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SUMMARIZING DATA GUIDED PRACTICE 2.10 The observation 64 thousand, a suspected outlier, was found to be an accurate observation. What would such an observation suggest about the nature of character counts in emails?6 GUIDED PRACTICE 2.11 Consider a data set that consists of the following numbers: 12, 12, 12, 12, 12, 1...
Looking for modes isn’t about ﬁnding a clear and correct answer about the number of modes in a distribution, which is why prominent is not rigorously deﬁned in this book. The important part of this examination is to better understand your data and how it might be structured. 12We suspect most households would have 0, 1...
center (mean and median) and the three measures of spread (standard deviation, interquartile range, and range). 2. Understand how the shape of a distribution aﬀects the relationship between the mean and the median. 3. Identify and apply the two rules of thumb for identify outliers (one involving standard deviation and...
eviation is the square root of the variance. It is roughly the “typical” distance of the observations from the mean. sX = 1 n − 1 (xi − ¯x)2 The variance is useful for mathematical reasons, but the standard deviation is easier to interpret because it has the same units as the data set. The units for variance will be th...
oint below this limit.26 The lower whisker stops at the lowest value, 33, since there are no additional data to reach. Outliers are each marked with a dot or asterisk. In a sense, the box is like the body of the box plot and the whiskers are like its arms trying to reach the rest of the data. 26You might wonder, isn’t ...
deleting, or changing one value, no matter how extreme that value, will have little eﬀect on their values. GUIDED PRACTICE 2.48 The distribution of vehicle prices tends to be right skewed, with a few luxury and sports cars lingering out into the right tail. If you were searching for a new car and cared about price, sh...
2 CHAPTER 2. SUMMARIZING DATA GUIDED PRACTICE 2.55 Do these graphs tell us about any association between income for the two groups?38 Looking at an association is diﬀerent than comparing distributions. When comparing distributions, we are interested in questions such as, “Which distribution has a greater average?” and ...
o calculate these statistics; simply state how the medians and IQRs compare. Make sure to explain your reasoning. (a) (1) 3, 5, 6, 7, 9 (2) 3, 5, 6, 7, 20 (b) (1) 3, 5, 6, 7, 9 (2) 3, 5, 7, 8, 9 (c) (1) 1, 2, 3, 4, 5 (2) 6, 7, 8, 9, 10 (d) (1) 0, 10, 50, 60, 100 (2) 0, 100, 500, 600, 1000 2.13 Means and SDs. For each p...
r total scores on the SAT and ACT. The distribution of SAT and ACT scores are both nearly normal. Suppose Ann scored 1300 on her SAT and Tom scored 24 on his ACT. Who performed better? As we saw in section 2.2.3, we can use Z-scores to compare observations from diﬀerent distributions. Using Ann’s SAT score, 1300 , alon...
bout 0.95 between Z = −2 and Z = 2.47 It is possible for a normal random variable to fall 4, 5, or even more standard deviations from the mean. However, these occurrences are very rare if the data are nearly normal. The probability of being further than 4 standard deviations from the mean is about 1-in-15,000. For 5 an...
(e.g. normcdf on a TI) or other technology to ﬁnd the area under the normal curve to the right/left of this Z-score; this is the estimate for the percent/probability. B: Find the x-value that corresponds to a given percentile. 1. Verify that the distribution of interest is approximately normal. 2. Find the Z-score tha...
ions in each group. 2.4.2 Row and column proportions Sometimes it is useful to understand the fractional breakdown of one variable in another, and we can modify our contingency table to provide such a view. Figure 2.37 shows the row proportions for Figure 2.34, which are computed as the counts divided by their row tota...
wer has a mortgage, and it was divided into individual loans (upper) and joint loans (lower). As another example, the bottom segment of the third column represents loans where the borrower owns their home and applied jointly, while the upper segment of this column represents borrowers who are homeowners and ﬁled indivi...
the truth was that the infection rates were independent of getting the vaccine. Additionally, with such small samples, perhaps it’s common to observe such large diﬀerences when we randomly split a group due to chance alone! Example 2.83 is a reminder that the observed outcomes in the data sample may not perfectly reﬂec...
ence is higher for the rosiglitazone group proves that rosiglitazone causes serious cardiovascular problems. iv. Based on the information provided so far, we cannot tell if the diﬀerence between the rates of incidences is due to a relationship between the two variables or due to chance. (b) What proportion of all patie...
