 97.6 98 98.3 97.8 98.2 98.4 97.8 98 98 98 99.3 99 99 98.5 98.5 99.2 99.1 98.6 99.5 99.1 98.6 99 99 99 98.2 98.4 97.8 98.3 99.2 98.7 98.2 98.4 97.9 98.4 99.4 99.1 98.2 98.4 98.2 98.5 98.2 98.6 98 98 98 98.4 98.6 98.6 99.9 99.3 100 99.4 100.8 Table 13.30 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 795 13.4 Test of Two V |
ariances 75. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded. Linda’s Rats Tuan’s Rats Javier’s Rats 43.5 39.4 41.3 46.0 38.2 Table 13.31 47.0 40.5 38.9 46.3 44.2 51.2 40.9 37.9 45.0 48.6 Determine whether the variance in weight gain is statistically the same between Javier’s and Linda’s rats. Test at a significance level of 10 percent. 76. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows. Working-Class Professional (middle incomes) Professional (wealthy) 17.8 26.7 49.4 9.4 65.4 47.1 19.5 51.2 Table 13.32 16.5 17.4 22.0 7.4 9.4 2.1 6.4 13.9 8.5 6.3 4.6 12.6 11.0 28.6 15.4 9.3 Determine whether the variance in mileage driven is statistically the same between the working class and professional (middle income) groups. Use a 5 percent significance level. Use the following information to answer the next two exercises. The following table lists the number of pages in four different types of magazines. Home Decorating News Health Computer 172 286 163 205 Table 13.33 87 94 123 106 82 153 87 103 104 136 98 207 796 Chapter 13 | F Distribution and One-way Anova Home Decorating News Health Computer 197 101 96 146 Table 13.33 77. Which two magazine types do you think have the same variance in length? 78. Which two magazine types do you think have different variances in length? 79. Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays |
at the mall? Suppose that Table 13.34 shows the results of a study. Saturday Sunday Saturday Sunday 75 18 150 94 62 73 Table 13.34 44 58 61 19 99 60 89 62 0 124 50 31 118 137 82 39 127 141 73 80. Are the variances for incomes on the East Coast and the West Coast the same? Suppose that Table 13.35 shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05. East West 71 126 42 51 44 90 88 38 47 30 82 75 52 115 67 Table 13.35 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 797 81. Thirty men in college were taught a method of finger tapping. They were randomly assigned to three groups of 10, with each receiving one of three doses of caffeine: 0 mg, 100 mg, or 200 mg. This is approximately the amount in zero, one, or two cups of coffee. Two hours after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how? Here are the data: 0 mg 100 mg 200 mg 0 mg 100 mg 200 mg 242 244 247 242 246 248 245 248 247 243 Table 13.36 246 250 248 246 245 245 248 248 244 242 246 247 250 246 244 248 252 250 248 250 82. King Manuel I Komnenos ruled the Byzantine Empire from Constantinople (Istanbul) during the years A.D. 1145–1170. The empire was very powerful during his reign but declined significantly afterward. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from the fourth. These spanned most of his reign. We have data on the silver content of the coins: First Coinage Second Coinage Third Coinage Fourth Coinage 6.9 9.0 6.6 8.1 9.3 9.2 8.6 4.9 5.5 4.6 4.5 5.3 5.6 5.5 5.1 6.2 5.8 5.8 5.9 6. |
8 6.4 7.0 6.6 7.7 7.2 6.9 6.2 Table 13.37 Did the silver content of the coins change over the course of Manuel’s reign? Here are the means and variances of each coinage. The data are unbalanced. First Second Third Fourth Mean 6.7444 8.2429 4.875 5.6143 Variance 0.2953 1.2095 0.2025 0.1314 Table 13.38 798 Chapter 13 | F Distribution and One-way Anova 83. The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in 2012 Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in the 2012 season, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same. Note that the data are not balanced, as two divisions had five teams, while one had only four. Division Team Wins East East East East East NY Yankees Baltimore Tampa Bay Toronto Boston 95 93 90 73 69 Table 13.39 Division Team Wins Central Detroit Central Chicago Sox Central Kansas City Central Cleveland Central Minnesota 88 85 72 68 66 Table 13.40 Division Team Wins West West West West Oakland Texas LA Angels Seattle 94 93 89 75 Table 13.41 REFERENCES 13.2 The F Distribution and the F Ratio Marist College School of Science. (n.d.). Tomato data (Unpublished student research). Marist College School of Science, Poughkeepsie, NY. 13.3 Facts About the F Distribution ESPN. (2012). MLB standings – 2012. Retrieved from http://espn.go.com/mlb/standings/_/year/2012. Hand, D. J. et al. (1994). A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 799 Hand, D. J |
. et al. (1994). A Handbook of Small Datasets. London: Chapman & Hall, p. 50. Hand. A Handbook of Small Datasets. p. 118. Mackowiak, P. A.,Wasserman, S. S., & Levine, M. M. (1992). A critical appraisal of 98.6 degrees F, the upper limit of the normal body temperature, and other legacies of Carl Reinhold August Wunderlich. Journal of the American Medical Association, 268, 1578–1580. Private K–12 school in San Jose, CA. (1994). Data from a fourth grade classroom. 13.4 Test of Two Variances ESPN. (2012). MLB standings – 2012. Retrieved from http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/ order/true. SOLUTIONS 1 Each population from which a sample is taken is assumed to be normal. 3 The populations are assumed to have equal standard deviations (or variances). 5 The response is a numerical value. 7 Ha: At least two of the group means μ1, μ2, μ3 are not equal. 9 4,939.2 11 2 13 2,469.6 15 3.7416 17 3 19 13.2 21 0.825 23 Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p value, so it is likely that we will reject the null hypothesis. 25 The curves approximate the normal distribution. 27 10 29 SS = 237.33; MS = 23.73 31 0.1614 33 two 35 SS = 5,700.4; MS = 2,850.2 37 3.6101 39 Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10 percent level. 43 The populations from which the two samples are drawn are normally distributed. 45 H0: σ1 = σ2 Ha: σ1 < σ2 or H0: σ1 47 4.11 2 = σ2 2 Ha: σ1 2 2 < σ2 49 0.7159 51 No, at the 10 percent level of significance, we do not reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker. 53 2.8 |
674 800 Chapter 13 | F Distribution and One-way Anova 55 Reject the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student. 57 0.7414 59 SSbetween = 26 SSwithin = 441 F = 0.2653 62 df(denom) = 15 64 a. H0: µL = µT = µJ b. Ha: at least any two of the means are different c. df(num) = 2; df(denom) = 12 d. F distribution e. 0.67 f. 0.5305 g. Check student’s solution. h. Decision: Do not reject null hypothesis. i. Conclusion: There is insufficient evidence to conclude that the means are different. 67 a. Ha: µc = µn = µh b. At least any two of the magazines have different mean lengths. c. df(num) = 2, df(denom) = 12 d. F distribtuion e. F = 15.28 f. p-value = 0.0005 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different. 69 a. H0: μo = μh = μf b. At least two of the means are different. c. df(n) = 2, df(d) = 13 d. F2,13 e. 0.64 f. 0.5437 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: The mean scores of different class delivery are not different. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 801 71 a. H0: μp = μm = μh b. At least any two of the means are different. c. df(n) = 2, df(d) = 12 d. F2,12 e. 3.13 f. 0.0807 g. Check student� |
�s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different. 73 The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups. H0: μ1 = μ2 = μ3 Ha: μi ≠ μj some i ≠ j Define μ1, μ2, μ3, as the population mean number of eggs laid by the three groups of fruitflies. F statistic = 8.6657 p-value = 0.0004 Figure 13.10 Decision: Since the p-value is less than the level of significance of 0.01, we reject the null hypothesis. Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different. Interestingly, if you perform a two sample t test to compare the RS and NS groups they are significantly different (p = 0.0013). Similarly, SS and NS are significantly different (p = 0.0006). However, the two selected groups, RS and SS are not significantly different (p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity. 75 a. H0 : σ1 2 2 = σ2 2 2 ≠ σ1 b. Ha : σ1 c. df(num) = 4; df(denom) = 4 d. F4, 4 802 e. 3.00 Chapter 13 | F Distribution and One-way Anova f. 2(0.1563) = 0.3126. Using the TI-83+/84+ function 2-SampFtest, you get the test statistic as 2.9986 and p-value directly as 0.3127. If you input the lists in a different order, you get a test statistic of 0.3335 but the p-value is the same because this is a two- |
tailed test. g. Check student's solution. h. Decision: Do not reject the null hypothesis. i. Conclusion: There is insufficient evidence to conclude that the variances are different. 78 The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances. 80 a. H0: = σ1 2 2 = σ2 2 2 ≠ σ1 b. Ha: σ1 c. df(n) = 7, df(d) = 6 d. F7,6 e. 0.8117 f. 0.7825 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the variances are different. 82 Here is a strip chart of the silver content of the coins: Figure 13.11 While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 13 | F Distribution and One-way Anova 803 Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (between) Error (within) Total Table 13.42 37.748 11.015 48.763 4 – 1 = 3 27 – 4 = 23 27 – 1 = 26 12.5825 0.4789 26.272 P(F > 26.272) = 0. Reject the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth. 83 Here is a stripchart of the number of wins for the 14 teams in the AL for the 2012 season. Figure 13.12 While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way ANOVA is robust. Here is the ANOVA table for the data: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square ( |
MS) F Factor (between) Error (within) Total Table 13.43 344.16 1,219.55 1,563.71 3 – 1 = 2 14 – 3 = 11 14 – 1 = 13 172.08 110.87 1.5521 P(F > 1.5521) = 0.2548 Since the p-value is so large, there is not good evidence against the null hypothesis of equal means. We decline to reject the null hypothesis. Thus, for 2012, there is not any good evidence of a significant difference in mean number of wins between the divisions of the American League. 804 Chapter 13 | F Distribution and One-way Anova This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 805 APPENDIX A: APPENDIX A REVIEW EXERCISES (CH 3–13) These review exercises are designed to provide extra practice on concepts learned before a particular chapter. For example, the review exercises for Chapter 3 cover material learned in Chapters 1 and 2. Chapter 3 Use the following information to answer the next six exercises. In a survey of 100 stocks on NASDAQ, the average percent increase for the past year was 9 percent for NASDAQ stocks. 1. The average increase for all NASDAQ stocks is the — A. population B. statistic C. parameter D. sample E. variable 2. All of the NASDAQ stocks are — A. population B. statistics C. parameter D. sample E. variable 3. Nine percent is — A. population B. statistics C. parameter D. sample E. variable 4. The 100 NASDAQ stocks in the survey are — A. population B. statistic C. parameter D. sample E. variable 806 Appendix A 5. The percent increase for one stock in the survey is — A. population B. statistic C. parameter D. sample E. variable 6. Would the data collected by qualitative, quantitative discrete, or quantitative continuous? Use the following information to answer the next two exercises. Thirty people spent two weeks around Mardi Gras in New Orleans. Their two-week weight gain is below. Note—a loss is shown by a negative weight gain. Weight Gain Frequency –2 –1 0 1 4 6 11 Table A1 3 5 2 4 13 2 1 7. Calculate the following values: A. The average weight gain for the two weeks B. The standard deviation C. The first, second, |
and third quartiles 8. Construct a histogram and box plot of the data. Chapter 4 Use the following information to answer the next two exercises. A recent poll concerning credit cards found that 35 percent of respondents use a credit card that gives them a mile of air travel for every dollar they charge. Thirty percent of the respondents charge more than $2,000 per month. Of those respondents who charge more than $2,000, 80 percent use a credit card that gives them a mile of air travel for every dollar they charge. 9. What is the probability that a randomly selected respondent will spend more than $2,000 and use a credit card that gives them a mile of air travel for every dollar they charge? A. B. C. D. (.30)(.35) (.80)(.35) (.80)(.30) (.80) 10. Are using a credit card that gives a mile of air travel for each dollar spent and charging more than $2,000 per month independent events? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A A. Yes B. No, and they are not mutually exclusive either C. No, but they are mutually exclusive D. Not enough information given to determine the answer 807 11. A sociologist wants to know the opinions of employed adult women about government funding for day care. She obtains a list of 520 members of a local business and professional women’s club and mails a questionnaire to 100 of these women selected at random. Sixty-eight questionnaires are returned. What is the population in this study? A. All employed adult women B. All the members of a local business and professional women’s club C. The 100 women who received the questionnaire D. All employed women with children Use the following information to answer the next two exercises. An article from the San Jose Mercury News was concerned with the racial mix of the 1,500 students at Prospect High School in Saratoga, CA. The table summarizes the results. Male and female values are approximate. Suppose one Prospect High School student is randomly selected. Gender/Ethnic Group White Asian Hispanic Black American Indian Male Female Table A2 400 440 468 132 115 140 35 40 16 14 12. Find the probability that a student is Asian or male. 13. Find the probability that a student is black given that the student is female. 14. A sample of pounds lost, in a certain month |
, by individual members of a weight reducing clinic produced the following statistics: • Mean = 5 lbs • Median = 4.5 lbs • Mode = 4 lbs • Standard deviation = 3.8 lbs • First quartile = 2 lbs • Third quartile = 8.5 lbs What is the correct statement? A. One fourth of the members lost exactly two pounds. B. The middle 50 percent of the members lost from two to 8.5 lbs. C. Most people lost 3.5 to 4.5 lbs. D. All of the choices above are correct. 15. What does it mean when a data set has a standard deviation equal to zero? A. All values of the data appear with the same frequency. B. The mean of the data is also zero. C. All of the data have the same value. 808 Appendix A D. There are no data to begin with. 16. Which statement describes the illustration? Figure A1 A. The mean is equal to the median. B. There is no first quartile. C. The lowest data value is the median. D. The median equals Q1 + Q3 2. 17. According to a recent article in the San Jose Mercury News the average number of babies born with significant hearing loss—deafness—is approximately 2 per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. 18. A friend offers you the following deal: For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift. • Ten of the coupons are for a free gift worth $6. • Eighty of the coupons are for a free gift worth $8. • Six of the coupons are for a free gift worth $12. • Four of the coupons are for a free gift worth $40. Based upon the financial gain or loss over the long run, should you play the game? A. Yes, I expect to come out ahead in money. B. No, I expect to come out behind in money. C. It doesn’t matter. I expect to break even. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the |
