A Confidence Interval When the Population Standard Deviation Is Known or Large Sample Size A confidence interval for a population mean, when the population standard deviation is known based on the conclusion of the Central Limit Theorem that the sampling distribution of the sample means follow an approximately normal distribution. Calculating the Confidence Interval Consider the standardizing formula for the sampling distribution developed in the discussion of the Central Limit Theorem: Z 1 = X – − μ X – σ X – = X – − μ σ n Notice that µ is substituted for µ x − because we know that the expected value of µ x − is µ from the Central Limit theorem and σ x − is replaced with σ n , also from the Central Limit Theorem. In this formula we know X – , σ x − and n, the sample size. (In actuality we do not know the population standard deviation, but we do have a point estimate for it, s, from the sample we took. More on this later.) What we do not know is μ or Z 1 . We can solve for either one of these in terms of the other. Solving for μ in terms of Z 1 gives: μ = X – ± Z 1 σ n Remembering that the Central Limit Theorem tells us that the distribution of the X ¯ 's, the sampling distribution for means, is normal, and that the normal distribution is symmetrical, we can rearrange terms thus: X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) This is the formula for a confidence interval for the mean of a population. Notice that Z α has been substituted for Z 1 in this equation. This is where a choice must be made by the statistician. The analyst must decide the level of confidence they wish to impose on the confidence interval. α is the probability that the interval will not contain the true population mean. The confidence level is defined as (1-α). Z α is the number of standard deviations X ¯ lies from the mean with a certain probability. If we chose Z α = 1.96 we are asking for the 95% confidence interval because we are setting the probability that the true mean lies within the range at 0.95. If we set Z α at 1.64 we are asking for the 90% confidence interval because we have set the probability at 0.90. These numbers can be verified by consulting the Standard Normal table. Divide either 0.95 or 0.90 in half and find that probability inside the body of the table. Then read on the top and left margins the number of standard deviations it takes to get this level of probability. In reality, we can set whatever level of confidence we desire simply by changing the Z α value in the formula. It is the analyst's choice. Common convention in Economics and most social sciences sets confidence intervals at either 90, 95, or 99 percent levels. Levels less than 90% are considered of little value. The level of confidence of a particular interval estimate is called by (1-α). A good way to see the development of a confidence interval is to graphically depict the solution to a problem requesting a confidence interval. This is presented in for the example in the introduction concerning the number of downloads from Apple Music. That case was for a 95% confidence interval, but other levels of confidence could have just as easily been chosen depending on the need of the analyst. However, the level of confidence MUST be pre-set and not subject to revision as a result of the calculations. μ = X ¯ ± Z α σ n = 2 ± 1 . 96 ( 0 . 1 ) = 2 ± 0 . 196 1 . 804 ≤ μ ≤ 2 . 196 For this example, let's say we know that the actual population mean number of Apple Music downloads is 2.1. The true population mean falls within the range of the 95% confidence interval. There is absolutely nothing to guarantee that this will happen. Further, if the true mean falls outside of the interval we will never know it. We must always remember that we will never ever know the true mean. Statistics simply allows us, with a given level of probability (confidence), to say that the true mean is within the range calculated. This is what was called in the introduction, the "level of ignorance admitted". Changing the Confidence Level or Sample Size Here again is the formula for a confidence interval for an unknown population mean assuming we know the population standard deviation: X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) It is clear that the confidence interval is driven by two things, the chosen level of confidence, Z α , and the standard deviation of the sampling distribution. The Standard deviation of the sampling distribution is further affected by two things, the standard deviation of the population and the sample size we chose for our data. Here we wish to examine the effects of each of the choices we have made on the calculated confidence interval, the confidence level and the sample size. For a moment we should ask just what we desire in a confidence interval. Our goal was to estimate the population mean from a sample. We have forsaken the hope that we will ever find the true population mean, and population standard deviation for that matter, for any case except where we have an extremely small population and the cost of gathering the data of interest is very small. In all other cases we must rely on samples. With the Central Limit Theorem we have the tools to provide a meaningful confidence interval with a given level of confidence, meaning a known probability of being wrong. By meaningful confidence interval we mean one that is useful. Imagine that you are asked for a confidence interval for the ages of your classmates. You have taken a sample and find a mean of 19.8 years. You wish to be very confident so you report an interval between 9.8 years and 29.8 years. This interval would certainly contain the true population mean and have a very high confidence level. However, it hardly qualifies as meaningful. The very best confidence interval is narrow while having high confidence. There is a natural tension between these two goals. The higher the level of confidence the wider the confidence interval as the case of the students' ages above. We can see this tension in the equation for the confidence interval. μ = x _ ± Z α ( σ n ) The confidence interval will increase in width as Z α increases, Z α increases as the level of confidence increases. There is a tradeoff between the level of confidence and the width of the interval. Now let's look at the formula again and we see that the sample size also plays an important role in the width of the confidence interval. The sample size, n , shows up in the denominator of the standard deviation of the sampling distribution. As the sample size increases, the standard deviation of the sampling distribution decreases and thus the width of the confidence interval, while holding constant the level of confidence. Again we see the importance of having large samples for our analysis although we then face a second constraint, the cost of gathering data. Suppose we are interested in the mean scores on an exam. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68 ( X ¯ = 68). In this example we have the unusual knowledge that the population standard deviation is 3 points. Do not count on knowing the population parameters outside of textbook examples. Find a confidence interval estimate for the population mean exam score (the mean score on all exams). Find a 90% confidence interval for the true (population) mean of statistics exam scores. The solution is shown step by step: The formula for a confidence interval for an unknown population mean assuming we know the population standard deviation is: X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) For a 90% confidence interval, visualize an area of 0.90 centered under the normal curve (See ). The remaining area for the two tails of the normal distribution is then 0.10, which indicates that the area in the left tail is one-half of 0.10, which is 0.05. The corresponding z-score that cuts off an area of 0.05 in the left tail is 1.645. In this example we are given that the population standard deviation σ = 3 . We are also given that the sample size n = 36 and the sample mean X ¯ = 68 . Substituting these values in the confidence interval formula results in the following: X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) 68 - 1 . 