awards for best actor and best actress were given out in 1929. The histograms below show the age distribution for all of the best actor and best actress winners from 1929 to 2018. Summary statistics for these distributions are also provided. Compare the distributions of ages of best actor and actress winners.67 Mean S...
andom processes include rolling a die and ﬂipping a coin. (a) Think of another random process. (b) Describe all the possible outcomes of that process. For instance, rolling a die is a random process with potential outcomes 1, 2, ..., 6. 1 What we think of as random processes are not necessarily random, but they may jus...
mails, 367 were spam, 2,827 contained some small numbers but no big numbers, and 168 had both characteristics. Create a Venn diagram for this setup.11 GUIDED PRACTICE 3.16 (a) Use your Venn diagram from Guided Practice 3.15 to determine the probability a randomly drawn email from the email data set is spam and had smal...
peated the process an inﬁnite number of times. Also, even when an outcome is not truly random, modeling it with probability can be useful. • The Law of Large Numbers states that the relative frequency, or proportion of times an outcome occurs after n repetitions, stabilizes around the true probability as n gets large. ...
obability In this section we will use conditional probabilities to answer the following questions: • What is the likelihood that a machine learning algorithm will misclassify a photo as being about fashion if it is not actually about fashion? • How much more likely are children to attend college whose parents attended ...
) and Guided Practice 3.28(b), compute P (mach learn is pred fashion \| truth is fashion) + P (mach learn is pred not \| truth is fashion) (c) Provide an intuitive argument to explain why the sum in (b) is 1.27 26(a) If the photo is about fashion and the ML algorithm prediction was correct, then the ML algorithm my have ...
., and 23/24 for the seventh. The probability you win no prize is the product of these separate probabilities: 23/30. That is, the probability of winning a prize is 1 − 23/30 = 7/30 = 0.233. (b) When the tickets are sampled with replacement, there are seven independent draws. Again we ﬁrst ﬁnd the probability of not wi...
ead of wanting to know P (lived \| inoculated), we might want to know P (inoculated \| lived). This is more challenging because it cannot be read directly from the tree diagram. In these instances we use Bayes’ Theorem. Let’s begin by looking at a new example. \| able to construct 42(a) The tree diagram is shown to the ri...
believing that the earth is warming and being a liberal Democrat mutually exclusive? (b) What is the probability that a randomly chosen respondent believes the earth is warming or is a liberal Democrat? (c) What is the probability that a randomly chosen respondent believes the earth is warming given that he is a liber...
C.1 on page 509. 3.3. SIMULATIONS 173 EXAMPLE 3.54 Mika’s favorite brand of cereal is running a special where 20% of the cereal boxes contain a prize. Mika really wants that prize. If her mother buys 6 boxes of the cereal over the next few months, what is the probability Mika will get a prize? To solve this problem us...
east one car will fail in a ﬂeet. (c) Compute the probability at least one car fails in a ﬂeet of seven. 3.30 To catch a thief. Suppose that at a retail store, 1/5th of all employees steal some amount of merchandise. The stores would like to put an end to this practice, and one idea is to use lie detector tests to catc...
The string holds the distribution at the mean to keep the system balanced. Figure 3.25: A continuous distribution can also be balanced at its mean. 52µX = xf (x)dx where f (x) represents a function for the density curve. 0137170117.85m 182 CHAPTER 3. PROBABILITY AND PROBABILITY DISTRIBUTIONS 3.4.4 Variability in rando...
chandise and the interest of the attendees. 59If X and Y are random variables, consider the following combinations: X 1+Y , X × Y , X/Y . In such cases, plugging in the average value for each random variable and computing the result will not generally lead to an accurate average value for the end result. 186 CHAPTER 3....