chance that he/she truly has the flu—and not just a nasty cold—is only about 4 percent. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. 19. Define the random variable and list its possible values. 20. State the distribution of X. 21. Find the probability that at least four of the 25 patients actually have the flu. 22. On average, for every 25 patients calling in, how many do you expect to have the flu? Use the following information to answer the next two exercises. Different types of writing can sometimes be distinguished by the number of letters in the words used. A student interested in this fact wants to study the number of letters of words used by Tom Clancy in his novels. She opens a Clancy novel at random and records the number of letters of the first 250 words on the page. 23. What kind of data was collected? A. Qualitative B. Quantitative continuous This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 809 C. Quantitative discrete 24. What is the population under study? Chapter 5 Use the following information to answer the next five exercises. A recent study of mothers of junior high school children in Santa Clara County reported that 76 percent of the mothers are employed in paid positions. Of those mothers who are employed, 64 percent work full-time—more than 35 hours per week—and 36 percent work part-time. However, out of all of the mothers in the population, 49 percent work full-time. The population under study is made up of mothers of junior high school children in Santa Clara County. Let E = employed and F = full-time employment. 25. A. Find the percent of all mothers in the population that are not employed. B. Find the percent of mothers in the population that are employed part-time. 26. The type of employment is considered to be what type of data? 27. Find the probability that a randomly selected mother works part-time given that she is employed. 28. Find the probability that a randomly selected person from the population will be employed or work full-time. 29. Being employed and working part-time— A. mutually exclusive events? Why or why not? B. independent events? Why or why not? Use the following additional information to answer the next two exercises. We randomly pick 10 mothers from the above population. We are interested |
in the number of the mothers that are employed. Let X = number of mothers that are employed. 30. State the distribution for X. 31. Find the probability that at least six are employed. 32. We expect the statistics discussion board to have, on average, 14 questions posted to it per week. We are interested in the number of questions posted to it per day. A. Define X. B. What are the values that the random variable may take on? C. State the distribution for X. D. Find the probability that from 10 to 14—inclusive—questions are posted to the listserv on a randomly picked day. 33. A person invests $1,000 into stock of a company that hopes to go public in one year. The probability that the person will lose all his money after one year, that is, his stock will be worthless, is 35 percent. The probability that the person’s stock will still have a value of $1,000 after one year, that is, no profit and no loss, is 60 percent. The probability that the person’s stock will increase in value by $10,000 after one year, that is, will be worth $11,000, is 5 percent. Find the expected profit after one year. 34. Rachel’s piano cost $3,000. The average cost for a piano is $4,000 with a standard deviation of $2,500. Becca’s guitar cost $550. The average cost for a guitar is $500 with a standard deviation of $200. Matt’s drums cost $600. The average cost for drums is $700 with a standard deviation of $100. Whose cost was lowest when compared to his or her own instrument? 810 Appendix A Figure A2 35. Explain why each statement is either true or false given the box plot in Figure A2. A. Twenty-five percent of the data are at most five. B. There is the same amount of data from 4–5 as there is from 5–7. C. There are no data values of three. D. Fifty percent of the data are four. Using the following information to answer the next two exercises. 64 faculty members were asked the number of cars they owned—including spouse and children’s cars. The results are given in the following graph. Figure A3 36. Find the approximate number of responses that were three. 37. Find the first, second, and third |
quartiles. Use them to construct a box plot of the data. Use the following information to answer the next three exercises. Table A3 shows data gathered from 15 girls on the Snow Leopard soccer team when they were asked how they liked to wear their hair. Supposed one girl from the team is randomly selected. Hair Style/Hair Color Blond Brown Black Ponytail Plain Table A3 3 2 2 2 5 1 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 811 38. Find the probability that the girl has black hair GIVEN that she wears a ponytail. 39. Find the probability that the girl wears her hair plain OR has brown hair. 40. Find the probability that the girl has blond hair AND that she wears her hair plain. Chapter 6 Use the following information to answer the next two exercises. X ~ U(3, 13) 41. Explain which of the following are false and which are true. A. f(x) = 1 10, 3 ≤ x ≤ 13 B. There is no mode. C. The median is less than the mean. D. P(x > 10) = P(x ≤ 6) 42. Calculate A. B. C. the mean, the median, and the 65th percentile. Figure A4 43. Which of the following is true for the box plot in Figure A4? A. Twenty-five percent of the data are at most five. B. There is about the same amount of data from 4–5 as there is from 5–7. C. There are no data values of three. D. Fifty percent of the data are four. 44. If P(G|H) = P(G), then which of the following is correct? A. G and H are mutually exclusive events. B. P(G) = P(H) C. Knowing that H has occurred will affect the chance that G will happen. D. G and H are independent events. 45. If P(J) =.3, P(K) =.63, and J and K are independent events, then explain which are correct and which are incorrect. A. P(J AND K) = 0 B. P(J OR K) =.9 C. P(J OR K) =.72 D. P(J) ≠ P(J|K) 812 Appendix A 46. On average |
, five students from each high school class get full scholarships to four-year colleges. Assume that most high school classes have about 500 students. X = the number of students from a high school class that get full scholarships to four-year schools. Which of the following is the distribution of X? A. P(5) B. B(500, 5) C. Exp ⎛ ⎝ ⎞ ⎠ 1 5 D. N ⎛ ⎝5, (.01)(.99) 500 ⎞ ⎠ Chapter 7 Use the following information to answer the next three exercises. Richard’s Furniture Company delivers furniture from 10 a.m. to 2 p.m. continuously and uniformly. We are interested in how long—in hours—past the 10 a.m. start time that individuals wait for their delivery. 47. X ~ ________ A. U(0, 4) B. U(10, 20) C. Exp(2) D. N(2, 1) 48. The average wait time is — A. one hour B. C. D. two hours two and a half hours four hours 49. Suppose that it is now past noon on a delivery day. The probability that a person must wait at least 1.5 more hours is — A. B. C. D. 1 4 1 2 3 4 3 8 50. Given X ~ Exp ⎛ ⎝ ⎞ ⎠ 1 3 A. Find P(x > 1). B. Calculate the minimum value for the upper quartile. C. Find P ⎛ ⎝x = 1 3 ⎞ ⎠ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 813 Appendix A 51. • Forty percent of full-time students took four years to graduate. • Thirty percent of full-time students took five years to graduate. • Twenty percent of full-time students took six years to graduate. • Ten percent of full-time students took seven years to graduate. The expected time for full-time students to graduate is — A. B. C. D. four years four and a half years five years five and a half years 52. Which of the following distributions is described by the following example? Many people can run a short distance of under two miles, but as the distance increases, fewer people can run |
that far. A. binomial B. uniform C. exponential D. normal 53. The length of time to brush one’s teeth is generally thought to be exponentially distributed with a mean of 3 4 minutes. Find the probability that a randomly selected person brushes his or her teeth less than 3 4 minutes. A..5 B. C. D. 3 4.43.63 54. Which distribution accurately describes the following situation? The chance that a teenage boy regularly gives his mother a kiss goodnight is about 20 percent. Fourteen teenage boys are randomly surveyed. Let X = the number of teenage boys that regularly give their mother a kiss goodnight. A. B(14,.20) B. P(2.8) C. N(2.8,2.24) D. Exp ⎛ ⎝ ⎞ ⎠ 1.20 55. A 2008 report on technology use states that approximately 20 percent of U.S. households have never sent an email. Suppose that we select a random sample of fourteen U.S. households. Let X = the number of households in a 2008 sample of 14 households that have never sent an email. A. B(14,.20) B. P(2.8) C. N(2.8,2.24) 814 D. Exp ⎛ ⎝ ⎞ ⎠ 1.20 Appendix A Chapter 8 Use the following information to answer the next three exercises. Suppose that a sample of 15 randomly chosen people were put on a special weight-loss diet. The amount of weight lost, in pounds, follows an unknown distribution with mean equal to 12 pounds and standard deviation equal to three pounds. Assume that the distribution for the weight loss is normal. 56. To find the probability that the mean amount of weight lost by 15 people is no more than 14 pounds, the random variable should be ________. A. number of people who lost weight on the special weight-loss diet B. C. D. the number of people who were on the diet the mean amount of weight lost by 15 people on the special weight-loss diet the total amount of weight lost by 15 people on the special weight-loss diet 57. Find the probability asked for in Question 56. 58. Find the 90th percentile for the mean amount of weight lost by 15 people. Using the following information to answer the next three exercises. The time of occurrence of the first accident during rushhour traffic at a major intersection |
is uniformly distributed between the three hour interval 4 p.m. to 7 p.m. Let X = the amount of time—hours—it takes for the first accident to occur. 59. What is the probability that the time of occurrence is within the first half-hour or the last hour of the period from 4 to 7 p.m.? A. It cannot be determined from the information given. B. C. D. 1 6 1 2 1 3 60. The 20th percentile occurs after how many hours? A. B. C..20.60.50 D. 1 61. Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. Let C = the total cumulative time. Then C follows which distribution? A. U(0,3) B. Exp(13) C. N(60, 5.477) D. N(1.5,.01875) 62. Using the information in Question 61, find the probability that the total time for all first accidents to occur is more than 43 hours. Use the following information to answer the next two exercises. The length of time a parent must wait for his children to This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 815 clean their rooms is uniformly distributed in the time interval from one to 15 days. 63. How long must a parent expect to wait for his children to clean their rooms? A. 8 days B. 3 days C. 14 days D. 6 days 64. What is the probability that a parent will wait more than six days given that the parent has already waited more than three days? A. B. C. D..5174.0174.7500.2143 Use the following information to answer the next five exercises. Twenty percent of the students at a local community college live in within five miles of the campus. Thirty percent of the students at the same community college receive some kind of financial aid. Of those who live within five miles of the campus, 75 percent receive some kind of financial aid. 65. Find the probability that a randomly chosen student at the local community college does not live within five miles of the campus. A. 80 percent B. 20 percent C. 30 percent D. Cannot be determined 66. Find the probability that a randomly chosen student at the local community college lives within five miles of the campus or receives some kind of financial aid. |
A. 50 percent B. 35 percent C. 27.5 percent D. 75 percent 67. Are living in student housing within five miles of the campus and receiving some kind of financial aid mutually exclusive? A. Yes B. No C. Cannot be determined 68. The interest rate charged on the financial aid is ________ data. A. Quantitative discrete B. Quantitative continuous C. Qualitative discrete D. Qualitative 816 Appendix A 69. The following information is about the students who receive financial aid at the local community college. • 1st quartile = $250 • 2nd quartile = $700 • 3rd quartile = $1,200 These amounts are for the school year. If a sample of 200 students is taken, how many are expected to receive $250 or more? A. 50 B. 250 C. 150 D. Cannot be determined Use the following information to answer the next two exercises. P(A) =.2, P(B) =.3; A and B are independent events. 70. P(A AND B) = — A. B..5.6 C. 0 D..06 71. P(A OR B) = — A. B. C..56.5.44 D. 1 72. If H and D are mutually exclusive events, P(H) =.25, P(D) =.15, then P(H|D). A. 1 B. 0 C. D..40.0375 Chapter 9 73. Rebecca and Matt are 14 year old twins. Matt’s height is two standard deviations below the mean for 14 year old boys’ height. Rebecca’s height is.10 standard deviations above the mean for 14 year old girls’ height. Interpret this. A. Matt is 2.1 inches shorter than Rebecca. B. Rebecca is very tall compared to other 14 year old girls. C. Rebecca is taller than Matt. D. Matt is shorter than the average 14 year old boy. 74. Construct a histogram of the IPO data (see Appendix C). Use the following information to answer the next three exercises. Ninety homeowners were asked the number of estimates they obtained before having their homes fumigated. Let X = the number of estimates. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 817 x Relative Frequency Cumulative Relative Frequency 1 2 4 5. |
3.2.4.1 Table A4 75. Complete the cumulative frequency column. 76. Calculate the sample mean (a), the sample standard deviation (b), and the percent of the estimates that fall at or below four (c). 77. Calculate the median, M, the first quartile, Q1, and the third quartile Q3. Then construct a box plot of the data. 78. The middle 50 percent of the data are between ________ and ________. Use the following information to answer the next three exercises. Seventy fifth and sixth graders were asked their favorite dinner. Pizza Hamburgers Spaghetti Fried Shrimp 5th Grader 15 6th Grader 15 6 7 Table A5 9 10 0 8 79. Find the probability that one randomly chosen child is in the 6th grade and prefers fried shrimp. A. B. C. D. 32 70 8 32 8 8 8 70 80. Find the probability that a child does not prefer pizza. A. B. C. 30 70 30 40 40 70 D. 1 81. Find the probability a child is in the fifth grade given that the child prefers spaghetti. A. B. 9 19 9 70 Appendix A 818 C. D. 9 30 19 70 82. A sample of convenience is a random sample. A. True B. False 83. A statistic is a number that is a property of the population. A. True B. False 84. You should always throw out any data that are outliers. A. True B. False 85. Lee bakes pies for a small restaurant in Felton, CA. She generally bakes 20 pies in a day, on average. Of interest is the number of pies she bakes each day. A. Define the random variable X. B. State the distribution for X. C. Find the probability that Lee bakes more than 25 pies in any given day. 86. Six different brands of Italian salad dressing were randomly selected at a supermarket. The grams of fat per serving are 7, 7, 9, 6, 8, and 5. Assume that the underlying distribution is normal. Calculate a 95 percent confidence interval for the population mean grams of fat per serving of Italian salad dressing sold in supermarkets. 87. Given: uniform, exponential, normal distributions. Match each to a statement below. A. mean = median ≠ mode B. mean > median > mode C. mean = median = mode Chapter 10 Use the following information to answer the next three |
