7.1 Inference for the Mean of a Population

Both confidence intervals and tests of significance for the mean μ of a Normal population are based on the sample mean x¯, which estimates the unknown μ. The sampling distribution of x¯ depends on population standard deviation σ. This fact causes no difficulty when σ is known. When σ is unknown, however, we must estimate σ even though we are primarily interested in μ.

In this section, we meet the sampling distribution of the standardized version of x¯ when we use the sample standard deviation s as our estimate of σ. It is this sampling distribution that is used to produce both confidence intervals and significance tests about the mean μ.

The t distributions

Suppose that we have a simple random sample (SRS) of size n from a Normally distributed population with mean μ and standard deviation σ. The sample mean x¯ is then Normally distributed, with mean μ and standard deviation σ/n. When σ is not known, we estimate it with the sample standard deviation s. We can then use s to estimate the standard deviation of x¯. This quantity, s/n, is called the standard error of the sample mean, and we denote it by SEx¯.

The term “standard error” is sometimes used for the actual standard deviation of a statistic. The estimated value is then called the “estimated standard error.” In this book, we will use the term “standard error” only when the standard deviation of a statistic is estimated from the data. The term has this meaning in the output of many statistical software packages and in research reports that apply statistical methods.

In the previous chapter, the standardized sample mean, or one-sample z statistic,

z=x¯-μσ/n

is the basis for inference about μ. This statistic has the standard Normal distribution N(0, 1). However, when we substitute the standard error s/n for the standard deviation of x¯, the statistic does not have a Normal distribution. It has a distribution that is new to us, called a t distribution.

A particular t distribution is specified by giving the degrees of freedom. We use t(k) to stand for the t distribution with k degrees of freedom. The degrees of freedom for this t statistic come from the sample standard deviation s in the denominator of t. We showed earlier that s has n−1 degrees of freedom. Thus, there is a different t distribution for each sample size n. We will meet many other t statistics with different degrees of freedom in this and later chapters.

The t distributions were discovered in 1908 by William S. Gosset. Gosset was a statistician employed by the Guinness brewing company, which prohibited its employees from publishing their discoveries that were brewing related. In this case, the company let him publish under the pen name “Student” using an example that did not involve brewing. The t distribution is often called “Student’s t” in his honor.

The density curves of the t(k) distributions are similar in shape to the standard Normal curve. That is, they are symmetric about 0 and are bell-shaped. Figure 7.1 compares the density curves of the standard Normal distribution and the t distributions with 5 and 15 degrees of freedom. The similarity in shape is apparent, as is the fact that the t distributions have more probability in the tails and, thus, less in the center.

Three symmetrical density curves.

Figure 7.1 Density curves for the standard Normal, t(5), and t(15) distributions. All are symmetric, with center 0.

This greater spread is due to the extra variability caused by substituting the random variable s for the fixed parameter σ. Comparing the two t curves in Figure 7.1, we see that as the degrees of freedom k increase, the t(k) curve gets closer to the N(0, 1) curve. This reflects the fact that s will be closer to σ (more precise) as the sample size increases.

Check-in

  1. 7.1 Two-bedroom apartment rates. Zillow.com contains several hundred postings for two-bedroom apartments in your city. You choose 16 at random and calculate a mean monthly rent of $906 and a standard deviation of $276.

    1. What is the standard error of the mean?

    2. What are the degrees of freedom for the one-sample t statistic?

  2. 7.2 Changing the sample size. Refer to the previous Check-in question. Suppose that instead of having an SRS of 16, you sampled 25 postings.

    1. Would you expect the standard error of the mean to be larger or smaller in this case? Explain your answer.

    2. State why you can’t be certain that the standard error for this new SRS will be larger or smaller.

With the t distributions to help us, we can now analyze an SRS from a Normal population with unknown σ or a large sample from a non-Normal population with unknown σ. We usually rely on software to do the necessary t calculations, but when software is not available, we can use tables, such as Table D in the back of the book, to get the critical values t*. For convenience, Table D labels the table entries both by the upper-tail probability p needed for significance tests and by the confidence level C (in percent) required for confidence intervals. The standard Normal critical values that we used in Chapter 6 are in the bottom row of entries and labeled z*.