0).65 65First ﬁnd the mean and standard deviation of Y − X. The mean of Y − X is µY −X = 5.8 − 5.6 = 0.2. The 0.170 = −1.18 and P (Z < −1.18) = .119. standard deviation is SDY −X = (0.13)2 + (0.11)2 = 0.170. Then Z = 0−0.2 There is an 11.9% chance that Friend 2 will complete the puzzle with a faster time than Friend 1...
s the probability that you pay more than $10 for a dinner consisting of a basket of fries and a fountain drink? 3.5. GEOMETRIC DISTRIBUTION 193 3.5 Geometric distribution How many times should we expect to roll a die until we get a 1? How many people should we expect to see at a hospital until we get someone with blood...
stribution with parameter p, then P (X = x) = (1 − p)x−1p, where x = 1, 2, 3 . . . . • The geometric distribution is always right skewed and, in fact, has no maximum value. The probabil- ities, though, decrease exponentially fast. • Even though the geometric distribution has an inﬁnite number of values, it has a well-d...
a success is p = 0.7. So the probability that 5 of 8 will not exceed the deductible and 3 will exceed the deductible is given by (0.7)5(1 − 0.7)8−5 = 8 5 8! 5!(5 − 3)! (0.7)5(1 − 0.7)8−5 = 8! 5!3! (0.7)5(0.3)3 Dealing with the factorial part: 8! 5!35 × 4 × 3 × 2 × 1)(3 × 2 × 1 = 56 Using (0.7)5(0.3)3 ≈ 0.00454, the ﬁna...
O+: 0, 1, 2, 3, 4. We can make a distribution table with these outcomes. Recall that the probability of a randomly sampled person being blood type O+ is about 0.35. The binomial distribution is used to describe the number of successes in a ﬁxed number of trials. This is diﬀerent from the geometric distribution, which d...
cutoﬀ value for the upper end are increased by 0.5. This correction is called the continuity correction and accounts for the fact that the binomial distribution is discrete. EXAMPLE 3.109 Use the method described to ﬁnd a more accurate estimate for the probability of observing 129, 130, or 131 people with blood type O...
he average, respectively, after many, many repetitions of the chance process. • A probability distribution can be summarized by its center (mean, median), spread (SD, IQR), and shape (right skewed, left skewed, approximately symmetric). • Adding a constant to every value in a probability distribution adds that value to...
this section, we investigate the distribution of the sample proportion ˆp. In the previous section, we saw that the number of people with blood type O+ in a random sample of size 40 follows a binomial distribution with n = 40 and p = 0.35 that is centered on 14 and has standard deviation 3.0. What does the distributio...
nd n = 25. n = 10n = 25p = 0.10.00.20.40.60.81.00.00.20.40.60.81.0p = 0.20.00.20.40.60.81.00.00.20.40.60.81.0p = 0.50.00.20.40.60.81.00.00.20.40.60.81.0p = 0.80.00.20.40.60.81.00.00.20.40.60.81.0p = 0.90.00.20.40.60.81.00.00.20.40.60.81.0 4.1. SAMPLING DISTRIBUTION FOR A SAMPLE PROPORTION 225 Figure 4.4: Sampling distr...
n17, which represents all 19,961 runners who ﬁnished the 2017 Cherry Blossom 10 mile run in Washington, DC. Part of this data set is shown in Figure 4.5, and the variables are described in Figure 4.6. ID 1 2 ... 16923 16924 time 92.25 106.35 ... 122.87 93.30 age 38.00 33.00 ... 37.00 27.00 gender state M MD M DC ... .....
the larger n needs to be for the sampling distribution for ¯x to be nearly normal. However, a good rule of thumb is that for almost all populations, the sampling distribution for ¯x will be approximately normal if n ≥ 30. This brings us to the Central Limit Theorem, the most fundamental theorem in Statistics. CENTRAL ...
here were no houses listed below $600,000 but a few houses above $3 million. (a) Is the distribution of housing prices in Topanga symmetric, right skewed, or left skewed? Hint: Sketch the distribution. (b) Would you expect most houses in Topanga to cost more or less than $1.3 million? (c) Can we estimate the probabilit...
= p1(1 − p1) n1 + p2(1 − p2) n2 . That is, given two independent random samples, the distribution of all possible values of ˆp1 − ˆp2 is centered on the true diﬀerence p1 − p2 and the typical distance or error of ˆp1 − ˆp2 from p1 − p2 is given by . Using µ for mean and σ for SD, we summarize this as follows. p1(1−p1)...
le means ¯x1 − ¯x2 follow a nearly normal distribution, we can ﬁnd the probability that the diﬀerence is greater than or less than a certain amount by ﬁnding a Z-score and using the normal approximation. 4.3. SAMPLING DISTRIBUTION FOR A DIFFERENCE OF PROPORTIONS OR MEANS 249 Exercises 4.23 Difference of proportions, Pa...