exercises. In a survey at Kirkwood Ski Resort the following information was recorded. 0–10 11–20 21–40 40+ Ski 10 Snowboard 6 12 17 30 12 8 5 Table A6 Suppose that one person from Table A6 was randomly selected. 88. Find the probability that the person was a skier or was age 11–20. 89. Find the probability that the person was a snowboarder given he or she was age 21–40. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 819 90. Explain which of the following are true and which are false. A. Sport and age are independent events. B. Ski and age 11–20 are mutually exclusive events. C. P(Ski AND age 21–40) < P(Ski|age 21–40) D. P(Snowboard OR age 0–10) < P(Snowboard|age 0–10) 91. The average length of time a person with a broken leg wears a cast is approximately six weeks. The standard deviation is about three weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the 30 people. 92. The distribution for X is uniform. What can we say for certain about the distribution for X ¯ when n = 1? ¯ A. The distribution for X ¯ B. The distribution for X ¯ C. The distribution for X ¯ D. The distribution for X is still uniform with the same mean and standard deviation as the distribution for X. is normal with the different mean and a different standard deviation as the distribution for X. is normal with the same mean but a larger standard deviation than the distribution for X. is normal with the same mean but a smaller standard deviation than the distribution for X. 93. The distribution for X is uniform. What can we say for certain about the distribution for ∑ X when n = 50? A. The distribution for ∑ X is still uniform with the same mean and standard deviation as the distribution for X. B. The distribution for ∑ X is normal with the same mean but a larger standard deviation as the distribution for X. C. The distribution for ∑ X is normal with a larger mean and a larger standard deviation than the distribution for X. D. The distribution for ∑ X is normal with the same mean but a smaller standard deviation than the distribution for X |
. Use the following information to answer the next three exercises. A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots. 94. State the approximate distribution for X ¯ ~ ________. X ¯, the distribution for the average lengths of baby carrots in 16 five-pound bags. 95. Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches. is between 2.0 and 2.25 inches. 96. Find the probability that x¯ Use the following information to answer the next three exercises. At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of five minutes and a standard deviation of two minutes. 97. Find the 90th percentile of waiting time in minutes. 98. Find the median waiting time for one student. 99. Find the probability that the average waiting time for 40 students is at least 4.5 minutes. Chapter 11 Use the following information to answer the next the time that owners keep their cars—purchased new—is normally distributed with a mean of seven years and a standard deviation of two years. We are interested in how long an individual keeps his car—purchased new. Our population is people who buy their cars new. four exercises. Suppose that 820 Appendix A 100. Sixty percent of individuals keep their cars at most how many years? 101. Suppose that we randomly survey one person. Find the probability that person keeps his or her car less than 2.5 years. 102. If we are to pick individuals 10 at a time, find the distribution for the mean car length ownership. 103. If we are to pick 10 individuals, find the probability that the sum of their ownership time is more than 55 years. 104. For which distribution is the median not equal to the mean? A. Uniform B. Exponential C. Normal D. Student t 105. Compare the standard normal distribution to the Student’s t distribution, centered at zero. Explain which of the following are true and which are false. A. As the number surveyed increases, the area to the left of –1 for the Student’s t distribution approaches the area for the standard normal distribution. B. As the degrees of freedom decrease, the graph of the Student |
’s t distribution looks more like the graph of the standard normal distribution. C. If the number surveyed is 15, the normal distribution should never be used. Use the following information to answer the next five exercises. We are interested in the checking account balance of 24-old college students. We randomly survey 16 20-year-old college students. We obtain a sample mean of $640 and a sample standard deviation of $150. Let X = checking account balance of an individual 20-year-old college student. 106. Explain why we cannot determine the distribution of X. 107. If you were to create a confidence interval or perform a hypothesis test for the population mean checking account balance of 20-year-old college students, what distribution would you use? 108. Find the 95 percent confidence interval for the true mean checking account balance of a 20-year-old college student. 109. What type of data is the balance of the checking account considered to be? 110. What type of data is the number of 20-year-olds considered to be? 111. On average, a busy emergency room gets a patient with a shotgun wound about once per week. We are interested in the number of patients with a shotgun wound the emergency room gets per 28 days. A. Define the random variable X. B. State the distribution for X. C. Find the probability that the emergency room gets no patients with shotgun wounds in the next 28 days. Use the following information to answer the next two exercises. The probability that a certain slot machine will pay back money when a quarter is inserted is.30. Assume that each play of the slot machine is independent from each other. A person puts in 15 quarters for 15 plays. 112. Is the expected number of plays of the slot machine that will pay back money greater than, less than, or the same as the median? Explain your answer. 113. Is it likely that exactly eight of the 15 plays would pay back money? Justify your answer numerically. 114. A game is played with the following rules: • It costs $10 to enter. • A fair coin is tossed four times. • • If you do not get four heads or four tails, you lose your $10. If you get four heads or four tails, you get back your $10, plus $30 more. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 821 Over the long run of |
playing this game, what are your expected earnings? 115. • The mean grade on a math exam in Rachel’s class was 74, with a standard deviation of five. Rachel earned an 80. • The mean grade on a math exam in Becca’s class was 47, with a standard deviation of two. Becca earned a 51. • The mean grade on a math exam in Matt’s class was 70, with a standard deviation of eight. Matt earned an 83. Find whose score was the best, compared to his or her own class. Justify your answer numerically. Use the following information to answer the next two exercises. A random sample of 70 compulsive gamblers were asked the number of days they go to casinos per week. The results are given in the following graph. Figure A5 116. Find the number of responses that were five. 117. Find the mean, standard deviation, the median, the first quartile, the third quartile, and the IQR. 118. Based upon research at De Anza College, it is believed that about 19 percent of the student population speaks a language other than English at home. Suppose that a study was done this year to see if that percent has decreased. Ninetyeight students were randomly surveyed with the following results: Fourteen said that they speak a language other than English at home. A. State an appropriate null hypothesis. B. State an appropriate alternative hypothesis. C. Define the random variable, P′. D. Calculate the test statistic. E. Calculate the p-value. F. At the 5 percent level of decision, what is your decision about the null hypothesis? G. What is the Type I error? H. What is the Type II error? 119. Assume that you are an emergency paramedic called in to rescue victims of an accident. You need to help a patient who is bleeding profusely. The patient is also considered to be a high risk for contracting a blood-borne illness. Assume that the null hypothesis is that the patient does not have the a blood-borne illness. What is a Type I error? 120. It is often said that Californians are more casual than the rest of Americans. Suppose that a survey was done to see if the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals. Fifty of each was surveyed with the following results: Fifteen Californians wear jeans to work and six non- |
822 Appendix A Californians wear jeans to work. Let C = Californian professional; NC = non-Californian professional A. State appropriate null and alternate hypotheses. B. Define the random variable. C. Calculate the test statistic and p-value. D. At the 5 percent significance level, what is your decision? E. What is the Type I error? F. What is the Type II error? Use the following information to answer the next two exercises. A group of statistics students have developed a technique that they feel will lower their anxiety level on statistics exams. They measured their anxiety level at the start of the quarter and again at the end of the quarter. Recorded is the paired data in that order: (1,000, 900); (1,200, 1,050); (600, 700); (1,300, 1,100); (1,000, 900); (900, 900). 121. This is a test of (pick the best answer) — A. B. large samples, and independent means small samples, and independent means C. dependent means 122. State the distribution to use for the test. Chapter 12 Use the following information to answer the next two exercises. A recent survey of U.S. teenagers was answered by 720 teenagers, age 15–18. Six percent of teenagers surveyed said they are planning on going to college in another country. We are interested in the true proportion of U.S. teens, ages 15–18, who are planning on going to college in another country. 123. Find the 95 percent confidence interval for the true proportion of U.S. teens, ages 15–19, who are planning to go to college in another country. 124. The report also stated that the results of the survey are accurate to within ±3.7 percent at the 95 percent confidence level. Suppose that a new study is to be done. It is desired to be accurate to within 2 percent of the 95 percent confidence level. What is the minimum number that should be surveyed? 125. Given X ~ Exp ⎛ ⎝ ⎞ ⎠ 1 3. Sketch the graph that depicts: P(x > 1). Use the following information to answer the next three exercises. The amount of money a customer spends in one trip to the supermarket is known to have an exponential distribution. Suppose the mean amount of money a customer spends in one trip to the supermarket is $72. 126. Find the probability that one customer spends less than $ |
72 in one trip to the supermarket? 127. Suppose five customers pool their money. How much money altogether would you expect the five customers to spend in one trip to the supermarket in dollars? 128. State the distribution to use if you want to find the probability that the mean amount spent by five customers in one trip to the supermarket is less than $60. Chapter 13 Use the following information to answer the next two exercises. Suppose that the probability of a drought in any independent year is 20 percent. Out of those years in which a drought occurs, the probability of water rationing is 10 percent. However, in any year, the probability of water rationing is 5 percent. 129. What is the probability of both a drought and water rationing occurring? 130. Out of the years with water rationing, find the probability that there is a drought. Use the following information to answer the next three exercises. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 823 Apple Pumpkin Pecan Female 40 Male 20 Table A7 10 30 30 10 131. Suppose that one individual is randomly chosen. Find the probability that the person’s favorite pie is apple or the person is male. 132. Suppose that one male is randomly chosen. Find the probability his favorite pie is pecan. 133. Conduct a hypothesis test to determine if favorite pie type and gender are independent. Use the following information to answer the next two exercises. Let’s say that the probability that an adult watches the news at least once per week is.60. 134. We randomly survey 14 people. On average, how many people do we expect to watch the news at least once per week? 135. We randomly survey 14 people. Of interest is the number that watch the news at least once per week. State the distribution of X. X ~ ________. 136. The following histogram is most likely to be a result of sampling from which distribution? Figure A6 A. Chi-square B. Geometric C. Uniform D. Binomial 137. The ages of De Anza evening students is known to be normally distributed with a population mean of 40 and a population standard deviation of six. A sample of six De Anza evening students reported their ages in years as: 28; 35; 47; 45; 30; 50. Find the probability that the mean of six ages of randomly chosen students is less than 35 years. Hint—Find the |
sample mean. 138. A math exam was given to all the fifth grade children attending Country School. Two random samples of scores were taken. The null hypothesis is that the mean math scores for boys and girls in fifth grade are the same. Conduct a hypothesis test. 824 Appendix A n ¯ x Boys 55 82 Girls 60 86 Table A8 s2 29 46 139. In a survey of 80 males, 55 had played an organized sport growing up. Of the 70 females surveyed, 25 had played an organized sport growing up. We are interested in whether the proportion for males is higher than the proportion for females. Conduct a hypothesis test. 140. Which of the following is preferable when designing a hypothesis test? A. Maximize α and minimize β B. Minimize α and maximize β C. Maximize α and β D. Minimize α and β Use the following information to answer the next three exercises. One hundred twenty people were surveyed as to their favorite beverage. The results are below. Beverage/Age 0–9 10–19 20–29 30+ Totals Milk Soda Juice Totals Table A9 14 3 7 10 8 12 24 330 6 26 12 44 0 15 7 22 30 52 38 120 141. Are the events of milk and 30+— A. independent events? Justify your answer. B. mutually exclusive events? Justify your answer. 142. Suppose that one person is randomly chosen. Find the probability that person is 10–19 given that he or she prefers juice. 143. Are Preferred Beverage and Age independent events? Conduct a hypothesis test. 144. Given the following histogram, which distribution is the data most likely to come from? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix A 825 Figure A7 A. Uniform B. Exponential C. Normal D. Chi-square Solutions Chapter 3 1. C Parameter 2. A Population 3. B Statistic 4. D Sample 5. E Variable 6. quantitative continuous 7. A. 2.27 B. 3.04 C. –1, 4, 4 8. Answers will vary. Chapter 4 9. C (.80)(.30) 10. B No, and they are not mutually exclusive either. 11. A All employed adult women 12..5773 13..0522 14. B The middle fifty percent of the members lost from 2 to 8.5 lbs. 15. C All of the data have the |