645 3 36 ≤ μ ≤ 68 + 1 . 645 3 36 68 - 0 . 8225 ≤ μ ≤ 68 + 0 . 8225 67 . 1775 ≤ μ ≤ 68 . 8225 We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Try It Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time. (34.1347, 37.8653) Suppose we change the original problem in by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score. μ = x _ ± Z α ( σ n ) μ = 68 ± 1.96 ( 3 36 ) 67.02 ≤ μ ≤ 68.98 σ = 3; n = 36; The confidence level is 95% ( CL = 0.95). CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05 Z α 2 = Z 0.025 = 1.96 Notice that the plus/minus term in the equation is larger for a 95% confidence level in the original problem. Comparing the results: The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider. This demonstrates a very important principle of confidence intervals. There is a trade off between the level of confidence and the width of the interval. Our desire is to have a narrow confidence interval, huge wide intervals provide little information that is useful. But we would also like to have a high level of confidence in our interval. This demonstrates that we cannot have both. Summary: Effect of Changing the Confidence Level Increasing the confidence level makes the confidence interval wider. Decreasing the confidence level makes the confidence interval narrower. And again here is the formula for a confidence interval for an unknown mean assuming we have the population standard deviation: X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) The standard deviation of the sampling distribution was provided by the Central Limit Theorem as σ n . While we infrequently get to choose the sample size it plays an important role in the confidence interval. Because the sample size is in the denominator of the equation, as n increases it causes the standard deviation of the sampling distribution to decrease and thus the width of the confidence interval to decrease. We have met this before as we reviewed the effects of sample size on the Central Limit Theorem. There we saw that as n increases the sampling distribution narrows until in the limit it collapses on the true population mean. Try It Refer back to the pizza-delivery Try It 8.1 exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time. Changing the Sample Size Suppose we change the original problem in to see what happens to the confidence interval if the sample size is changed. Leave everything the same except the sample size. Use the original 90% confidence level. What happens to the confidence interval if we increase the sample size and use n = 100 instead of n = 36? What happens if we decrease the sample size to n = 25 instead of n = 36? μ = x _ ± Z α ( σ n ) μ = 68 ± 1.645 ( 3 100 ) 67.5065 ≤ μ ≤ 68.4935 If we increase the sample size n to 100, we decrease the width of the confidence interval relative to the original sample size of 36 observations. μ = x _ ± Z α ( σ n ) μ = 68 ± 1.645 ( 3 25 ) 67.013 ≤ μ ≤ 68.987 If we decrease the sample size n to 25, we increase the width of the confidence interval by comparison to the original sample size of 36 observations. Summary: Effect of Changing the Sample Size Increasing the sample size makes the confidence interval narrower. Decreasing the sample size makes the confidence interval wider. Try It Refer back to the pizza-delivery Try It 8.1 exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time. We have already seen this effect when we reviewed the effects of changing the size of the sample, n , on the Central Limit Theorem. See to see this effect. Before we saw that as the sample size increased the standard deviation of the sampling distribution decreases. This was why we choose the sample mean from a large sample as compared to a small sample, all other things held constant. Thus far we assumed that we knew the population standard deviation. This will virtually never be the case. We will have the sample standard deviation, s , however. This is a point estimate for the population standard deviation and can be substituted into the formula for confidence intervals for a mean under certain circumstances. We just saw the effect the sample size has on the width of confidence interval and the impact on the sampling distribution for our discussion of the Central Limit Theorem. We can invoke this to substitute the point estimate for the standard deviation if the sample size is large "enough". Simulation studies indicate that 30 observations or more will be sufficient to eliminate any meaningful bias in the estimated confidence interval. Spring break can be a very expensive holiday. A sample of 80 students is surveyed, and the average amount spent by students on travel and beverages is $593.84. The sample standard deviation is approximately $369.34. Construct a 92% confidence interval for the population mean amount of money spent by spring breakers. We begin with the confidence interval for a mean. We use the formula for a mean because the random variable is dollars spent and this is a continuous random variable. The point estimate for the population standard deviation, s , has been substituted for the true population standard deviation because with 80 observations there is no concern for bias in the estimate of the confidence interval. μ = x ¯ ± [ Z ( a / 2 ) s n ] Substituting the values into the formula, we have: μ = 593.84 ± [ 1.75 369.34 80 ] Z ( a / 2 ) is found on the standard normal table by looking up 0.46 in the body of the table and finding the number of standard deviations on the side and top of the table; 1.75. The solution for the interval is thus: μ = 593.84 ± 72.2636 = ( 521.57 , 666.10 ) $ 521.58 ≤ μ ≤ $ 666.10 Try It The price of a chair is a large range of cost. The average cost of 25 chairs in a store is $100. The sample standard deviation is $50. Construct a 92% confidence interval for the population mean of the cost of chairs. Since there is no bias in the estimate of the confidence interval, the confidence interval is obtained as follows: μ = x ¯ ± Z ( α / 2 ) s n μ = 100 ± 1 . 4 50 25 μ = 100 ± 14 So, the confidence interval is obtained as ($86, $114). References “American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/nav/jsf/pages/searchresults.xhtml?refresh=t (accessed July 2, 2013). “Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013). “Headcount Enrollment Trends by Student Demographics Ten-Year Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm (accessed September 30,2013). Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. Grummer-Strawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/2000growthchart-us.pdf (accessed July 2, 2013). La, Lynn, Kent German. "Cell Phone Radiation Levels." c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013). “Mean Income in the Past 12 Months (in 2011 Inflation-Adjusted Dollars): 2011 American Community Survey 1-Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table (accessed July 2, 2013). “Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013). “National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013). Formula Review The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by X ¯ − Z α ( σ n ) ≤ μ ≤ X ¯ + Z α ( σ n ) This formula is used when the population standard deviation is known. CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter α = 1 – CL = the proportion of confidence intervals that will not contain the population parameter z α 2 = the z -score with the property that the area to the right of the z-score is ∝ 2 this is the z -score used in the calculation where α = 1 – CL . Confidence Level (CL) the percent expression for the probability that the confidence interval contains the true population parameter; for example, if the CL = 90%, then in 90 out of 100 samples the interval estimate will enclose the true population parameter. Error Bound for a Population Mean ( EBM ) the margin of error; depends on the confidence level, sample size, and known or estimated population standard deviation.