Statistical inference is primarily concerned with understanding and quantifying the uncertainty of parameter estimates. While the equations and details change depending on the setting, the foundations for inference are the same throughout all of statistics. We start with a familiar topic: the idea of using a sample pr...
The typical error when p = 0.88 and n = 50 would be larger than the error we would expect when n = 1000. Example 5.3 highlights an important property we will see again and again: a bigger sample tends to provide a more precise point estimate than a smaller sample. Remember though, that this is only true for random sam...
d not cover a $400 unexpected expense without borrowing money or going into debt. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic that we can use to measure the uncertainty of the p...
5.5: The area between -z and z increases as \|z\| becomes larger. If the conﬁdence level is 99%, we choose z such that 99% of the normal curve is between -z and z, which corresponds to 0.5% in the lower tail and 0.5% in the upper tail: z = 2.58. 6This is equivalent to asking how often the Z-score will be larger than -2.5...
with a particular conﬁdence, how much we expect our point estimate to deviate from the true population value due to chance. • The margin of error depends on the conﬁdence level ; the standard error does not. Other things being constant, a higher conﬁdence level leads to a larger margin of error. • For a ﬁxed conﬁdence ...
an eﬀect. Our job as data scientists is to play the skeptic: before we buy into the alternative hypothesis, we need to see strong supporting evidence. EXAMPLE 5.24 Identify the null and alternative claim regarding the consultant’s complication rate. H0: The true complication rate for the consultant’s clients is the sam...
hypothesis is unreasonable.17 DOUBLE NEGATIVES CAN SOMETIMES BE USED IN STATISTICS In many statistical explanations, we use double negatives. For instance, we might say that the null hypothesis is not implausible or we failed to reject the null hypothesis. Double negatives are used to communicate that while we are not...
r conviction from “beyond a reasonable doubt” to “beyond a conceivable doubt” so fewer people would be wrongly convicted. However, this would also make it more diﬃcult to convict the people who are actually guilty, so we would make more Type II Errors. GUIDED PRACTICE 5.35 How could we reduce the Type II Error rate in ...
ulate the test statistic. The test statistic tells us how many standard errors the point estimate (sample value) is from the null value (i.e. the value hypothesized for the parameter in the null hypothesis). When investigating a single mean or proportion or a diﬀerence of means or proportions, the test statistic is cal...
e reasonable), and vice-versa. – Values that fall outside a 95% conﬁdence interval (implying they are not reasonable) will be rejected by a test at the 5% signiﬁcance level (implying they are not reasonable), and vice-versa. – When the conﬁdence level and the signiﬁcance level add up to 100%, the conclusions of the two...
andom sample of size n from a population with a true proportion p, is nearly normal when 1. the sample observations are independent and 2. np ≥ 10 and n(1 − p) ≥ 10. This is called the success-failure condition. If these conditions are met, then the sampling distribution for ˆp is nearly normal with mean µ ˆp = p and s...
on Z-interval. Calculate: We will calculate the interval: point estimate ± z × SE of estimate The point estimate is the sample proportion: ˆp = 0.68 The SE of the sample proportion is: ˆp(1− ˆp) n 0.68(1−0.68) 1033 = = 0.015. z is found using the t-table at row ∞ and conﬁdence level C%. For a 95% conﬁdence level, z = 1...
RVALS VERSUS HYPOTHESIS TESTS FOR A SINGLE PROPORTION 1-proportion Z-interval Check: nˆp ≥ 10 and n(1 − ˆp) ≥ 10 SE = 1-proportion Z-test Check: np0 ≥ 10 and n(1 − p0) ≥ 10 SE = ˆp(1 − ˆp) n p0(1 − p0) n 6.1. INFERENCE FOR A SINGLE PROPORTION 303 EXAMPLE 6.15 (Continues previous example). Deborah Toohey’s campaign mana...
ws. Conﬁdence interval: point estimate ± z × SE of estimate Test statistic: Z = point estimate − null value SE of estimate Here the point estimate is the sample proportion ˆp. The SE of estimate is the SE of the sample proportion. – For an Interval, use SE = ˆp(1− ˆp) n . – For a Test with H0: p = p0, use SE = p0(1−p0)...