same value. 16. C The lowest data value is the median. 17..279 826 Appendix A 18. B No, I expect to come out behind in money. 19. X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2, …25 20. B(25,.04) 21..0165 22. 1 23. C Quantitative discrete 24. all words used by Tom Clancy in his novels Chapter 5 25. A. 24 percent B. 27 percent 26. qualitative 27..36 28..7636 29. A. no B. no 30. B(10,.76) 31..9330 32. A. X = the number of questions posted to the statistics listserv per day. B. X = 0, 1, 2,… C. X ~ P(2) D. 0 33. $150 34. Matt 35. A. False B. True C. False D. False 36. 16 37. first quartile: 2 second quartile: 2 third quartile: 3 38. 0.5 39. 7 15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 827 Appendix A 40. 2 15 Chapter 6 41. A. True B. True C. False – the median and the mean are the same for this symmetric distribution. D. True 42. A. 8 B. 8 C. P(x < k) = 0.65 = (k – 3) ⎛ ⎝ ⎞ ⎠ 1 10. k = 9.5 43. A. False – 3 4 of the data are at most five. B. True – each quartile has 25 percent of the data. C. False – that is unknown. D. False – 50 percent of the data are four or less. 44. D G and H are independent events. 45. A. False – J and K are independent so they are not mutually exclusive which would imply dependency (meaning P(J AND K) is not 0). B. False – see answer c. C. True – P(J OR K) = P(J) + P(K) – P(J AND K) = P(J) + P(K) – P(J)P(K) =.3 +.6 – (.3)(.6 |
) =.72. Note the P(J AND K) = P(J)P(K) because J and K are independent. D. False – J and K are independent so P(J) = P(J|K). 46. A P(5) Chapter 7 47. A U(0, 4) 48. B 2 hours 49. A 1 4 50. A..7165 B. 4.16 C. 0 51. C 5 years 828 52. C exponential 53..63 54. A B(14,.20) 55. A B(14,.20) Chapter 8 56. C The mean amount of weight lost by 15 people on the special weight-loss diet. Appendix A 57..9951 58. 12.99 59. C 1 2 60. B.60 61. C N(60, 5.477) 62..9990 63. A eight days 64. C.7500 65. A 80 percent 66. B 35 percent 67. B no 68. B Quantitative continuous 69. C 150 70. D.06 71. C.44 72. B 0 Chapter 9 73. D Matt is shorter than the average 14 year old boy. 74. Answers will vary. 75. x Relative Frequency Cumulative Relative Frequency 1 2 4 5.3.2.4.1 Table A10.3.2.4.1 76. A. 2.8 B. 1.48 C. 90 percent 77. M = 3; Q1 = 1; Q3 = 4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 829 Appendix A 78. 1 and 4 79. D 8 70 80. C 40 70 81. A 9 19 82. B False 83. B False 84. B False 85. A. X = the number of pies Lee bakes every day. B. P(20) C..1122 86. CI: (5.25, 8.48) 87. A. uniform B. exponential C. normal Chapter 10 88. 77 100 89. 12 42 90. A. False B. False C. True D. False 91. N(180, 16.43) ¯ 92. A The distribution for X is still uniform with the same mean and standard deviation as the distribution for X. 93. C The distribution for ∑ X is normal with a larger mean and a larger |
standard deviation than the distribution for X. 94. N ⎛ ⎝2,.25 16 ⎞ ⎠ 95. Answers will vary. 96..5000 97. 7.6 98. 5 99..9431 Appendix A 830 Chapter 11 100. 7.5 101..0122 102. N(7,.63) 103..9911 104. B exponential 105. A. True B. False C. False 106. Answers will vary. 107. Student’s t with df = 15 108. (560.07, 719.93) 109. quantitative continuous data 110. quantitative discrete data 111. A. X = the number of patients with a shotgun wound the emergency room gets per 28 days. B. P(4) C..0183 112. greater than 113. no; P(x = 8) =.0348 114. You will lose $5. 115. Becca 116. 14 117. sample mean = 3.2 sample standard deviation = 1.85 median = 3 Q1 = 2 Q3 = 5 IQR = 3 118. d. z = –1.19 e..1171 f. Do not reject the null hypothesis. 119. We conclude that the patient does have the illness when, in fact, the patient does not. 120. c. z = 2.21; p =.0136 d. Reject the null hypothesis. e. We conclude that the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals when, in fact, it is not greater. f. We cannot conclude that the proportion of Californian professionals that wear jeans to work is greater than the proportion of non-Californian professionals when, in fact, it is greater. 121. C dependent means 122. t5 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 831 Appendix A Chapter 12 123. (.0424,.0770) 124. 2,401 125. Check student's solution. 126..6321 127. $360 128. N ⎛ ⎝72, ⎞ ⎠ 72 5 Chapter 13 129..02 130..40 131. 100 140 132. 10 60 133. p-value = 0; reject the null hypothesis; conclude that they are dependent events 134. 8.4 135. B |
(14,.60) 136. D Binomial 137..3669 138. p-value =.0006; reject the null hypothesis; conclude that the averages are not equal 139. p-value = 0; reject the null hypothesis; conclude that the proportion of males is higher 140. minimize α and β 141. A. no B. yes, P(M AND 30+) = 0 142. 12 38 143. no; p-value = 0 144. A uniform References Baran, D. (2010). Twenty percent of Americans have never used email. Retrieved from http://www.webguild.org/20080519/ 20-percent-of-americans-have-never-used-email. Parade Magazine. (n.d.). Retrieved from https://parade.com/. San Jose Mercury News. (n.d.). Retrieved from http://www.mercurynews.com/. 832 Appendix A This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 833 APPENDIX B: APPENDIX B PRACTICE TESTS (1–4) AND FINAL EXAMS Practice Test 1 1.1: Definitions of Statistics, Probability, and Key Terms Use the following information to answer the next three exercises. A grocery store is interested in how much money, on average, their customers spend each visit in the produce department. Using their store records, they draw a sample of 1,000 visits and calculate each customer’s average spending on produce. 1. Identify the population, sample, parameter, statistic, variable, and data for this example. A. population B. sample C. parameter D. statistic E. variable F. data 2. What kind of data is amount of money spent on produce per visit? A. Qualitative B. Quantitative-continuous C. Quantitative-discrete 3. The study finds that the mean amount spent on produce per visit by the customers in the sample is $12.84. This is an example of a A. Population B. Sample C. Parameter D. Statistic E. Variable 1.2: Data, Sampling, and Variation in Data and Sampling Use the following information to answer the next two exercises. A health club is interested in knowing how many times a typical member uses the club in a week. They decide to ask every tenth customer on a specified day to complete a |
short survey, including information about how many times they have visited the club in the past week. 4. What kind of a sampling design is this? A. Cluster B. Stratified 834 C. Simple random D. Systematic 5. Number of visits per week is what kind of data? A. Qualitative B. Quantitative-continuous C. Quantitative-discrete Appendix B 6. Describe a situation in which you would calculate a parameter, rather than a statistic. 7. The U.S. federal government conducts a survey of high school seniors concerning their plans for future education and employment. One question asks whether they are planning to attend a four-year college or university in the following year. Fifty percent answer yes to this question. That 50 percent is a A. Parameter B. Statistic C. Variable D. Data 8. Imagine that the U.S. federal government had the means to survey all high school seniors in the United States concerning their plans for future education and employment, and found that 50 percent were planning to attend a four-year college or university in the following year. This 50 percent is an example of a A. Parameter B. Dtatistic C. Variable D. Data Use the following information to answer the next three exercises. A survey of a random sample of 100 nurses working at a large hospital asked how many years they had been working in the profession. Their answers are summarized in the following (incomplete) table. 9. Fill in the blanks in the table and round your answers to two decimal places for the Relative Frequency and Cumulative Relative Frequency cells. # of years Frequency Relative Frequency Cumulative Relative Frequency < 5 5–10 > 10 Table B1 25 30 empty 10. What proportion of nurses have five or more years of experience? 11. What proportion of nurses have 10 or fewer years of experience? 12. Describe how you might draw a random sample of 30 students from a lecture class of 200 students. 13. Describe how you might draw a stratified sample of students from a college, where the strata are the students’ class standing (freshman, sophomore, junior, or senior). 14. A manager wants to draw a sample, without replacement, of 30 employees from a workforce of 150. Describe how the chance of being selected will change over the course of drawing the sample. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 8 |
35 15. The manager of a department store decides to measure employee satisfaction by selecting four departments at random, and conducting interviews with all the employees in those four departments. What type of survey design is this? A. Cluster B. Stratified C. Simple random D. Systematic 16. A popular American television sports program conducts a poll of viewers to see which team they believe will win the National Football League (NFL) championship this year. Viewers vote by calling a number displayed on the television screen and telling the operator which team they think will win. Do you think that those who participate in this poll are representative of all football fans in America? 17. Two researchers studying vaccination rates independently draw samples of 50 children, aged three–18 months, from a large urban area, and determine if they are up to date on their vaccinations. One researcher finds that 84 percent of the children in her sample are up to date, and the other finds that 86 percent in his sample are up to date. Assuming both followed proper sampling procedures and did their calculations correctly, what is a likely explanation for this discrepancy? 18. A high school increased the length of the school day from 6.5 to 7.5 hours. Students who wished to attend this high school were required to sign contracts pledging to put forth their best effort on their school work and to obey the school rules; if they did not wish to do so, they could attend another high school in the district. At the end of one year, student performance on statewide tests had increased by 10 percentage points over the previous year. Does this prove that a longer school day improves student achievement? 19. You read a newspaper article reporting that eating almonds leads to increased life satisfaction. The study was conducted by the Almond Growers Association, and was based on a randomized survey asking people about their consumption of various foods, including almonds, and also about their satisfaction with different aspects of their life. Does anything about this poll lead you to question its conclusion? 20. Why is non-response a problem in surveys? 1.3: Frequency, Frequency Tables, and Levels of Measurement 21. Compute the mean of the following numbers, and report your answer using one more decimal place than is present in the original data: 14, 5, 18, 23, 6 1.4: Experimental Design and Ethics 22. A psychologist is interested in whether the size of tableware (bowls, plates, etc.) influences how much college students eat. He randomly assigns 100 college students to one of two groups. |
The first is served a meal using normal-sized tableware, while the second is served the same meal but using tableware that it 20 percent smaller than normal. He records how much food is consumed by each group. Identify the following components of this study. A. population B. sample C. experimental units D. explanatory variable E. F. treatment response variable 23. A researcher analyzes the results of the Scholastic Aptitude Test (SAT) over a five-year period and finds that male students on average score higher on the math section, and female students on average score higher on the verbal section. She concludes that these observed differences in test performance are due to genetic factors. Explain how lurking variables could offer an alternative explanation for the observed differences in test scores. 24. Explain why it would not be possible to use random assignment to study the health effects of exercise. 25. A professor conducts a telephone survey of a city’s population by drawing a sample of numbers from the phone book and having her student assistants call each of the selected numbers once to administer the survey. What are some sources of 836 bias with this survey? Appendix B 26. A professor offers extra credit to students who take part in her research studies. What is an ethical problem with this method of recruiting subjects? 2.1: Stem-and Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs Use the following information to answer the next four exercises. The midterm grades on a chemistry exam, graded on a scale of 0 to 100, were 62, 64, 65, 65, 68, 70, 72, 72, 74, 75, 75, 75, 76, 78, 78, 81, 83, 83, 84, 85, 87, 88, 92, 95, 98, 98, 100, 100, 740 27. Do you see any outliers in this data? If so, how would you address the situation? 28. Construct a stem plot for this data, using only the values in the range zero–100. 29. Describe the distribution of exam scores. 2.2: Histograms, Frequency Polygons, and Time Series Graphs 30. In a class of 35 students, seven students received scores in the 70–79 range. What is the relative frequency of scores in this range? Use the following information to answer the next three exercises. You conduct a poll of 30 students to see how many classes they are taking this term. Your results are 1; |
1; 1; 1 2; 2; 2; 2; 2 3; 3; 3; 3; 3; 3; 3; 3 4; 4; 4; 4; 4; 4; 4; 4; 4 5; 5; 5; 5 31. You decide to construct a histogram of this data. What will be the range of your first bar, and what will be the central point? 32. What will be the widths and central points of the other bars? 33. Which bar in this histogram will be the tallest, and what will be its height? 34. You get data from the U.S. Census Bureau on the median household income for your city, and decide to display it graphically. Which is the better choice for this data, a bar graph or a histogram? 35. You collect data on the color of cars driven by students in your statistics class, and want to display this information graphically. Which is the better choice for this data, a bar graph or a histogram? 2.3: Measures of the Location of the Data 36. Your daughter brings home test scores showing that she scored in the 80th percentile in math and the 76th percentile in reading for her grade. Interpret these scores. 37. You have to wait 90 minutes in the emergency room of a hospital before you can see a doctor. You learn that your wait time was in the 82nd percentile of all wait times. Explain what this means, and whether you think it is good or bad. 2.4: Box Plots Use the following information to answer the next three exercises. 1; 1; 2; 3; 4; 4; 5; 5; 6; 7; 7; 8; 9 38. What is the median for this data? 39. What is the first quartile for this data? 40. What is the third quartile for this data? Use the following information to answer the next four exercises. This box plot represents scores on the final exam for a physics class. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 837 Figure B1 41. What is the median for this data, and how do you know? 42. What are the first and third quartiles for this data, and how do you know? 43. What is the interquartile range for this data? 44. What is the range for this data? 2.5: Measures of |