One-sample t confidence interval

The one-sample t confidence interval is similar in both reasoning and computational detail to the z confidence interval of Chapter 6. There, the margin of error for the population mean was z*σ/n. When σ is unknown, we replace it with its estimate s and switch from z* to t*. This means that the margin of error for the population mean when we use the data to estimate σ is t*s/n.

Example 7.1 Average delivery time for robot service.

Data set icon for botdel.

Your university recently launched robot food delivery services that works in conjunction with your meal plan. An advertisement states that you can order food and drinks to be delivered anywhere on campus, within minutes.1 You decide to investigate this claim and record the delivery times for your weekday lunch orders from Cosi. Let’s construct a 95% confidence interval for the average delivery time (in minutes) from Cosi to your dormitory based on the following n=8 samples:

13.7 26.3 20.0 45.3 8.5 43.6 10.1 17.3

The sample mean is

x¯=13.7+26.3+⋯+17.38=23.1

and the standard deviation is

s=(13.7-23.1)2+(26.3-23.1)2+⋯+(17.3-23.1)28-1=14.328

with degrees of freedom n−1=7. The standard error is

SEx¯=s/n=14.328/8=5.066
df=7
t* 1.895 2.365 2.517
C 0.90 0.95 0.96

From Table D, we find t*=2.365. The 95% confidence interval is

x¯±t*sn=23.1±2.36514.3288=23.1±(2.365)(5.066)=23.1±11.981=(11.1, 35.1)

We are 95% confident that the average delivery time from Cosi to your dormitory is between 11.1 and 35.1 minutes.

In this example, we have given the actual interval (11.1, 35.1) minutes as our answer. Sometimes, we prefer to report the mean and margin of error: the mean time is 23.1 minutes, with a margin of error of 12.0 minutes. This is a large margin of error in relation to size of the estimated mean. In Section 7.3, we will return to this example and discuss determining an appropriate sample size for a smaller margin of error, such as ±5 minutes.

Valid interpretation of the t confidence interval in Example 7.1 rests on assumptions that appear reasonable here. First, we assume that our sample of delivery times is an SRS from the population of all weekday delivery times from Cosi to your dormitory. Because your decision to order from Cosi on a weekday is fairly random, this is likely reasonable. If you only ordered from Cosi on Thursdays, generalizing to any weekday may be suspect as the campus foot traffic and robot delivery demands may vary day to day.

We also assume that the distribution of delivery times is Normal. With only eight observations, this assumption cannot be effectively checked. We can, however, check if the data suggest a severe departure from Normality. Figure 7.2 shows the Normal quantile plot, and we can see that there are no extreme outliers, and there is no severe skewness. Deciding whether to use the t procedures for inference about μ is often a judgment call. We provide some guidelines to assist in this decision later in this section.

A normal quantile plot of delivery times and scores.

Figure 7.2 Normal quantile plot of the data, Example 7.1.

Check-in

  1. 7.3 More on apartment rents. Recall Check-in question 7.1 (page 386). Assume that the rents are Normally distributed.

    1. Construct a 95% confidence interval for the mean monthly rent of two-bedroom apartments.

    2. Can we consider that this interval applies to the mean of all two-bedroom apartment rents in your city? Explain your answer.

  2. 7.4 Finding critical t*-values. What critical value t* from Table D should be used to construct

    1. a 90% confidence interval when n=25?

    2. a 95% confidence interval when n=11?

    3. a 99% confidence interval when n=61?

The one-sample t test

Significance tests of the mean μ using the standard error of x¯ are also very similar to the z test described in the previous chapter. We still carry out the four steps common to all significance tests, but because we use s in place of σ, we use a t distribution, instead of the standard Normal distribution, to find the P-value. Here are the details.

Example 7.2 Significance test for robot delivery service.

Data set icon for botdel.

If you were to pick up an order from Cosi, it would take you 15 minutes to walk there and back. Focusing just on the time it takes to get an order, we might want to test whether the average robot delivery time differs from this 15 minutes. Specifically, we want to test

H0:  μ=15.0Ha:  μ≠15.0

at the 0.05 significance level. Recall that n=8, x¯=23.1, and s=14.328. The one-sample t statistic is

t=x¯−μ0s/n=23.1−15.014.328/8=1.599

This means that the sample mean x¯=23.1 is slightly more than 1.5 standard deviations above the null hypothesized value μ=15.0.

df=7
p 0.10 0.05
t* 1.415 1.895

Because the degrees of freedom are n−1=7, this t statistic has the t(7) distribution. Figure 7.3 shows that the P-value is 2P(T≥| 1.599 |), where T has the t(7) distribution. From Table D, we see that P(T≥1.415)=0.10 and P(T≥1.895)=0.05.