evidence that these people are able to detect the diﬀerence between diet and regular soda, in other words, are the results signiﬁcantly better than just random guessing? Carry out an appropriate test and include all steps of the Identify, Choose, Check, Calculate, Conclude framework. (b) Interpret the p-value in this c...
resulting in happier customers and fewer warranty claims. However, management must be convinced that the more expensive gears are worth the conversion before they approve the switch. The quality control engineer collects a sample of gears from each supplier, examining 1000 gears from each company, and ﬁnds that 879 gea...
proportion ˆpc = 0.407, n1 = 771, and n2 = 732. The data do come from a randomized experiment, where the treatments are the two diﬀerent orderings of the question regarding healthcare (because this is an experiment, the 10% condition does not need to be checked). Also, the success-failure condition (minimums of 10) eas...
d what people do when they see a very obviously bruised woman getting picked on by her boyfriend. On two diﬀerent occasions at the same restaurant, the same couple was depicted. In one scenario the woman was dressed “provocatively” and in the other scenario the woman was dressed “conservatively”. The table below shows ...
instead of simply yes/no. This allows us to answer questions such as the following: • Are juries representative of the population in terms of race/ethnicity, or is there a bias in jury selection? • Is the color distribution of actual M&M’s consistent with what was reported on the Mars website? • Do people choose rock, ...
a calculator or statistical software allows us to get more precise areas under the chi-square curve than we can get from the table alone. 0510152025Degrees of Freedom249 330 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA (a) (c) (e) (b) (d) (f) Figure 6.8: (a) Chi-square distribution with 3 degrees of freedom, area above 6...
the stated distribution on the website in 2008? Use the ﬁve step framework to organize your work. Identify: We will test the following hypotheses at the α = 0.05 signiﬁcance level. H0: The distribution of M&M colors is the same as the stated distribution in 2008. HA: The distribution of M&M colors is diﬀerent than the ...
the professor expect to buy the book, print the book, and read the book exclusively online? (c) List the conditions required for the chi-square goodness of ﬁt test and discuss whether they are satisﬁed. (d) Assume conditions are suﬃciently met. Calculate the chi-square statistic, the degrees of freedom asso- ciated wi...
ION METHODS FOR 2-BY-2 CONTINGENCY TABLES When analyzing 2-by-2 contingency tables, use the two-proportion methods introduced in Section 6.2. 38Recall: in the one-way table, the degrees of freedom was the number of groups minus 1. 344 CHAPTER 6. INFERENCE FOR CATEGORICAL DATA Figure 6.13: Computing the p-value for the ...
Figure 6.14, if the null hypothesis is true, that is, if group and rating are independent. The expected count for row one, column one is found by multiplying the row one total (2119) and column one total (1458), then dividing by the table total (4223): 2119×1458 = 731.6. Similarly for the ﬁrst column and the second row...
cance level using a χ2χ2χ2 test for independence. • In addition to the independence/random condition, all expected counts must be at least 5 for the test statistic to follow a chi-square distribution. • The chi-square statistic and associated df are found as follows: test statistic: χ2 = (observed − expected)2 expected...
ct a hypothesis test to determine if these data provide strong evidence that the majority of the Americans think the Civil War is still relevant. (b) Interpret the p-value in this context. (c) Calculate a 90% conﬁdence interval for the proportion of Americans who think the Civil War is still relevant. Interpret the int...
a certain city, county, or state? • What is the average mercury content in various types of ﬁsh? • Are people’s run times getting faster or slower, on average? • How does the sample size aﬀect the expected error in our estimates? • When is it reasonable to model the sample mean ¯x using a normal distribution, and when...
). 2. Select DIST (F5), then t (F2), and then tcd (F2). 3. If needed, set Data to Variable (Var option, which is F2). 4. Enter the Lower t-score and the Upper t-score. Set the degrees of freedom (df). • If ﬁnding just a lower tail area, set Lower to -100. • For an upper tail area, set Upper to 100. 5. Hit EXE, which wi...
iﬁc). We want to estimate this at the 95% conﬁdence level. Choose: Because the parameter to be estimated is a single mean, we will use a 1-sample t-interval. Check: We must check that the sampling distribution for the mean can be modeled using a normal distribution. We will assume that the sample constitutes a random s...