the Center of the Data 45. In a marathon, the median finishing time was 3:35:04 (three hours, 35 minutes, and four seconds). You finished in 3:34:10. Interpret the meaning of the median time, and discuss your time in relation to it. Use the following information to answer the next three exercises. The values, in thousands of dollars, for houses on a block, are 45; 47; 47.5; 51; 53.5; 125. 46. Calculate the mean for this data. 47. Calculate the median for this data. 48. Which do you think better reflects the average value of the homes on this block? 2.6: Skewness and the Mean, Median, and Mode 49. In a left-skewed distribution, which is greater? A. The mean B. The media C. The mode 50. In a right-skewed distribution, which is greater? A. The mean B. The median C. The mode 51. In a symmetrical distribution, what will be the relationship among the mean, median, and mode? 2.7: Measures of the Spread of the Data Use the following information to answer the next four exercises. 10; 11; 15; 15; 17; 22 52. Compute the mean and standard deviation for this data; use the sample formula for the standard deviation. 53. What number is two standard deviations above the mean of this data? 54. Express the number 13.7 in terms of the mean and standard deviation of this data. 55. In a biology class, the scores on the final exam were normally distributed, with a mean of 85 and a standard deviation of five. Susan got a final exam score of 95. Express her exam result as a z score, and interpret its meaning. 3.1: Terminology Use the following information to answer the next two exercises. You have a jar full of marbles: 50 are red, 25 are blue, and 15 are yellow. Assume you draw one marble at random for each trial and replace it before the next trial. Let P(R) = the probability of drawing a red marble. Let P(B) = the probability of drawing a blue marble. 838 Appendix B Let P(Y) = the probability of drawing a yellow marble. 56. Find P(B). 57. Which is more likely, drawing a red marble or a yellow marble? Justify your answer numerically. Use the following information to answer |
the next two exercises. The following are probabilities describing a group of college students. Let P(M) = the probability that the student is male Let P(F) = the probability that the student is female Let P(E) = the probability the student is majoring in education Let P(S) = the probability the student is majoring in science 58. Write the symbols for the probability that a student, selected at random, is both female and a science major. 59. Write the symbols for the probability that the student is an education major, given that the student is male. 3.2: Independent and Mutually Exclusive Events 60. Events A and B are independent. If P(A) = 0.3 and P(B) = 0.5, find P(A AND B). 61. C and D are mutually exclusive events. If P(C) = 0.18 and P(D) = 0.03, find P(C OR D). 3.3: Two Basic Rules of Probability 62. In a high school graduating class of 300, 200 students are going to college, 40 are planning to work full-time, and 80 are taking a gap year. Are these events mutually exclusive? Use the following information to answer the next two exercises. An archer hits the center of the target (the bullseye) 70 percent of the time. However, she is a streak shooter, and if she hits the center on one shot, her probability of hitting it on the shot immediately following is 0.85. Written in probability notation P(A) = P(B) = P(hitting the center on one shot) = 0.70 P(B|A) = P(hitting the center on a second shot, given that she hit it on the first) = 0.85 63. Calculate the probability that she will hit the center of the target on two consecutive shots. 64. Are P(A) and P(B) independent in this example? 3.4: Contingency Tables Use the following information to answer the next three exercises. The following contingency table displays the number of students who report studying at least 15 hours per week, and how many made the honor roll in the past semester. Honor Roll No Honor Roll Total Study at least 15 hours/week Study less than 15 hours/week 125 200 193 Total Table B2 1,000 65. Complete the table. 66. Find P (honor roll|study at least 15 hours per |
week). 67. What is the probability a student studies less than 15 hours per week? 68. Are the events study at least 15 hours per week and makes the honor roll independent? Justify your answer numerically. 3.5: Tree and Venn Diagrams 69. At a high school, some students play on the tennis team and some play on the soccer team, but neither plays both tennis and soccer. Draw a Venn diagram illustrating this. 70. At a high school, some students play tennis, some play soccer, and some play both. Draw a Venn diagram illustrating this. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 839 Practice Test 1 Solutions 1.1: Definitions of Statistics, Probability, and Key Terms 1. A. population: all the shopping visits by all the store’s customers B. sample: the 1,000 visits drawn for the study C. parameter: the average expenditure on produce per visit by all the store’s customers D. statistic: the average expenditure on produce per visit by the sample of 1,000 E. variable: the expenditure on produce for each visit F. data: the dollar amounts spent on produce; for instance, $15.40, $11.53, etc. 2. C 3. D 1.2: Data, Sampling, and Variation in Data and Sampling 4. D 5. C 6. Answers will vary. Sample Answer: Any solution in which you use data from the entire population is acceptable. For instance, a professor might calculate the average exam score for her class: Because the scores of all members of the class were used in the calculation, the average is a parameter. 7. B 8. A 9. 10. 0.75 11. 0.55 # of years Frequency Relative Frequency Cumulative Relative Frequency < 5 5–10 > 10 Table B3 25 30 45 0.25 0.30 0.45 0.25 0.55 1 12. Answers will vary. Sample Answer: One possibility is to obtain the class roster and assign each student a number from 1 to 200. Then, use a random number generator or table of random number to generate 30 numbers between 1 and 200, and select the students matching the random numbers. It would also be acceptable to write each student’s name on a card, shuffle them in a box, and draw 30 names at random. 13. One |
possibility would be to obtain a roster of students enrolled in the college, including the class standing for each student. Then, you would draw a proportionate random sample from within each class. For instance, if 30 percent of the students in the college are freshman, then 30 percent of your sample would be drawn from the freshman class. 14. For the first person picked, the chance of any individual being selected is one in 150. For the second person, it is one in 149, for the third it is one in 148, and so on. For the 30th person selected, the chance of selection is one in 121. 15. A 16. No. There are at least two chances for bias. First, the viewers of this particular program may not be representative of American football fans as a whole. Second, the sample will be self-selected, because people have to make a phone call in order to take part, and those people are probably not representative of the American football fan population as a whole. 840 Appendix B 17. These results (84 percent in one sample, 86 percent in the other) are probably due to sampling variability. Each researcher drew a different sample of children, and you would not expect them to get exactly the same result, although you would expect the results to be similar, as they are in this case. 18. No. The improvement could also be due to self-selection: Only motivated students were willing to sign the contract, and they would have done well even in a school with 6.5 hour days. Because both changes were implemented at the same time, it is not possible to separate out their influence. 19. At least two aspects of this poll are troublesome. The first is that it was conducted by a group who would benefit by the result—almond sales are likely to increase if people believe that eating almonds will make them happier. The second is that this poll found that almond consumption and life satisfaction are correlated, but it does not establish that eating almonds causes satisfaction. It is equally possible, for instance, that people with higher incomes are more likely to eat almonds and are also more satisfied with their lives. 20. You want the sample of people who take part in a survey to be representative of the population from which they are drawn. People who refuse to take part in a survey often have different views than those who do participate, and so even a random sample may produce biased results if a large percentage of those selected refuse to participate in a survey. 1.3: Frequency, Frequency Tables, |
and Levels of Measurement 21. 13.2 1.4: Experimental Design and Ethics 22. A. population: all college students B. sample: the 100 college students in the study C. experimental units: each individual college student who participated D. explanatory variable: the size of the tableware E. F. treatment: tableware that is 20 percent smaller than normal response variable: the amount of food eaten 23. There are many lurking variables that could influence the observed differences in test scores. Perhaps the boys, on average, have taken more math courses than the girls, and the girls have taken more English classes than the boys. Perhaps the boys have been encouraged by their families and teachers to prepare for a career in math and science, and thus have put more effort into studying math, while the girls have been encouraged to prepare for fields like communication and psychology that are more focused on language use. A study design would have to control for these and other potential lurking variables (anything that could explain the observed difference in test scores, other than the genetic explanation) in order to draw a scientifically sound conclusion about genetic differences. 24. To use random assignment, you would have to be able to assign people to either exercise or not exercise. Because exercise has many beneficial effects, this would not be an ethical experiment. We will study people who chose to exercise and compare them to people who chose not to exercise, and try to control for the other ways those two groups may differ (lurking variables). 25. Sources of bias include the fact that not everyone has a telephone, that cell phone numbers are often not listed in published directories, and that an individual might not be at home at the time of the phone call; all these factors make it likely that the respondents to the survey will not be representative of the population as a whole. 26. Research subjects should not be coerced into participation, and offering extra credit in exchange for participation could be construed as coercion. In addition, this method will result in a volunteer sample, which cannot be assumed to be representative of the population as a whole. 2.1: Stem-and Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs 27. The value 740 is an outlier, because the exams were graded on a scale of zero to 100, and 740 is far outside that range. It may be a data entry error, with the actual score being 74, so the professor should check that exam again to see what the actual score was. 28 |
. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 841 Stem Leaf 10 0 0 Table B4 29. Most scores on this exam were in the range of 70–89, with a few scoring in the 60–69 range, and a few in the 90–100 range. 2.2: Histograms, Frequency Polygons, and Time Series Graphs 30. RF = 7 35 = 0.2 31. The range will be 0.5–1.5, and the central point will be 1. 32. Range 1.5–2.5, central point 2; range 2.5–3.5, central point 3; range 3.5–4.5, central point 4; range 4.5–5.5, central point 5. 33. The bar from 3.5 to 4.5, with a central point of 4, will be tallest; its height will be nine, because there are nine students taking four courses. 34. The histogram is a better choice, because income is a continuous variable. 35. A bar graph is the better choice, because this data is categorical rather than continuous. 2.3: Measures of the Location of the Data 36. Your daughter scored better than 80 percent of the students in her grade on math and better than 76 percent of the students in reading. Both scores are very good, and place her in the upper quartile, but her math score is slightly better in relation to her peers than her reading score. 37. You had an unusually long wait time, which is bad: 82 percent of patients had a shorter wait time than you, and only 18 percent had a longer wait time. 2.4: Box Plots 38. 5 39. 3 40. 7 41. The median is 86, as represented by the vertical line in the box. 42. The first quartile is 80, and the third quartile is 92, as represented by the left and right boundaries of the box. 43. IQR = 92 – 80 = 12 44. Range = 100 – 75 = 25 2.5: Measures of the Center of the Data 45. Half the runners who finished the marathon ran a time faster than 3:35:04, and half ran a time slower than 3:35:04. Your time is faster than the median time, so you did better than more than half of the runners in this race |
. 46. 61.5, or $61,500 47. 49.25, or $49,250 48. The median, because the mean is distorted by the high value of one house. 2.6: Skewness and the Mean, Median, and Mode 49. C Appendix B 842 50. A 51. They will all be fairly close to one another. 2.7: Measures of the Spread of the Data 52. Mean: 15 Standard deviation: 4.3 μ = 10 + 11 + 15 + 15 + 17 + 22 6 = 15 2 ∑ ⎛ ⎝x − x¯ ⎞ ⎠ n − 1 s = = 94 5 = 4.3 53. 15 + (2)(4.3) = 23.6 54. 13.7 is one standard deviation below the mean of this data, because 15 – 4.3 = 10.7 55. z = 95 − 85 5 = 2.0 Susan’s z score was 2.0, meaning she scored two standard deviations above the class mean for the final exam. 3.1: Terminology 56. P(B) = 25 90 = 0.28 57. Drawing a red marble is more likely. P(R) = 50 80 P(Y) = 15 80 = 0.19 = 0.62 58. P(F AND S) 59. P(E|M) 3.2: Independent and Mutually Exclusive Events 60. P(A AND B) = (0.3)(0.5) = 0.15 61. P(C OR D) = 0.18 + 0.03 = 0.21 3.3: Two Basic Rules of Probability 62. No, they cannot be mutually exclusive, because they add up to more than 300. Therefore, some students must fit into two or more categories (e.g., both going to college and working full time). 63. P(A and B) = (P(B|A))(P(A)) = (0.85)(0.70) = 0.595 64. No. If they were independent, P(B) would be the same as P(B|A). We know this is not the case, because P(B) = 0.70 and P(B|A) = 0.85. 3.4: Contingency Tables 65. Honor roll No honor roll Total Study at least 15 hours/week 48 |
2 Study less than 15 hours/week 125 Total Table B5 607 200 193 393 682 318 1,000 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 843 66. P(honor roll|study at least 15 hours word per week) = 482 1,000 = 0.482 67. P(study less than 15 hours word per week) = 125 + 193 1,000 = 0.318 68. Let P(S) = study at least 15 hours per week Let P(H) = make the honor roll From the table, P(S) = 0.682, P(H) = 0.607, and P(S AND H) = 0.482. If P(S) and P(H) were independent, then P(S AND H) would equal (P(S))(P(H)). However, (P(S))(P(H)) = (0.682)(0.607) = 0.414, while P(S AND H) = 0.482. Therefore, P(S) and P(H) are not independent. 3.5: Tree and Venn Diagrams 69. Figure B2 70. Figure B3 Practice Test 2 4.1: Probability Distribution Function (PDF) for a Discrete Random Variable Use the following information to answer the next five exercises. You conduct a survey among a random sample of students at a particular university. The data collected includes their major, the number of classes they took the previous semester, and the amount of money they spent on books purchased for classes in the previous semester. 844 Appendix B 1. If X = student’s major, then what is the domain of X? 2. If Y = the number of classes taken in the previous semester, what is the domain of Y? 3. If Z = the amount of money spent on books in the previous semester, what is the domain of Z? 4. Why are X, Y, and Z in the previous example random variables? 5. After collecting data, you find that, for one case, z = –7. Is this a possible value for Z? 6. What are the two essential characteristics of a discrete probability distribution? Use this discrete probability distribution represented in this table to answer the following six questions. The university library records the number of books checked out by each patron over the course |