Business


A Confidence Interval When the Population Standard Deviation Is Unknown and Small Sample Case In practice, we rarely know the population standard deviation . In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close enough results. This is what we did in above. The point estimate for the standard deviation, s, was substituted in the formula for the confidence interval for the population standard deviation. In this case the 80 observations are well above the suggested 30 observations to eliminate any bias from a small sample. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval. William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution . The name comes from the fact that Gosset wrote under the pen name "A Student." Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student's t-distribution only for sample sizes of at most 30 observations. If you draw a simple random sample of size n from a population with mean μ and unknown population standard deviation σ and calculate the t -score t = x – – μ ( s n ) , then the t -scores follow a Student's t-distribution with n – 1 degrees of freedom . The t -score has the same interpretation as the z -score . It measures how far in standard deviation units x – is from its mean μ . For each sample size n , there is a different Student's t-distribution. The degrees of freedom , n – 1 , come from the calculation of the sample standard deviation s . Remember when we first calculated a sample standard deviation we divided the sum of the squared deviations by n − 1, but we used n deviations ( x – x ¯ values ) to calculate s . Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df) in recognition that one is lost in the calculations. The effect of losing a degree of freedom is that the t-value increases and the confidence interval increases in width. Properties of the Student's t-Distribution The graph for the Student's t-distribution is similar to the standard normal curve and at infinite degrees of freedom it is the normal distribution. You can confirm this by reading the bottom line at infinite degrees of freedom for a familiar level of confidence, e.g. at column 0.05, 95% level of confidence, we find the t-value of 1.96 at infinite degrees of freedom. The mean for the Student's t-distribution is zero and the distribution is symmetric about zero, again like the standard normal distribution. The Student's t-distribution has more probability in its tails than the standard normal distribution because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student's t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution. The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's t-distribution becomes more like the graph of the standard normal distribution. The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ . This assumption comes from the Central Limit theorem because the individual observations in this case are the x ¯ s of the sampling distribution. The size of the underlying population is generally not relevant unless it is very small. If it is normal then the assumption is met and doesn't need discussion. A probability table for the Student's t-distribution is used to calculate t-values at various commonly-used levels of confidence. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). When using a t -table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails. Notice that at the bottom the table will show the t-value for infinite degrees of freedom. Mathematically, as the degrees of freedom increase, the t -distribution approaches the standard normal distribution. You can find familiar Z-values by looking in the relevant alpha column and reading value in the last row. A Student's t table (See ) gives t -scores given the degrees of freedom and the right-tailed probability. The Student's t -distribution has one of the most desirable properties of the normal distribution: it is symmetrical. What the Student's t -distribution does is spread out the horizontal axis so it takes a larger number of standard deviations to capture the same amount of probability. In reality there are an infinite number of Student's t -distributions, one for each adjustment to the sample size. As the sample size increases, the Student's t -distribution become more and more like the normal distribution. When the sample size reaches 30 the normal distribution is usually substituted for the Student's t because they are so much alike. This relationship between the Student's t -distribution and the normal distribution is shown in . This is another example of one distribution limiting another one, in this case the normal distribution is the limiting distribution of the Student's t when the degrees of freedom in the Student's t approaches infinity. This conclusion comes directly from the derivation of the Student's t -distribution by Mr. Gosset. He recognized the problem as having few observations and no estimate of the population standard deviation. He was substituting the sample standard deviation and getting volatile results. He therefore created the Student's t -distribution as a ratio of the normal distribution and Chi squared distribution. The Chi squared distribution is itself a ratio of two variances, in this case the sample variance and the unknown population variance. The Student's t -distribution thus is tied to the normal distribution, but has degrees of freedom that come from those of the Chi squared distribution. The algebraic solution demonstrates this result. Development of Student's t-distribution: t = z χ 2 v where z is the standard normal variable and χ 2 is the chi-squared distribution with v degrees of freedom. Substitute values and simplify: t = ( x ¯ − μ ) σ s 2 ( n − 1 ) σ 2 ( n − 1 ) = ( x ¯ − μ ) σ s 2 σ 2 = ( x ¯ − μ ) σ s σ = x ¯ − μ s n t = s n Restating the formula for a confidence interval for the mean for cases when the sample size is smaller than 30 and we do not know the population standard deviation, σ: x ¯ - t v,α ( s n ) ≤ μ ≤ x ¯ + t v,α ( s n ) Here the point estimate of the population standard deviation, s has been substituted for the population standard deviation, σ, and t ν ,α has been substituted for Z α . The Greek letter ν (pronounced nu) is placed in the general formula in recognition that there are many Student t v distributions, one for each sample size. ν is the symbol for the degrees of freedom of the distribution and depends