cance level. However, as the p-value of 0.02 > 0.01, there is not suﬃcient evidence at the 1% signiﬁcance level. EXAMPLE 7.16 Would you expect the hypothesized value of 93.3 to fall inside or outside of a 95% conﬁdence interval? What about a 99% conﬁdence interval? Because the hypothesized value of 93.3 was rejected by...
h an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at α = 0.05. (a) n = 11, T = 1.91 (b) n = 17, T = −3.45 (c) n = 7, T = 0.83 (d) n = 28, T = 2.13 7.4 Find the p-value, Part II. A random sample is selected from an appr...
is a common and useful way to analyze paired data. GUIDED PRACTICE 7.20 The ﬁrst diﬀerence shown in Figure 7.9 is computed as: 47.97 − 47.45 = 0.52. What does this diﬀerence tell us about the price for this textbook on Amazon versus the UCLA bookstore?8 7.2.2 Hypothesis tests for a mean of differences To analyze a pair...
val with paired data. Consider again the table summarizing data on: UCLA Bookstore price − Amazon price, for each of the 68 books sampled. ndiﬀ 68 ¯xdiﬀ 3.58 sdiﬀ 13.42 Figure 7.13: Summary statistics for the price diﬀerences. There were 68 books, so there are 68 diﬀerences. We construct a 95% conﬁdence interval for th...
size is less than 10% of the population size. 2. Large sample or normal population: ndiﬀ ≥ 30 or population of diﬀerences nearly normal. - If the number of diﬀerences is less than 30 and it is not known that the population of diﬀerences is nearly normal, we argue that the population of diﬀerences could be nearly normal...
nce on a diﬀerence of means, µ1 − µ2. We will use the t-distribution just as we did when carrying out inference on a single mean. In order to use the t-distribution, we need the sampling distribution for ¯x1 − ¯x2 to be nearly normal. We check whether this assumption is reasonable by verifying the following conditions....
se a 95% conﬁdence interval to estimate the diﬀerence in average score: version A - version B. Version A B n 30 30 ¯x 79.4 74.1 s min max 100 45 100 32 14 20 Identify: The parameter we want to estimate is µ1 − µ2, which is the true average score under Version A − the true average score under Version B. We will estimate...
est to test the claim that two means µ1 and µ2 are equal to each other, Identify: Identify the hypotheses and the signiﬁcance level, α. H0: µ1 = µ2 HA: µ1 = µ2; HA: µ1 > µ2; or HA: µ1 < µ2 Choose: Choose the appropriate test procedure and identify it by name. To test hypotheses about the diﬀerence of means we use a 2-s...
at diamonds. Include all steps of the Identify, Choose, Check, Calculate, Conclude framework. Mean SD n 0.99 carats $44.51 $13.32 23 1 carat $56.81 $16.13 23 Chicken farming is a multi-billion dollar industry, and any 7.25 Chicken diet and weight, Part I. methods that increase the growth rate of young chicks can reduce...
y students at this college? What about the IQR? (c) Is a load of 16 credits unusually high for this college? What about 18 credits? Explain your reasoning. (d) The college counselor takes another random sample of 100 students and this time ﬁnds a sample mean of 14.02 units. Should she be surprised that this sample stat...
rmation about how much ﬁnancial support a college may oﬀer a prospective student (y). However, the prediction would be far from perfect, since other factors play a role in ﬁnancial support beyond a family’s income. It is rare for all of the data to fall perfectly on a straight line. Instead, it’s more common for data t...
ws a pattern in the residuals. There is some curvature in the scatterplot, which is more obvious in the residual plot. We should not use a straight line to model these data. Instead, a more advanced technique should be used. The last plot shows very little upwards trend, and the residuals also show no obvious patterns....
llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(e)lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll(f) 8.1. LINE FITTING, RESIDUALS, AND CORRELATION 433 8.4 Identify relationships, Part II. For each of the six plots...
nes are ﬁt to the data, the solid line being the least squares line. Family Income ($1000s)0501001502002500102030Gift Aid From University ($1000s) 438 CHAPTER 8. INTRODUCTION TO LINEAR REGRESSION We begin by thinking about what we mean by “best”. Mathematically, we want a line that has small residuals. Perhaps our crit...