of one day, with the following result: x P(x) 0 1 2 3 4 0.20 0.45 0.20 0.10 0.05 Table B6 7. Define the random variable X for this example. 8. What is P(x > 2)? 9. What is the probability a patron will check out at least one book? 10. What is the probability a patron will take out no more than three books? 11. If the table listed P(x) as 0.15, how would you know that there was a mistake? 12. What is the average number of books taken out by a patron? 4.2: Mean or Expected Value and Standard Deviation Use the following information to answer the next four exercises. Three jobs are open in a company: one in the accounting department, one in the human resources department, and one in the sales department. The accounting job receives 30 applicants, and the human resources and sales department 60 applicants. 13. If X = the number of applications for a job, use this information to fill in Table B7. x P(x) xP(x) Table B7 14. What is the mean number of applicants? 15. What is the PDF for X? 16. Add a fourth column to the table, for (x – μ)2P(x). 17. What is the standard deviation of X? 4.3: Binomial Distribution 18. In a binomial experiment, if p = 0.65, what does q equal? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 845 19. What are the required characteristics of a binomial experiment? 20. Joe conducts an experiment to see how many times he has to flip a coin before he gets four heads in a row. Does this qualify as a binomial experiment? Use the following information to answer the next three exercises. In a particular community, 65 percent of households include at least one person who has graduated from college. You randomly sample 100 households in this community. Let X = the number of households including at least one college graduate. 21. Describe the probability distribution of X. 22. What is the mean of X? 23. What is the standard deviation of X? Use the following information to answer the next four exercises. Joe is the star of his school’s baseball team. His batting average is 0.400, meaning that for every |
10 times he comes to bat (an at-bat), four of those times he gets a hit. You decide to track his batting performance for his next 20 at-bats. 24. Define the random variable X in this experiment. 25. Assuming Joe’s probability of getting a hit is independent and identical across all 20 at-bats, describe the distribution of X. 26. Given this information, what number of hits do you predict Joe will get? 27. What is the standard deviation of X? 4.4: Geometric Distribution 28. What are the three major characteristics of a geometric experiment? 29. You decide to conduct a geometric experiment by flipping a coin until it comes up heads. This takes five trials. Represent the outcomes of this trial, using H for heads and T for tails. 30. You are conducting a geometric experiment by drawing cards from a normal 52-card pack, with replacement, until you draw the Queen of Hearts. What is the domain of X for this experiment? 31. You are conducting a geometric experiment by drawing cards from a normal 52-card deck, without replacement, until you draw a red card. What is the domain of X for this experiment? Use the following information to answer the next three exercises. In a particular university, 27 percent of students are engineering majors. You decide to select students at random until you choose one that is an engineering major. Let X = the number of students you select until you find one that is an engineering major. 32. What is the probability distribution of X? 33. What is the mean of X? 34. What is the standard deviation of X? 4.5: Hypergeometric Distribution 35. You draw a random sample of 10 students to participate in a survey, from a group of 30, consisting of 16 boys and 14 girls. You are interested in the probability that seven of the students chosen will be boys. Does this qualify as a hypergeometric experiment? List the conditions and whether or not they are met. 36. You draw five cards, without replacement, from a normal 52-card deck of playing cards, and are interested in the probability that two of the cards are spades. What are the group of interest, size of the group of interest, and sample size for this example? 4.6: Poisson Distribution 37. What are the key characteristics of the Poisson distribution? Use the following information to answer the next three exercises. The number of drivers to arrive at a toll booth in an hour can be |
modeled by the Poisson distribution. 38. If X = the number of drivers, and the average numbers of drivers per hour is four, how would you express this distribution? 39. What is the domain of X? 40. What are the mean and standard deviation of X? 846 Appendix B 5.1: Continuous Probability Functions 41. You conduct a survey of students to see how many books they purchased the previous semester, the total amount they paid for those books, the number they sold after the semester was over, and the amount of money they received for the books they sold. Which variables in this survey are discrete, and which are continuous? 42. With continuous random variables, we never calculate the probability that X has a particular value, but we always speak in terms of the probability that X has a value within a particular range. Why is this? 43. For a continuous random variable, why are P(x < c) and P(x ≤ c) equivalent statements? 44. For a continuous probability function, P(x < 5) = 0.35. What is P(x > 5), and how do you know? 45. Describe how you would draw the continuous probability distribution described by the function f (x) = 1 10 for 0 ≤ x ≤ 10. What type of a distribution is this? 46. For the continuous probability distribution described by the function f (x) = 1 10 for 0 ≤ x ≤ 10. what is the P(0 < x < 4)? 5.2: The Uniform Distribution 47. For the continuous probability distribution described by the function f (x) = 1 10 for 0 ≤ x ≤ 10, what is the P(2 < x < 5)? Use the following information to answer the next four exercises. The number of minutes that a patient waits at a medical clinic to see a doctor is represented by a uniform distribution between zero and 30 minutes, inclusive. 48. If X equals the number of minutes a person waits, what is the distribution of X? 49. Write the probability density function for this distribution. 50. What is the mean and standard deviation for waiting time? 51. What is the probability that a patient waits less than 10 minutes? 5.3: The Exponential Distribution 52. The distribution of the variable X, representing the average time to failure for an automobile battery, can be written as X ~ Exp(m). Describe this distribution in words. 53. If the value of m for an exponential distribution is 10, what are |
the mean and standard deviation for the distribution? 54. Write the probability density function for a variable distributed as X ~ Exp(0.2). 6.1: The Standard Normal Distribution 55. Translate this statement about the distribution of a random variable X into words: X ~ (100, 15). 56. If the variable X has the standard normal distribution, express this symbolically. Use the following information for the next six exercises. According to the World Health Organization, distribution of height in centimeters for girls aged five years and zero months has the distribution X ~ N(109, 4.5). 57. What is the z score for a height of 112 inches? 58. What is the z score for a height of 100 centimeters? 59. Find the z score for a height of 105 centimeters and explain what that means in the context of the population. 60. What height corresponds to a z score of 1.5 in this population? 61. Using the empirical rule, we expect about 68 percent of the values in a normal distribution to lie within one standard deviation above or below the mean. What does this mean, in terms of a specific range of values, for this distribution? 62. Using the empirical rule, about what percentage of heights in this distribution do you expect to be between 95.5 cm and 122.5 cm? 6.2: Using the Normal Distribution Use the following information to answer the next four exercises. The distributor of raffle tickets claims that 20 percent of the tickets are winners. You draw a sample of 500 tickets to test this proposition. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 847 63. Can you use the normal approximation to the binomial for your calculations? Why or why not. 64. What are the expected mean and standard deviation for your sample, assuming the distributor’s claim is true? 65. What is the probability that your sample will have a mean greater than 100? 66. If the z score for your sample result is –2, explain what this means, using the empirical rule. 7.1: The Central Limit Theorem for Sample Means (Averages) 67. What does the central limit theorem state with regard to the distribution of sample means? 68. The distribution of results from flipping a fair coin is uniform: Heads and tails are equally likely on any flip, and over a large number of trials, you expect about the same number of heads and tails. Yet |
if you conduct a study by flipping 30 coins and recording the number of heads, and repeat this 100 times, the distribution of the mean number of heads will be approximately normal. How is this possible? 69. The mean of a normally-distributed population is 50, and the standard deviation is four. If you draw 100 samples of size 40 from this population, describe what you would expect to see in terms of the sampling distribution of the sample mean. 70. X is a random variable with a mean of 25 and a standard deviation of two. Write the distribution for the sample mean of samples of size 100 drawn from this population. 71. Your friend is doing an experiment drawing samples of size 50 from a population with a mean of 117 and a standard deviation of 16. This sample size is large enough to allow use of the central limit theorem, so he says the standard deviation of the sampling distribution of sample means will also be 16. Explain why this is wrong, and calculate the correct value. 72. You are reading a research article that refers to the standard error of the mean. What does this mean, and how is it calculated? Use the following information to answer the next six exercises. You repeatedly draw samples of n = 100 from a population with a mean of 75 and a standard deviation of 4.5. 73. What is the expected distribution of the sample means? 74. One of your friends tries to convince you that the standard error of the mean should be 4.5. Explain what error your friend made. 75. What is the z score for a sample mean of 76? 76. What is the z score for a sample mean of 74.7? 77. What sample mean corresponds to a z score of 1.5? 78. If you decrease the sample size to 50, will the standard error of the mean be smaller or larger? What would be its value? Use the following information to answer the next two questions. We use the empirical rule to analyze data for samples of size 60 drawn from a population with a mean of 70 and a standard deviation of 9. 79. What range of values would you expect to include 68 percent of the sample means? 80. If you increased the sample size to 100, what range would you expect to contain 68 percent of the sample means, applying the empirical rule? 7.2: The Central Limit Theorem for Sums 81. How does the central limit theorem apply to sums of random variables? 82. Explain how the rules applying the central limit theorem to sample means, |
and to sums of a random variable, are similar. 83. If you repeatedly draw samples of size 50 from a population with a mean of 80 and a standard deviation of four, and calculate the sum of each sample, what is the expected distribution of these sums? Use the following information to answer the next four exercises. You draw one sample of size 40 from a population with a mean of 125 and a standard deviation of seven. 84. Compute the sum. What is the probability that the sum for your sample will be less than 5,000? 85. If you drew samples of this size repeatedly, computing the sum each time, what range of values would you expect to contain 95 percent of the sample sums? 86. What value is one standard deviation below the mean? 87. What value corresponds to a z score of 2.2? 848 Appendix B 7.3: Using the Central Limit Theorem 88. What does the law of large numbers say about the relationship between the sample mean and the population mean? 89. Applying the law of large numbers, which sample mean would you expect to be closer to the population mean: a sample of size 10 or a sample of size 100? Use this information for the next three questions. A manufacturer makes screws with a mean diameter of 0.15 cm (centimeters) and a range of 0.10 cm to 0.20 cm; within that range, the distribution is uniform. 90. If X = the diameter of one screw, what is the distribution of X? 91. Suppose you repeatedly draw samples of size 100 and calculate their mean. Applying the central limit theorem, what is the distribution of these sample means? 92. Suppose you repeatedly draw samples of 60 and calculate their sum. Applying the central limit theorem, what is the distribution of these sample sums? Practice Test 2 Solutions Probability Distribution Function (PDF) for a Discrete Random Variable 1. The domain of X = {English, Mathematics,...}, i.e., a list of all the majors offered at the university, plus undeclared. 2. The domain of Y = {0, 1, 2,...}; i.e., the integers from zero to the upper limit of classes allowed by the university. 3. The domain of Z = any amount of money from zero upwards. 4. Because they can take any value within their domain, and their value for any particular case is not known until the survey is completed. 5. No, because the domain of |
Z includes only positive numbers (you cannot spend a negative amount of money). Possibly the value –7 is a data entry error, or a special code to indicate that the student did not answer the question. 6. The probabilities must sum to 1.0, and the probabilities of each event must be between 0 and 1, inclusive. 7. Let X = the number of books checked out by a patron. 8. P(x > 2) = 0.10 + 0.05 = 0.15 9. P(x ≥ 0) = 1 – 0.20 = 0.80 10. P(x ≤ 3) = 1 – 0.05 = 0.95 11. The probabilities would sum to 1.10, and the total probability in a distribution must always equal 1.0. 12. x¯ = 0(0.20) + 1(0.45) + 2(0.20) + 3(0.10) + 4(0.05) = 1.35 Mean or Expected Value and Standard Deviation 13. x P(x) xP(x) 30 0.33 9.90 40 0.33 13.20 60 0.33 19.80 Table B8 14. x¯ = 9.90 + 13.20 + 19.80 = 42.90 15. P(x = 30) = 0.33 P(x = 40) = 0.33 P(x = 60) = 0.33 16. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 849 x P(x) xP(x) (x – μ)2P(x) 30 0.33 9.90 (30 – 42.90)2(0.33) = 54.91 40 0.33 13.20 (40 – 42.90)2(0.33) = 2.78 60 0.33 19.90 (60 – 42.90)2(0.33) = 96.49 Table B9 17. σ x = 54.91 + 2.78 + 96.49 = 12.42 Binomial Distribution 18. q = 1 – 0.65 = 0.35 19. 1. There are a fixed number of trials. 2. There are only two possible outcomes, and they add up to one. 3. The trials are independent and conducted under identical conditions. 20. No |