on the size of the sample. Often df is used to abbreviate degrees of freedom. For this type of problem, the degrees of freedom is ν = n-1, where n is the sample size. To look up a probability in the Student's t table we have to know the degrees of freedom in the problem. The average earnings per share (EPS) for 10 industrial stocks randomly selected from those listed on the Dow-Jones Industrial Average was found to be X ¯ = 1.85 with a standard deviation of s=0.395. Calculate a 99% confidence interval for the average EPS of all the industrials listed on the DJIA. x ¯ - t v,α ( s n ) ≤ μ ≤ x ¯ + t v,α ( s n ) To help visualize the process of calculating a confident interval we draw the appropriate distribution for the problem. In this case this is the Student’s t because we do not know the population standard deviation and the sample is small, less than 30. To find the appropriate t-value requires two pieces of information, the level of confidence desired and the degrees of freedom. The question asked for a 99% confidence level. On the graph this is shown where (1-α) , the level of confidence , is in the unshaded area. The tails, thus, have .005 probability each, α/2. The degrees of freedom for this type of problem is n-1= 9. From the Student’s t table, at the row marked 9 and column marked .005, is the number of standard deviations to capture 99% of the probability, 3.2498. These are then placed on the graph remembering that the Student’s t is symmetrical and so the t-value is both plus or minus on each side of the mean. Inserting these values into the formula gives the result. These values can be placed on the graph to see the relationship between the distribution of the sample means, X ¯ 's and the Student’s t -distribution. μ = X ¯ ± t α/2,df=n-1 s n = 1.851 ± 3.2498 0.395 10 = 1.8551 ± 0.406 1.445 ≤ μ ≤ 2.257 We state the formal conclusion as : With 99% confidence level, the average EPS of all the industries listed at DJIA is from $1.44 to $2.26. Try It You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data. 8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5 (8.1634, 9.8032) References “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013). Data from Microsoft Bookshelf . Data from http://www.businessweek.com/. Data from http://www.forbes.com/. “Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013). “Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013). “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013). Chapter Review In many cases, the researcher does not know the population standard deviation, σ , of the measure being studied. In these cases, it is common to use the sample standard deviation, s , as an estimate of σ . The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula: t = x – - μ s n The t -score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with x ¯ ± ( t α 2 ) s n where t α 2 is the t -score with area to the right equal to α 2 , s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find t α 2 for a given α . Formula Review s = the standard deviation of sample values. t = x – − μ s n is the formula for the t -score which measures how far away a measure is from the population mean in the Student’s t-distribution df = n - 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample T ~ t df the random variable, T , has a Student’s t-distribution with df degrees of freedom The general form for a confidence interval for a single mean, population standard deviation unknown, and sample size less than 30 Student's t is given by: x ¯ - t v,α ( s n ) ≤ μ ≤ x ¯ + t v,α ( s n ) Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours. Identify the following: x – =_______ s x =_______ n =_______ n – 1 =_______ Define the random variables X and X – in words. X is the number of hours a patient waits in the emergency room before being called back to be examined. X – is the mean wait time of 70 patients in the emergency room. Which distribution should you use for this problem? Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound. CI: (1.3808, 1.6192) EBM = 0.12 Explain in complete sentences what the confidence interval means. Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal. Identify the following: x – =_______ s x =_______ n =_______ n – 1 =_______ x – = 151 s x = 32 n = 108 n – 1 = 107 Define the random variable X in words. Define the random variable X – in words. X – is the mean number of hours spent watching television per month from a sample of 108 Americans. Which distribution should you use for this problem? Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound. CI: (142.92, 159.08) EBM = 8.08 Why would the error bound change if the confidence level were lowered to 95%? Use the following information to answer the next 13 exercises: The data in are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. X Freq. 1 1 2 7 3 18 4 7 5 6 Calculate the following: x – =______ s x =______ n =______ 3.26 1.02 39 Define the random variable X – in words. What is x – estimating? μ Is σ x known? As a result of your answer to , state the exact distribution to use when calculating the confidence interval. t 38 Construct a 95% confidence interval for the true mean number of colors on national flags. How much area is in both tails (combined)? How much area is in each tail? 0.025 Calculate the following: lower limit upper limit error bound The 95% confidence interval is_____. (2.93, 3.59) Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean. In one complete sentence, explain what the interval means. We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors. Using the same x – , s x , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know? The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean. Using the same x - , s x , and n = 39, how would the error bound change if the confidence level were reduced to 90%? Why? Homework In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice Calculate p ′. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag. State the confidence interval. Sketch the graph. Calculate the error bound. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not? A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal. x – = __________ s x = __________ n = __________ n – 1 = __________ Define the random variables X and X – in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States. State the confidence interval. Sketch the graph. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why? 8629 6944 35 34 t 34 CI: (6244, 11,014) It will become smaller Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours. x – = __________ s x = __________ n = __________ n – 1 = __________ Define the random variables X and X – in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean time wasted. State the confidence interval. Sketch the graph. Explain in a complete sentence what the confidence interval means. A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4. x ¯ = __________ s x = __________ n = __________ n – 1 = __________ Define the random variable X in words. Define the random variable X - in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean length of time. State the confidence interval. Sketch the graph. What does it mean to be “95% confident” in this problem? x – = 2.51 s x = 0.318 n = 9 n - 1 = 8 the effective length of time for a tranquilizer the mean effective length of time of tranquilizers from a sample of nine patients We need to use a Student’s-t distribution, because we do not know the population standard deviation. CI: (2.27, 2.76) Answers may vary. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time. Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal. x - = __________ s x = __________ n = __________ n – 1 = __________ Define the random variable X in words. Define the random variable X – in words. Which distribution should you use for this problem? Explain your choice. Construct a 99% confidence interval for the population mean length of time using training wheels. State the confidence interval. Sketch the graph. Why would the error bound change if the confidence level were lowered to 90%? The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns. The FEC has reported financial information for 556 Leadership PACs that were operating during a certain election cycle. The following table shows the total receipts during this cycle for a random selection of 30 Leadership PACs. $46,500.00 $0 $40,966.50 $105,887.20 $5,175.00 $29,050.00 $19,500.00 $181,557.20 $31,500.00 $149,970.80 $2,555,363.20 $12,025.00 $409,000.00 $60,521.70 $18,000.00 $61,810.20 $76,530.80 $119,459.20 $0 $63,520.00 $6,500.00 $502,578.00 $705,061.10 $708,258.90 $135,810.00 $2,000.00 $2,000.00 $0 $1,287,933.80 $219,148.30 x ¯ = $ 251 , 854.23 s = $ 521 , 130.41 Use this sample data to construct a 95% confidence interval for the mean amount of money raised by all Leadership PACs during that specific election cycle. Use the Student's t-distribution. x ¯ = $ 251 , 854.23 s = $ 521 , 130.41 Note that we are not given the population standard deviation, only the standard deviation of the sample. There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29 CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04 α 2 = 0.02 t α 2 = t 0.02 = 2.150 E B M = t α 2 ( s n ) = 2.150 ( 521 , 130.41 30 ) ~ $ 204 , 561.66 x - - EBM = $251,854.23 - $204,561.66 = $47,292.57 x - + EBM = $251,854.23+ $204,561.66 = $456,415.89 We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during that specific election cycle lies between $47,292.57 and $456,415.89. Forbes magazine published data on the best small firms in a certain year. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The shows the ages of the corporate CEOs for a random sample of these firms. 48 58 51 61 56 59 74 63 53 50 59 60 60 57 46 55 63 57 47 55 57 43 61 62 49 67 67 55 55 49 Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. x ¯ = __________ s x = __________ n = __________ n -1 = __________ Define the random variables X and X ¯ in words. Which distribution should you use for this problem? Explain your choice. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight. State the confidence interval. Sketch the graph. x ¯ = 11.6 s x = 4.1 n = 225 n - 1 = 224 X is the number of unoccupied seats on a single flight. X – is the mean number of unoccupied seats from a sample of 225 flights. We will use a Student’s-t distribution, because we do not know the population standard deviation. CI: (11.12 , 12.08) Answers may vary. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal. Which distribution should you use for this problem? Explain your choice. Define the random variable X - in words. Construct a 95% confidence interval for the population mean cost of a used car. State the confidence interval. Sketch the graph. Explain what a “95% confidence interval” means for this study. Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal. Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets. State the confidence interval. Sketch the graph. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done? Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies. Calculate the mean. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not? CI: (7.64 , 9.36) The sample should have been increased. Answers will vary. Answers will vary. Answers will vary. A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. x - = __________ s x = __________ n = __________ n -1 = __________ Define the random variables X and X - in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population mean worth of coupons. State the confidence interval. Sketch the graph. If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why. Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. Find the 95% Confidence Interval for the true population mean for the amount of soda served. (12.42, 14.18) (12.32, 14.29) (12.50, 14.10) Impossible to determine b Degrees of Freedom ( df ) the number of objects in a sample that are free to vary Normal Distribution a continuous random variable (RV) with pdf f ( x ) = 1 σ 2 π e – ( x – μ ) 2 / 2 σ 2 , where μ is the mean of the distribution and σ is the standard deviation, notation: X ~ N ( μ , σ ). If μ = 0 and σ = 1, the RV is called the standard normal distribution . Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t -Distribution investigated and reported by William S. Gossett in 1908 and published under the pseudonym Student; the major characteristics of this random variable (RV) are: It is continuous and assumes any real values. The pdf is symmetrical about its mean of zero. It approaches the standard normal distribution as n get larger. There is a "family" of t–distributions: each representative of the family is completely defined by the number of degrees of freedom, which depends upon the application for which the t is being used.