6th it was 10 degrees. Today it hit almost 80. At this rate, by August it will be 220 degrees. So clearly folks the climate debate rages on. Stephen Colbert April 6th, 2010 11 Linear models can be used to approximate the relationship between two variables. However, these models have real limitations. Linear regression ...
he line around the primary cloud doesn’t appear to ﬁt very well. (4) There is a primary cloud and then a small secondary cloud of four outliers. The secondary cloud appears to be inﬂuencing the line somewhat strongly, making the least squares line ﬁt poorly almost everywhere. There might be an interesting explanation f...
ince Starbucks only lists the number of calories on the display items, we are interested in predicting the amount of carbs a menu item has based on its calorie content. (a) Describe the relationship between number of calories and amount of carbohydrates (in grams) that Starbucks food menu items contain. (b) In this sce...
del is good. Another model might be much better. When assessing the appropriateness of a linear model, we should look at the residual plot. The ∪-pattern in the residual plot tells us the original data is curved. If we inspect the two plots, we can see that for small and large values of x we systematically underestimat...
uld be nearly normal. When this assumption is found to be unreasonable, it is usually because of outliers or concerns about inﬂuential points. An example which suggestions non-normal residuals is shown in the second panel of Figure 8.26. If the sample size n ≥ 30, then this assumption is not necessary. Constant variabi...
al value t using a t-table (the t value is not the same as the T -statistic for the hypothesis test). Using the t-table at row df = 100 (round down since 102 is not on the table) and conﬁdence level 95%, we get t = 1.984. So the 95% conﬁdence interval is given by: 0.57290 ± 1.984 × 0.05933 (0.456, 0.691) Conclude: We a...
the scatterplot does not show a curved trend and that the residual plot shows no ∪-shape pattern. 3. Constant variability: Use the residual plot to check that the standard deviation of the residuals is constant across all x-values. 4. Normality: The population of residuals is nearly normal or the sample size is ≥ 30. ...
arity: Check that the scatterplot does not show a curved trend and that the residual plot shows no ∪-shape pattern. 3. Constant variability: Use the residual plot to check that the standard deviation of the residuals is constant across all x-values. 3. Normality: The population of residuals is nearly normal or the samp...
ing whether body weight is associated with heart weight in cats? (b) State the conclusion of the hypothesis test from part (a) in context of the data. (c) Calculate a 95% conﬁdence interval for the slope of body weight, and interpret it in context of the data. (d) Do your results from the hypothesis test and the conﬁde...
performance. (b) Light level: ﬂuorescent overhead lighting, yellow overhead lighting, no overhead lighting (only desk lamps). (c) Sex: man, woman. 1.31 (a) Experiment. (b) Light level (overhead lighting, yellow overhead lighting, no overhead lighting) and noise level (no noise, construction noise, and human chatter no...
We see the order of the categories and the relative frequencies in the bar plot. (b) There are no features that are apparent in the pie chart but not in the bar plot. (c) We usually prefer to use a bar plot as we can also see the relative frequencies of the categories in this graph. 2.39 The vertical locations at which...
3) Only two outcomes at each trial: Consumed or did not consume alcohol. (4) Probability of a success is the same for each trial: p = 0.697. (b) 0.203. (c) 0.203. (d) 0.167. (e) 0.997. 3.53 (a) 1 − 0.753 = 0.5781. (b) 0.1406. (c) 0.4219. (d) 1 − 0.253 = 0.9844. 3.55 (a) µ = 35, σ = 3.24 (b) Z = 45−35 3.24 = 3.09. 45 is...
60/15 = 4 min√ 15, Z = 1.31 → 0.0951. utes. Using SD¯x = 1.63/ Option 2Since the population distribution is not normal, a small sample size may not be suﬃcient to yield a nearly normal sampling distribution. Therefore, we cannot estimate the probability using the tools we (c) We can now be conﬁdent have learned so far...
diﬀerent from 0.5, and since the data provide a point estimate above 0.5, we have strong evidence to support this claim by the TV pundit. (b) No. Generally we expect a hypothesis test and a conﬁdence interval to align, so we would expect the conﬁdence interval to show a range of plausible values entirely above 0.5. Ho...
= 1.015, then use t 35 = 2.03 n in the formula for margin of error and SE = s/ to identify s = 3. √ 7.7 (a) H0: µ = 8 (New Yorkers sleep 8 hrs per night on average.) HA: µ = 8 (New Yorkers sleep less or more than 8 hrs per night on average.) (b) Independence: The sample is random. The min/max suggest there are no conce...

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