, because there are not a fixed number of trials 21. X ~ B(100, 0.65) 22. μ = np = 100(0.65) = 65 23. σ x = npq = 100(0.65)(0.35) = 4.77 24. X = Joe gets a hit in one at-bat (in one occasion of his coming to bat) 25. X ~ B(20, 0.4) 26. μ = np = 20(0.4) = 8 27. σ x = npq = 20(0.40)(0.60) = 2.19 4.4: Geometric Distribution 28. 1. A series of Bernoulli trials are conducted until one is a success, and then the experiment stops. 2. At least one trial is conducted, but there is no upper limit to the number of trials. 3. The probability of success or failure is the same for each trial. 29. T T T T H 30. The domain of X = {1, 2, 3, 4, 5,... n}. Because you are drawing with replacement, there is no upper bound to the number of draws that may be necessary. 31. The domain of X = {1, 2, 3, 4, 5, 6, 7, 8., 9, 10, 11, 12,... 27}. Because you are drawing without replacement, and 26 of the 52 cards are red, you have to draw a red card within the first 17 draws. 32. X ~ G(0.24) 33. μ = 1 p = 1 0.27 = 3.70 34. σ = 1 − p p2 = 1 − 0.27 0.272 = 3.16 4.5: Hypergeometric Distribution 35. Yes, because you are sampling from a population composed of two groups (boys and girls), have a group of interest (boys), and are sampling without replacement (hence, the probabilities change with each pick, and you are not performing Bernoulli trials). 850 Appendix B 36. The group of interest is the cards that are spades, the size of the group of interest is 13, and the sample size is five. 4.6: Poisson Distribution 37. A Poisson distribution models the number of events occurring in a fixed interval of time or space, when the events are independent and the average rate of the events is known. 38. X ~ P(4 |
) 39. The domain of X = {0, 1, 2, 3,...}; i.e., any integer from 0 upwards. 40.1: Continuous Probability Functions 41. The discrete variables are the number of books purchased, and the number of books sold after the end of the semester. The continuous variables are the amount of money spent for the books, and the amount of money received when they were sold. 42. Because for a continuous random variable, P(x = c) = 0, where c is any single value. Instead, we calculate P(c < x < d); i.e., the probability that the value of x is between the values c and d. 43. Because P(x = c) = 0 for any continuous random variable. 44. P(x > 5) = 1 – 0.35 = 0.65, because the total probability of a continuous probability function is always 1. 45. This is a uniform probability distribution. You would draw it as a rectangle with the vertical sides at 0 and 20, and the horizontal sides at 1 10 and 0. ⎛ 46. P(0 < x < 4) = (4 − 0) ⎝ ⎞ ⎠ = 0.4 1 10 5.2: The Uniform Distribution ⎛ 47. P(2 < x < 5) = (5 − 2) ⎝ ⎞ ⎠ = 0.3 1 10 48. X ~ U(0, 15) 49. f (x) = 1 b − a for (a ≤ x ≤ b) so f (x) = 1 30 for (0 ≤ x ≤ 30) 50. μ = a + b 2 = 0 + 30 5 = 15.0 σ = (b − a)2 12 = (30 − 0)2 12 = 8.66 ⎛ 51. P(x < 10) = (10) ⎝ ⎞ ⎠ = 0.33 1 30 5.3: The Exponential Distribution 52. X has an exponential distribution with decay parameter m and mean and standard deviation 1 m. In this distribution, there will be relatively large numbers of small values, with values becoming less common as they become larger. 53. μ = σ = 1 m = 1 10 54. f(x) = 0.2e–0.2x where x ≥ 0. = 0.1 6.1: The Standard |
Normal Distribution 55. The random variable X has a normal distribution with a mean of 100 and a standard deviation of 15. 56. X ~ N(0,1) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 851 57. z = 58 so z = 112 − 109 4.5 = 0.67 so z = 100 − 109 4.5 = − 2.00 59. z = 105 − 109 4.5 = −0.89 This girl is shorter than average for her age, by 0.89 standard deviations. 60. 109 + (1.5)(4.5) = 115.75 cm 61. We expect about 68 percent of the heights of girls aged five years and zero months to be between 104.5 cm and 113.5 cm. 62. We expect 99.7 percent of the heights in this distribution to be between 95.5 cm and 122.5 cm, because that range represents the values three standard deviations above and below the mean. 6.2: Using the Normal Distribution 63. Yes, because both np and nq are greater than five. np = (500)(0.20) = 100 and nq = 500(0.80) = 400 64. μ = np = (500)(0.20) = 100 σ = npq = 500(0.20)(0.80) = 8.94 65. Fifty percent, because in a normal distribution, half the values lie above the mean. 66. The results of our sample were two standard deviations below the mean, suggesting it is unlikely that 20 percent of the raffle tickets are winners, as claimed by the distributor, and that the true percentage of winners is lower. Applying the Empirical Rule, if that claim were true, we would expect to see a result this far below the mean only about 2.5 percent of the time. 7.1: The Central Limit Theorem for Sample Means (Averages) 67. The central limit theorem states that if samples of sufficient size are drawn from a population, the distribution of sample means will be normal, even if the distribution of the population is not normal. 68. The sample size of 30 is sufficiently large in this example to apply the central limit theorem. This theorem states that, for samples of sufficient size drawn from a population, the sampling distribution of the sample mean will approach normality, regardless of the distribution of the population from |
which the samples were drawn. 69. You would not expect each sample to have a mean of 50, because of sampling variability. However, you would expect the sampling distribution of the sample means to cluster around 50, with an approximately normal distribution, so that values close to 50 are more common than values further removed from 50. ¯ 70. X ¯ ∼ N(25, 0.2) because X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ 71. The standard deviation of the sampling distribution of the sample means can be calculated using the formula ⎛ ⎝ σ x n ⎞ ⎠, which in this case is ⎛ ⎝ ⎞ ⎠ 16 50 is therefore 2.26.. The correct value for the standard deviation of the sampling distribution of the sample means 72. The standard error of the mean is another name for the standard deviation of the sampling distribution of the sample ⎞ mean. Given samples of size n drawn from a population with standard deviation σx, the standard error of the mean is ⎛ ⎠. ⎝ σ x n 73. X ~ N(75, 0.45) 74. Your friend forgot to divide the standard deviation by the square root of n. 75. z = x¯ − μ x σ x = 76 − 75 4.5 = 2.2 852 Appendix B 76. z = x¯ − μ x σ x = 74.7 − 75 4.5 = −0.67 77. 75 + (1.5)(0.45) = 75.675 78. The standard error of the mean will be larger, because you will be dividing by a smaller number. The standard error of the mean for samples of size n = 50 is ⎛ ⎝ = 0.64 σ x n ⎞ ⎠ = 4.5 50 79. You would expect this range to include values up to one standard deviation above or below the mean of the sample means. In this case: 70 + 9 60 = 68.84 so you would expect 68 percent of the sample means to be between 68.84 and = 71.16 and 70 − 9 60 71.16. 80. 70 + 9 = 70.9 and 70 − 9 = 69.1 so you would expect 68 percent of the sample means to be between 69.1 100 and 70.9. Note that |
this is a narrower interval due to the increased sample size. 100 7.2: The Central Limit Theorem for Sums 81. For a random variable X, the random variable ΣX will tend to become normally distributed as the size n of the samples used to compute the sum increases. 82. Both rules state that the distribution of a quantity (the mean or the sum) calculated on samples drawn from a population will tend to have a normal distribution as the sample size increases, regardless of the distribution of population from which the samples are drawn. 83. ΣX ∼ N ⎛ ⎝nμ x, ( n)(σ x)⎞ ⎠ so ΣX ∼ N(4,000, 28.3) 84. The probability is 0.50, because 5,000 is the mean of the sampling distribution of sums of size 40 from this population. Sums of random variables computed from a sample of sufficient size are normally distributed, and in a normal distribution, half the values lie below the mean. 85. Using the empirical rule, you would expect 95 percent of the values to be within two standard deviations of the mean. Using the formula for the standard deviation is for a sample sum ( n)(σ x) = ⎛ ⎠(7) = 44.3, so you would expect 95 percent of the values to be between 5,000 + (2)(44.3) and 5,000 – (2)(44.3), or between 4,911.4 and 588.6. ⎝ 40⎞ 86. μ − ( n)(σ x) = 5,000 − ⎛ ⎝ 40⎞ ⎠(7) = 4,955.7 87. 5,000 + (2.2)⎛ ⎝ 40⎞ ⎠(7) = 5097.4 7.3: Using the Central Limit Theorem 88. The law of large numbers says that, as sample size increases, the sample mean tends to get nearer and nearer to the population mean. 89. You would expect the mean from a sample of size 100 to be nearer to the population mean, because the law of large numbers says that, as sample size increases, the sample mean tends to approach the population mean. 90. X ~ N(0.10, 0.20) ¯ 91. X ∼ N ⎛ ⎝μ x, |
σ x n ⎞ ⎠ and the standard deviation of a uniform distribution is b − a 12. In this example, the standard deviation of the distribution is b − a 12 = 0.10 12 = 0.03 ¯ so X ∼ N(0.15, 0.003) 92. ΣX ∼ N((n)(μ x), ( n)(σ x)) so ΣX ∼ N(9.0, 0.23) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 853 Practice Test 3 8.1: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal Use the following information to answer the next seven exercises. You draw a sample of size 30 from a normally distributed population with a standard deviation of four. 1. What is the standard error of the sample mean in this scenario, rounded to two decimal places? 2. What is the distribution of the sample mean? 3. If you want to construct a two-sided 95 percent confidence interval, how much probability will be in each tail of the distribution? 4. What is the appropriate z score and error bound or margin of error (EBM) for a 95 percent confidence interval for this data? 5. Rounding to two decimal places, what is the 95 percent confidence interval if the sample mean is 41? 6. What is the 90 percent confidence interval if the sample mean is 41? Round to two decimal places 7. Suppose the sample size in this study had been 50, rather than 30. What would the 95 percent confidence interval be if the sample mean is 41? Round your answer to two decimal places. 8. For any given data set and sampling situation, which would you expect to be wider: a 95 percent confidence interval or a 99 percent confidence interval? 8.2: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s t 9. Comparing graphs of the standard normal distribution (z distribution) and a t distribution with 15 degrees of freedom (df), how do they differ? 10. Comparing graphs of the standard normal distribution (z distribution) and a t distribution with 15 degrees of freedom (df), how are they similar? Use the following information to answer the next five exercises. Body temperature is known to be distributed normally among healthy adults. Because you do not know the population standard deviation, you use the t distribution to study body temperature. |
You collect data from a random sample of 20 healthy adults and find that your sample temperatures have a mean of 98.4 and a sample standard deviation of 0.3 (both in degrees Fahrenheit). 11. What are the degrees of freedom (df) for this study? 12. For a two-tailed 95 percent confidence interval, what is the appropriate t value to use in the formula? 13. What is the 95 percent confidence interval? 14. What is the 99 percent confidence interval? Round to two decimal places. 15. Suppose your sample size had been 30 rather than 20. What would the 95 percent confidence interval be then? Round to two decimal places 8.3: Confidence Interval for a Population Proportion Use this information to answer the next four exercises. You conduct a poll of 500 randomly selected city residents, asking them if they own an automobile. Of the respondents, 280 say they own an automobile, and 220 say they do not. 16. Find the sample proportion and sample standard deviation for this data. 17. What is the 95 percent two-sided confidence interval? Round to four decimal places. 18. Calculate the 90 percent confidence interval. Round to four decimal places. 19. Calculate the 99 percent confidence interval. Round to four decimal places. Use the following information to answer the next three exercises. You are planning to conduct a poll of community members aged 65 and older, to determine how many own mobile phones. You want to produce an estimate whose 95 percent confidence interval will be within four percentage points (plus or minus) of the true population proportion. Use an estimated population proportion of 0.5. 20. What sample size do you need? 854 Appendix B 21. Suppose you knew from prior research that the population proportion was 0.6. What sample size would you need? 22. Suppose you wanted a 95 percent confidence interval within three percentage points of the population. Assume the population proportion is 0.5. What sample size do you need? 9.1: Null and Alternate Hypotheses 23. In your state, 58 percent of registered voters in a community are registered as republicans. You want to conduct a study to see if this also holds up in your community. State the null and alternative hypotheses to test this. 24. You believe that at least 58 percent of registered voters in a community are registered as republicans. State the null and alternative hypotheses to test this. 25. The mean household value in a city is $268,000. You believe that the mean household value |
in a particular neighborhood is lower than the city average. Write the null and alternative hypotheses to test this. 26. State the appropriate alternative hypothesis to this null hypothesis: H0: μ = 107 27. State the appropriate alternative hypothesis to this null hypothesis: H0: p < 0.25 9.2: Outcomes and the Type I and Type II Errors 28. If you reject H0 when H0 is correct, what type of error is this? 29. If you fail to reject H0 when H0 is false, what type of error is this? 30. What is the relationship between the Type II error and the power of a test? 31. A new blood test is being developed to screen patients for cancer. Positive results are followed up by a more accurate (and expensive) test. It is assumed that the patient does not have cancer. Describe the null hypothesis and the Type I and Type II errors for this situation, and explain which type of error is more serious. 32. Explain in words what it means that a screening test for TB has an α level of 0.10. The null hypothesis is that the patient does not have TB. 33. Explain in words what it means that a screening test for TB has a β level of 0.20. The null hypothesis is that the patient does not have TB. 34. Explain in words what it means that a screening test for TB has a power of 0.80. 9.3: Distribution Needed for Hypothesis Testing 35. If you are conducting a hypothesis test of a single population mean, and you do not know the population variance, what test will you use if the sample size is 10 and the population is normal? 36. If you are conducting a hypothesis test of a single population mean, and you know the population variance, what test will you use? 37. If you are conducting a hypothesis test of a single population proportion, with np and nq greater than or equal to five, what test will you use, and with what parameters? 38. Published information indicates that, on average, college students spend less than 20 hours studying per week. You draw a sample of 25 students from your college and find the sample mean to be 18.5 hours, with a standard deviation of 1.5 hours. What distribution will you use to test whether study habits at your college are the same as the national average, and why? 39. A published study says that 95 percent of American children are vaccinated against a disease, with a standard deviation |