Business


A Confidence Interval for A Population Proportion During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43. Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval for a population proportion is similar to that for the population mean, but the formulas are a bit different although conceptually identical. While the formulas are different, they are based upon the same mathematical foundation given to us by the Central Limit Theorem. Because of this we will see the same basic format using the same three pieces of information: the sample value of the parameter in question, the standard deviation of the relevant sampling distribution, and the number of standard deviations we need to have the confidence in our estimate that we desire. How do you know you are dealing with a proportion problem? First, the underlying distribution has a binary random variable and therefore is a binomial distribution . (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B ( n , p ) where n is the number of trials and p is the probability of a success. To form a sample proportion, take X , the random variable for the number of successes and divide it by n , the number of trials (or the sample size). The random variable p ′ (read "P prime") is the sample proportion, P ′ = X n (Sometimes the random variable is denoted as P ^ , read "P hat".) p ′ = the estimated proportion of successes or sample proportion of successes ( p ′ is a point estimate for p , the true population proportion, and thus q is the probability of a failure in any one trial.) x = the number of successes in the sample n = the size of the sample The formula for the confidence interval for a population proportion follows the same format as that for an estimate of a population mean. Remembering the sampling distribution for the proportion from The Central Limit Theorem , the standard deviation was found to be: σ p' = p ( 1 − p ) n The confidence interval for a population proportion, therefore, becomes: p = p ′ ± [ Z ( a 2 ) p ′ ( 1 − p ′ ) n ] Z ( a 2 ) is set according to our desired degree of confidence and p ′ ( 1 − p ′ ) n is the standard deviation of the sampling distribution. The sample proportions p ′ and q ′ are estimates of the unknown population proportions p and q . The estimated proportions p ′ and q ′ are used because p and q are not known. Remember that as p moves further from 0.5 the binomial distribution becomes less symmetrical. Because we are estimating the binomial with the symmetrical normal distribution the further away from symmetrical the binomial becomes the less confidence we have in the estimate. This conclusion can be demonstrated through the following analysis. Proportions are based upon the binomial probability distribution. The possible outcomes are binary, either “success” or “failure”. This gives rise to a proportion, meaning the percentage of the outcomes that are “successes”. It was shown that the binomial distribution could be fully understood if we knew only the probability of a success in any one trial, called p. The mean and the standard deviation of the binomial were found to be: μ = np σ = np q It was also shown that the binomial could be estimated by the normal distribution if BOTH np AND nq were greater than 5. From the discussion above, it was found that the standardizing formula for the binomial distribution is: Z = p' - p ( p q n ) which is nothing more than a restatement of the general standardizing formula with appropriate substitutions for μ and σ from the binomial. We can use the standard normal distribution, the reason Z is in the equation, because the normal distribution is the limiting distribution of the binomial. This is another example of the Central Limit Theorem. We have already seen that the sampling distribution of means is normally distributed. Recall the extended discussion in The Central Limit Theorem concerning the sampling distribution of proportions and the conclusions of the Central Limit Theorem. We can now manipulate this formula in just the same way we did for finding the confidence intervals for a mean, but to find the confidence interval for the binomial population parameter, p. p ′ - Z α 2 p ′ q ′ n ≤ p ≤ p ′ + Z α 2 p ′ q ′ n Where p ′ = x/n, the point estimate of p taken from the sample. Notice that p ′ has replaced p in the formula. This is because we do not know p , indeed, this is just what we are trying to estimate. Unfortunately, there is no correction factor for cases where the sample size is small so np ′ and nq ′ must always be greater than 5 to develop an interval estimate for p . Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have smartphones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have smartphones. Of the 500 people sampled, 421 responded yes - they own smartphones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have smartphones. The solution step-by-step. Let X = the number of people in the sample who have smartphones. X is binomial: the random variable is binary, people either have a smartphone or they do not. To calculate the confidence interval, we must find p ′ , q ′ . n = 500 x = the number of successes in the sample = 421 p ′ = x n = 421 500 = 0.842 p ′ = 0.842 is the sample proportion; this is the point estimate of the population proportion. q ′ = 1 – p ′ = 1 – 0.842 = 0.158 Since the requested confidence level is CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 ( α 2 ) = 0.025. Then z α 2 = z 0.025 = 1.96 This can be found using the Standard Normal probability table in . This can also be found in the students t table at the 0.025 column and infinity degrees of freedom because at infinite degrees of freedom the students t -distribution becomes the standard normal distribution, Z. The confidence interval for the true binomial population proportion is p ′ - Z α p ′ q ′ n ≤ p ≤ p ′ + Z α p ′ q ′ n Substituting in the values from above we find the confidence interval is : 0.810 ≤ p ≤ 0.874 Interpretation: We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have smartphones. Explanation of 95% Confidence Level: Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have smartphones. Try It Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets. (0.3315, 0.4525) The Dundee Dog Training School has a larger than average proportion of clients who compete in competitive professional events. A confidence interval for the population proportion of dogs that compete in professional events from 150 different training schools is constructed. The lower limit is determined to be 0.08 and the upper limit is determined to be 0.16. Determine the level of confidence used to construct the interval of the population proportion of dogs that compete in professional events. We begin with the formula for a confidence interval for a proportion because the random variable is binary; either the client competes in professional competitive dog events or they don't. p = p ′ ± [ Z ( a 2 ) p ′ ( 1 − p ′ ) n ] Next we find the sample proportion: p ′ = 0.08 + 0.16 2 = 0.12 The ± that makes up the confidence interval is thus 0.04; 0.12 + 0.04 = 0.16 and 0.12 − 0.04 = 0.08, the boundaries of the confidence interval. Finally, we solve for Z . [ Z ⋅ 0.12 ( 1 − 0.12 ) 150 ] = 0.04 , therefore Z = 1.51 And then look up the probability for 1.51 standard deviations on the standard normal table. p ( Z = 1.51 ) = 0.4345 , p ( Z ) ⋅ 2 = 0.8690 or 86.90 % . Try It A student polls their school to see if students in the school district are for or against the new legislation regarding school uniforms. They survey 600 students and finds that 480 are against the new legislation. a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval. b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone. A financial officer for a company wants to estimate the percent of accounts receivable that are more than 30 days overdue. They survey 500 accounts and find that 300 are more than 30 days overdue. Compute a 90% confidence interval for the true percent of accounts receivable that are more than 30 days overdue, and interpret the confidence interval. The solution is step-by-step: x = 300 and n = 500 p ′ = x n = 300 500 = 0.600 q ′ = 1 - p ′ = 1 - 0.600 = 0.400 Since confidence level = 0.90, then α = 1 – confidence level = (1 – 0.90) = 0.10 ( α 2 ) = 0.05 Z α 2 = Z 0.05 = 1.645 This Z-value can be found using a standard normal probability table. The student's t-table can also be used by entering the table at the 0.05 column and reading at the line for infinite degrees of freedom. The t-distribution is the normal distribution at infinite degrees of freedom. This is a handy trick to remember in finding Z-values for commonly used levels of confidence. We use this formula for a confidence interval for a proportion: p' - Z α p'q' n ≤ p ≤ p' + Z α p'q' n Substituting in the values from above we find the confidence interval for the true binomial population proportion is 0.564 ≤ p ≤ 0.636 Interpretation: We estimate with 90% confidence that the true percent of all accounts receivable overdue 30 days is between 56.4% and 63.6%. Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL accounts are overdue 30 days. Explanation of 90% Confidence Level: Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of accounts receivable that are overdue 30 days. Try It A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation. a. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval. (0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%. b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone. Solution Sixty-eight percent (68%) of students own an iPod and a smart phone. p ′ = 0.68 q ′ = 1 – p ′ = 1 – 0.68 = 0.32 Since CL = 0.97, we know α = 1 – 0.97 = 0.03 and α 2 = 0.015. The area to the left of z 0.015 is 0.015, and the area to the right of z 0.015 is 1 – 0.015 = 0.985. Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1), z 0.015 = 2.17 E P B = ( z α 2 ) p ′ q ′ n = 2.17 0.68 ( 0.32 ) 300 ≈ 0.0584 p ′ – EPB = 0.68 – 0.0584 = 0.0584 p ′ + EPB = 0.68 + 0.0584 = 0.0584 We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6216 and 0.7384. References Jensen, Tom. “Democrats, Republicans Divided on Opinion of Music Icons.” Public Policy Polling. Available online at http://www.publicpolicypolling.com/Day2MusicPoll.pdf (accessed July 2, 2013). Madden, Mary, Amanda Lenhart, Sandra Coresi, Urs Gasser, Maeve Duggan, Aaron Smith, and Meredith Beaton. “Teens, Social Media, and Privacy.” PewInternet, 2013. Available online at http://www.pewinternet.org/Reports/2013/Teens-Social-Media-And-Privacy.aspx (accessed July 2, 2013). Prince Survey Research Associates International. “2013 Teen and Privacy Management Survey.” Pew Research Center: Internet and American Life Project. Available online at http://www.pewinternet.org/~/media//Files/Questionnaire/2013/Methods%20and%20Questions_Teens%20and%20Social%20Media.pdf (accessed July 2, 2013). Saad, Lydia. “Three in Four U.S. Workers Plan to Work Pas Retirement Age: Slightly more say they will do this by choice rather than necessity.” Gallup® Economy, 2013. Available online at http://www.gallup.com/poll/162758/three-four-workers-plan-work-past-retirement-age.aspx (accessed July 2, 2013). The Field Poll. Available online at http://field.com/fieldpollonline/subscribers/ (accessed July 2, 2013). Zogby. “New SUNYIT/Zogby Analytics Poll: Few Americans Worry about Emergency Situations Occurring in Their Community; Only one in three have an Emergency Plan; 70% Support Infrastructure ‘Investment’ for National Security.” Zogby Analytics, 2013. Available online at http://www.zogbyanalytics.com/news/299-americans-neither-worried-nor-prepared-in-case-of-a-disaster-sunyit-zogby-analytics-poll (accessed July 2, 2013). “52% Say Big-Time College Athletics Corrupt Education Process.” Rasmussen Reports, 2013. Available online at http://www.rasmussenreports.com/public_content/lifestyle/sports/may_2013/52_say_big_time_college_athletics_corrupt_education_process (accessed July 2, 2013). Chapter Review Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning. Let p ′ represent the sample proportion, x/n , where x represents the number of successes and n represents the sample size. Let q ′ = 1 – p ′ . Then the confidence interval for a population proportion is given by the following formula: p' - Z α p'q' n ≤ p ≤ p' + Z α p'q' n Formula Review p′= x n where x represents the number of successes in a sample and n represents the sample size. The variable p ′ is the sample proportion and serves as the point estimate for the true population proportion. q ′ = 1 – p ′ The variable p ′ has a binomial distribution that can be approximated with the normal distribution shown here. The confidence interval for the true population proportion is given by the formula: p' - Z α p'q' n ≤ p ≤ p' + Z α p'q' n n = Z α 2 2 p ′ q ′ e 2 provides the number of observations needed to sample to estimate the population proportion, p , with confidence 1 - α and margin of error e . Where e = the acceptable difference between the actual population proportion and the sample proportion. Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 90% confident that the population proportion is estimated to within 0.05? If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why? The sample size needed would increase. As the confidence level increases, α decreases and z ( a 2 ) increases. To maintain the same error bound, the size of the sample needs to increase. Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions. Identify the following: x = ______ n = ______ p ′ = ______ Define the random variables X and p ′ in words. X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P ′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household. Which distribution should you use for this problem? Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound. CI: (0.5321, 0.6679) EBM : 0.0679 List two difficulties the company might have in obtaining random results, if this survey were done by email. Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks. We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables X and P ′ in words. X is the number of “successes” where an executive prefers a truck. P ′ is the percentage of executives sampled who prefer a truck. Which distribution should you use for this problem? Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound. CI: (0.19432, 0.33068) Suppose we want to lower the sampling error. What is one way to accomplish that? The sampling error given in the survey is ±2%. Explain what the ±2% means. The sampling error means that the true mean can be 2% above or below the sample mean. Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important. Define the random variable X in words. Define the random variable P ′ in words. P ′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election. Which distribution should you use for this problem? Construct a 90% confidence interval, and state the confidence interval and the error bound. CI: (0.62735, 0.67265) EBM : 0.02265 What would happen to the confidence interval if the level of confidence were 95%? Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. What is being counted? The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class. In words, define the random variable X . Calculate the following: x = _______ n = _______ p ′ = _______ x = 64 n = 80 p ′ = 0.8 State the estimated distribution of X . X ~________ Define a new random variable P ′. What is p ′ estimating? p In words, define the random variable P ′. State the estimated distribution of P ′. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet. P ′ ~ N ( 0.8 , ( 0.8 ) ( 0.2 ) 80 ) . (0.72171, 0.87829). How much area is in both tails (combined)? How much area is in each tail? 0.04 Calculate the following: lower limit upper limit error bound The 92% confidence interval is _______. (0.72; 0.88) Fill in the blanks on the graph with the areas, upper and lower limits of the confidence interval, and the sample proportion. In one complete sentence, explain what the interval means. With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%. Using the same p ′ and level of confidence, suppose that n were increased to 100. Would the error bound become larger or smaller? How do you know? Using the same p ′ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why? The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error. If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of confidence)? Homework Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03? If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why? 1,068 The sample size would need to be increased since the critical value increases as the confidence level increases. Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up. x = __________ n = __________ p ′ = __________ Define the random variables X and P ′, in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion who claim they always buckle up. State the confidence interval. Sketch the graph. If this survey were done by telephone, list three difficulties the companies might have in obtaining random results. According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job. State the confidence interval. Sketch the graph. X = the number of people who feel that the president is doing an acceptable job; P ′ = the proportion of people in a sample who feel that the president is doing an acceptable job. N ( 0.61 , ( 0.61 ) ( 0.39 ) 1200 ) CI: (0.59, 0.63) Answers may vary. An article regarding interracial dating and marriage recently appeared in the Washington Post . Of the 1,709 randomly selected adults, 315 identified themselves as Hispanic/Latino, 323 identified themselves as Black, 254 identified themselves as Asian, and 779 identified themselves as White. In this survey, 86% of Black people said that they would welcome a White person into their families. Among Asians, 77% would welcome a White person into their families, 71% would welcome a Hispanic/Latino, and 66% would welcome a Black person. We are interested in finding the 95% confidence interval for the percent of all Black adults who would welcome a White person into their families. Define the random variables X and P ′, in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval. State the confidence interval. Sketch the graph. Refer to the information in . Construct three 95% confidence intervals. percent of all Asians who would welcome a White person into their families. percent of all Asians who would welcome a Hispanic/Latino into their families. percent of all Asians who would welcome a Black person into their families. Even though the three point estimates are different, do any of the confidence intervals overlap? Which? For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions? For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions? (0.72, 0.82) (0.65, 0.76) (0.60, 0.72) Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Hispanic/Latino person into their families. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Black person into their families. Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period. State the confidence interval. Sketch the graph. Explain what a “97% confidence interval” means for this study. A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine . One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem. State the confidence interval. Sketch the graph. Suppose we want to lower the sampling error. What is one way to accomplish that? The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. X = the number of adult Americans who feel that crime is the main problem; p ′ = the proportion of adult Americans who feel that crime is the main problem Since we are estimating a proportion, given p ′ = 0.2 and n = 1000, the distribution we should use is N ( 0.2 , ( 0.2 ) ( 0.8 ) 1000 ) . CI: (0.18, 0.22) Answers may vary. One way to lower the sampling error is to increase the sample size. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%. Refer to . Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools. Define the random variables X and P ′ in words. Which distribution should you use for this problem? Explain your choice. Construct a 95% confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools. State the confidence interval. Sketch the graph. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents. Use the following information to answer the next three exercises: According to a Field Poll, 79% of Georgia adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing Georgia. We wish to construct a 90% confidence interval for the true proportion of Georgia adults who feel that education and the schools is one of the top issues facing Georgia. A point estimate for the true population proportion is: 0.90 1.27 0.79 400 c A 90% confidence interval for the population proportion is _______. (0.761, 0.820) (0.125, 0.188) (0.755, 0.826) (0.130, 0.183) Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not. Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness. (0.2975, 0.3796) (0.6270, 0.6959) (0.3041, 0.3730) (0.6204, 0.7025) The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______. 0.6614 0.3386 173 338 a In a specific year, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a ±3% margin of error. Determine the estimated proportion from the sample. Determine the sample size. Identify CL and α . Calculate the error bound based on the information provided. Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the values. Create a confidence interval for the results of this study. A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience? A national survey of 1,000 adults was conducted by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education. Find the point estimate and the error bound for this confidence interval. Can we (with 95% confidence) conclude that more than half of all American adults believe this? Use the point estimate from part a and n = 1,000 to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education. Can we (with 75% confidence) conclude that at least half of all American adults believe this? p ′ = (0 .55 + 0 .49) 2 = 0.52; EBP = 0.55 - 0.52 = 0.03 No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this. CL = 0.75, so α = 1 – 0.75 = 0.25 and α 2 = 0.125 z α 2 = 1.150 . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.) E B P = ( 1.150 ) 0.52 ( 0.48 ) 1 , 000 ≈ 0.018 ( p ′ - EBP , p ′ + EBP ) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538) Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence. Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music. Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music. This survey was conducted through automated telephone interviews. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error. Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%. You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2020 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview? Binomial Distribution a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, n , of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B ( n , p ). The mean is μ = np and the standard deviation is σ = n p q . The probability of exactly x successes in n trials is P ( X = x ) = n x p x q n − x . Error Bound for a Population Proportion (EBP) the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes.

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