of 1.5 percent. You draw a sample of 100 children from your community and check their vaccination records to see if the vaccination rate in your community is the same as the national average. What distribution will you use for this test, and why? 9.4: Rare Events, the Sample, Decision, and Conclusion 40. You are conducting a study with an α level of 0.05. If you get a result with a p-value of 0.07, what will be your decision? 41. You are conducting a study with α = 0.01. If you get a result with a p-value of 0.006, what will be your decision? Use the following information to answer the next five exercises. According to the World Health Organization, the average height of a one-year-old child is 29”. You believe children with a particular disease are smaller than average, so you draw a sample of 20 children with this disease and find a mean height of 27.5” and a sample standard deviation of 1.5”. 42. What are the null and alternative hypotheses for this study? 43. What distribution will you use to test your hypothesis, and why? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 855 44. What is the test statistic and the p-value? 45. Based on your sample results, what is your decision? 46. Suppose the mean for your sample was 25. Redo the calculations and describe what your decision would be. 9.5: Additional Information and Full Hypothesis Test Examples 47. You conduct a study using α = 0.05. What is the level of significance for this study? 48. You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses: H0: μ = 35.5 Ha: μ ≠ 35.5 Will you conduct a one-tailed or two-tailed test? 49. You conduct a study, based on a sample drawn from a normally distributed population with a known variance, with the following hypotheses: H0: μ ≥ 35.5 Ha: μ < 35.5 Will you conduct a one-tailed or two-tailed test? Use the following information to answer the next three exercises. Nationally, 80 percent of adults own an automobile. You are interested in whether the same proportion in your community own cars. You draw a sample of |
100 and find that 75 percent own cars. 50. What are the null and alternative hypotheses for this study? 51. What test will you use, and why? 10.1: Comparing Two Independent Population Means with Unknown Population Standard Deviations 52. You conduct a poll of political opinions, interviewing both members of 50 married couples. Are the groups in this study independent or matched? 53. You are testing a new drug to treat insomnia. You randomly assign 80 volunteer subjects to either the experimental (new drug) or control (standard treatment) conditions. Are the groups in this study independent or matched? 54. You are investigating the effectiveness of a new math textbook for high school students. You administer a pretest to a group of students at the beginning of the semester, and a posttest at the end of a year’s instruction using this textbook, and compare the results. Are the groups in this study independent or matched? Use the following information to answer the next two exercises. You are conducting a study of the difference in time at two colleges for undergraduate degree completion. At College A, students take an average of 4.8 years to complete an undergraduate degree, while at College B, they take an average of 4.2 years. The pooled standard deviation for this data is 1.6 years. 55. Calculate Cohen’s d and interpret it. 56. Suppose the mean time to earn an undergraduate degree at College A was 5.2 years. Calculate the effect size and interpret it. 57. You conduct an independent-samples t test with sample size 10 in each of two groups. If you are conducting a two-tailed hypothesis test with α = 0.01, what p-values will cause you to reject the null hypothesis? 58. You conduct an independent samples t test with sample size 15 in each group, with the following hypotheses: H0: μ ≥ 110 Ha: μ < 110 If α = 0.05, what t values will cause you to reject the null hypothesis? 10.2: Comparing Two Independent Population Means with Known Population Standard Deviations Use the following information to answer the next six exercises. College students in the sciences often complain that they must spend more on textbooks each semester than students in the humanities. To test this, you draw random samples of 50 science and 50 humanities students from your college, and record how much each spent last semester on textbooks. Consider the science students to be group one, and the humanities students to be group two. 856 Appendix B |
59. What is the random variable for this study? 60. What are the null and alternative hypotheses for this study? 61. If the 50 science students spent an average of $530 with a sample standard deviation of $20, and the 50 humanities students spent an average of $380 with a sample standard deviation of $15, would you not reject or reject the null hypothesis? Use an alpha level of 0.05. What is your conclusion? 62. What would be your decision, if you were using α = 0.01? 10.3: Comparing Two Independent Population Proportions Use the information to answer the next six exercises. You want to know if the proportion of homes with cable television service differs between Community A and Community B. To test this, you draw a random sample of 100 for each and record whether they have cable service. 63. What are the null and alternative hypotheses for this study? 64. If 65 households in Community A have cable service, and 78 households in Community B, what is the pooled proportion? 65. At α = 0.03, will you reject the null hypothesis? What is your conclusion? Sixty-five households in Community A have cable service, and 78 households in community B. One hundred households in each community were surveyed. 66. Using an alpha value of 0.01, would you reject the null hypothesis? What is your conclusion? Sixty-five households in Community A have cable service, and 78 households in Community B. One hundred households in each community were surveyed. 10.4: Matched or Paired Samples Use the following information to answer the next five exercises. You are interested in whether a particular exercise program helps people run a mile faster. You conduct a study in which you weigh the participants at the start of the study, and again at the conclusion, after they have participated in the exercise program for six months. You compare the results using a matched-pairs t test, in which the data is {time to run a mile at conclusion, time at start}. You believe that, on average, the participants will be able to run a mile faster after six months on the exercise program. 67. What are the null and alternative hypotheses for this study? 68. Calculate the test statistic, assuming that x¯ d = –5, sd = 6, and n = 30 (pairs). 69. What are the degrees of freedom for this statistic? 70. Using α = 0.05, what is your decision regarding the effectiveness of this |
program in improving running speed? What is the conclusion? 71. What would it mean if the t statistic had been 4.56, and what would have been your decision in that case? 11.1: Facts About the Chi-Square Distribution 72. What is the mean and standard deviation for a chi-square distribution with 20 degrees of freedom? 11.2: Goodness-of-Fit Test Use the following information to answer the next four exercises. Nationally, about 66 percent of high school graduates enroll in higher education. You perform a chi-square goodness of fit test to see if this same proportion applies to your high school’s most recent graduating class of 200. Your null hypothesis is that the national distribution also applies to your high school. 73. What are the expected numbers of students from your high school graduating class enrolled and not enrolled in higher education? 74. Fill out the rest of this table. Observed (O) Expected (E) O – E (O – E)2 (O − E)2 z Enrolled 145 Table B10 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 857 Observed (O) Expected (E) O – E (O – E)2 (O − E)2 z Not enrolled 55 Table B10 75. What are the degrees of freedom for this chi-square test? 76. What is the chi-square test statistic and the p-value? At the five percent significance level, what do you conclude? 77. For a chi-square distribution with 92 degrees of freedom, the curve _____________. 78. For a chi-square distribution with five degrees of freedom, the curve is ______________. 11.3: Test of Independence Use the following information to answer the next four exercises. You are considering conducting a chi-square test of independence for the data in this table, which displays data about cell phone ownership for freshman and seniors at a high school. Your null hypothesis is that cell phone ownership is independent of class standing. 79. Compute the expected values for the cells. Cell = Yes Cell = No Freshman 100 Senior 200 Table B11 150 50 80. Compute (O − E)2 z for each cell, where O = observed and E = expected. 81. What is the chi-square statistic and degrees of freedom for this study? 82. At the α = 0.5 significance level, what |
is your decision regarding the null hypothesis? 11.4: Test of Homogeneity 83. You conduct a chi-square test of homogeneity for data in a five-by-two table. What are the degrees of freedom for this test? 11.5: Comparison Summary of the Chi-Square Tests: Goodness-of-Fit, Independence and Homogeneity 84. A 2013 poll in the State of California surveyed people about a tax. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the five percent significance level. Ethnic Group/Response Type Favor Oppose No Opinion Row Total White/Non-Hispanic Latino African American Asian American Column Total Table B12 234 147 24 54 433 106 41 48 459 628 43 19 6 16 84 710 272 71 118 1171 858 Appendix B 85. In a test of homogeneity, what must be true about the expected value of each cell? 86. Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence? 87. Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity? 11.6: Test of a Single Variance 88. A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypotheses to test this? Practice Test 3 Solutions 8.1: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal 1. σ n = 4 30 = 0.73 2. normal 3. 0.025 or 2.5 percent; A 95 percent confidence interval contains 95 percent of the probability, and excludes 5 percent, and the 5 percent excluded is split evenly between the upper and lower tails of the distribution. 4. z score = 1.96; EBM = 1.96)(0.73) = 1.4308 5. 41 ± 1.43 = (39.57, 42.43); using the calculator function ZInterval, answer is (40.74, 41.26). Answers differ due to rounding. 6. The z-value for a 90 percent confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085. The 90 percent confidence interval is 41 ± 1.20 = (39.80, 42.20). The calculator |
function ZInterval answer is (40.78, 41.23). Answers differ due to rounding. 7. The standard error of measurement is σ n = 4 50 = 0.57 EBM = 1.96)(0.57) = 1.12 The 95 percent confidence interval is 41 ± 1.12 = (39.88, 42.12). The calculator function ZInterval answer is (40.84, 41.16). Answers differ due to rounding. 8. The 99 percent confidence interval, because it includes all but one percent of the distribution. The 95 percent confidence interval will be narrower, because it excludes five percent of the distribution. 8.2: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s t 9. The t distribution will have more probability in its tails (thicker tails) and less probability near the mean of the distribution (shorter in the center). 10. Both distributions are symmetrical and centered at zero. 11. df = n – 1 = 20 – 1 = 19 12. You can get the t value from a probability table or a calculator. In this case, for a t distribution with 19 degrees of freedom and a 95 percent two-sided confidence interval, the value is 2.093; i.e., t α 2 = 2.093. The calculator function is invT(0.975, 19). ⎛ ⎝ ⎛ ⎞ ⎠ = (2.093) ⎝ 13. EBM = t α 2 98.4 ± 0.14 = (98.26, 98.54). The calculator function TInterval answer is (98.26, 98.54). ⎞ ⎠ = 0.140 0.3 20 s n 14. t α 2 = 2.861. The calculator function is invT(0.995, 19). EBM = 2.861) ⎝ ⎞ ⎠ = 0.192 0.3 20 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Appendix B 859 98.4 ± 0.19 = (98.21, 98.59). The calculator function TInterval answer is (98.21, 98.59). 15. df = n – 1 = 30 – 1 = 29. t α 2 = 2.045 EBM = z |
t ⎛ ⎝ s n ⎛ ⎞ ⎠ = (2.045) ⎝ ⎞ ⎠ = 0.112 0.3 30 98.4 ± 0.11 = (98.29, 98.51). The calculator function TInterval answer is (98.29, 98.51). 8.3: Confidence Interval for a Population Proportion 16. p′ = 280 500 = 0.56 q′ = 1 − p′ = 1 − 0.56 = 0.44 s = pq n = 0.56(0.44) 500 = 0.0222 17. Because you are using the normal approximation to the binomial, z α 2 = 1.96. Calculate the error bound for the population (EBP): EBP = z a 2 pq n = 1.96(0.222) = 0.0435 Calculate the 95 percent confidence interval: 0.56 ± 0.0435 = (0.5165, 0.6035). The calculator function 1-PropZint answer is (0.5165, 0.6035). 18. z α 2 = 1.64 EBP = z a 2 pq n = 1.64(0.0222) = 0.0364 0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965). 19. z α 2 = 2.58 EBP = z a 2 pq n = 2.58(0.0222) = 0.0573 0.56 ± 0.05 = (0.5127, 0.6173). The calculator function 1-PropZint answer is (0.5028, 0.6172). 20. EBP = 0.04 (because 4 percent = 0.04) z α 2 = 1.96 for a 95 percent confidence interval. n = z2 pq EBP2 = 1.962 (0.5)(0.5) 0.042 = 0.9604 0.0016 = 600.25 You need 601 subjects (rounding upward from 600.25). 21. n = 1.962 (0.6)(0.4) 0.042 You need 577 subjects (rounding upward from 576.24). n2 p |
q EBP2 = = 0.9220 0.0016 = 576.24 22. n = 1.962 (0.5)(0.5) 0.032 You need 1,068 subjects (rounding upward from 1,067.11). n2 pq EBP2 = = 0.9604 0.0009 = 1067.11 9.1: Null and Alternate Hypotheses 23. H0: p = 0.58 Ha: p ≠ 0.58 24. H0: p ≥ 0.58 Ha: p < 0.58 25. H0: μ ≥ $268,000 860 Appendix B Ha: μ < $268,000 26. Ha: μ ≠ 107 27. Ha: p ≥ 0.25 9.2: Outcomes and the Type I and Type II Errors 28. a Type I error 29. a Type II error 30. Power = 1 – β = 1 – P(Type II error). 31. The null hypothesis is that the patient does not have cancer. A Type I error would be detecting cancer when it is not present. A Type II error would be not detecting cancer when it is present. A Type II error is more serious, because failure to detect cancer could keep a patient from receiving appropriate treatment. 32. The screening test has a 10 percent probability of a Type I error, meaning that 10 percent of the time, it will detect TB when it is not present. 33. The screening test has a 20 percent probability of a Type II error, meaning that 20 percent of the time, it will fail to detect TB when it is in fact present. 34. Eighty percent of the time, the screening test will detect TB when it is actually present. 9.3: Distribution Needed for Hypothesis Testing 35. The Student’s t test. 36. The normal distribution or z test. 37. The normal distribution with μ = p and σ = pq n 38. t24. You use the t distribution because you do not know the population standard deviation, and the degrees of freedom are 24 because df = n – 1. ¯ 39. X ~ N ⎛ ⎝0.95, 0.051 100 ⎞ ⎠ Because you know the population standard deviation and have a large sample, you can use the normal distribution. 9.4: Rare Events, the Sample, Decision, and Conclusion 40. Fail to reject the |
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