A normal distribution curve.

Figure 7.3 Sketch of the P-value calculation, Example 7.2.

Therefore, we conclude that the P-value is between 2×0.05=0.10 and 2×0.10=0.20. In Excel, the exact P-value for the two-sided alternative can be obtained using the function =T.DIST.2T(1.599,7), where the first input is the absolute value of the test statistic and the second input is the degrees of freedom. It gives P=0.1539.

These data are compatible with an average of 15.0 minutes. Under H0, a difference this large or larger would occur roughly one time in six simply due to chance. There is not enough evidence to reject the null hypothesis at the 0.05 level.

In this example, we tested the null hypothesis μ=15.0 minutes against the two-sided alternative μ≠15 minutes because we had no prior suspicion that the average would be larger or smaller. If we had suspected that the average would be larger (for example, the delivery robots are more cautious crossing streets), we would have used a one-sided test.

Example 7.3 One-sided test for robot delivery service.

Data set icon for botdel.

For the problem described in the previous example, we want to test whether the robot average delivery time is larger than the stated 15.0 minutes. Here we test

H0:  μ=15.0

versus

Ha:  μ>15.0

The t test statistic does not change: t=1.599. As Figure 7.4 illustrates, however, the P-value is now P(T≥1.599), half of the value in the previous example. From Table D, we can determine that 0.05<P<0.10; again rounding to four decimal places, Excel gives P=0.0769, using the function =T.DIST.RT(1.599,7). Similar to the two-sided alternative, there is not enough evidence to reject the null hypothesis in favor of the alternative at the 0.05 significance level.

A normal distribution curve.

Figure 7.4 Sketch of the P-value calculation, Example 7.3.

For the robot delivery example, our conclusion did not depend on the choice between one-sided and two-sided alternatives. Sometimes, however, this choice will affect the conclusion, and so this choice needs to be made prior to analysis. If in doubt, always use a two-sided test. caution It is wrong to examine the data first and then decide to do a one-sided test in the direction indicated by the data. Often, a significant result for a two-sided test can be used to justify a one-sided test for another sample from the same population.

Check-in

  1. 7.5 Significant? A two-sided t test for the mean μ of a Normal population gives t=2.137.

    1. The sample size is 28. Is the test result significant at the 5% level? Explain how you obtained your answer.

    2. The sample size is 10. Is the test result significant at the 5% level? Explain how you obtained your answer.

    3. Sketch the two t distributions to illustrate your answers.

  2. 7.6 Significance test for apartment rents. Refer to Check-in questions 7.1 (page 386) and 7.3 (page 388). Do the data give good reason to believe that the mean rent of all two-bedroom apartments is greater than $750? State the hypotheses, find the t statistic and its P-value, and state your conclusion using the 5% significance level.

Using software

Data set icon for Vtm.

For small data sets, such as the one in Example 7.1 (page 387), it is easy to perform the computations for confidence intervals and significance tests with an ordinary calculator and Table D. For larger data sets, however, we prefer to use software.

Example 7.4 Stock portfolio diversification?

Data set icon for stock.

An investor with a stock portfolio worth several hundred thousand dollars sued his broker and brokerage firm because lack of diversification in his portfolio led to poor performance. TABLE 7.1 gives the rates of return for the 39 months that the account was managed by the broker. An arbitration panel compared these returns with the average of the Standard & Poor’s 500-stock index for the same period.2 Let’s do the same using the one-sample t test.

Table 7.1 Monthly rates of return on a portfolio (%)

−8.36 1.63 −2.27 −2.93 −2.70 −2.93 −9.14 −2.64
6.82 −2.35 −3.58 6.13 7.00 −15.25 −8.66 −1.03
−9.16 −1.25 −1.22 −10.27 −5.11 −0.80 −1.44 1.28
−0.65 4.34 12.22 −7.21 −0.09 7.34 5.04 −7.24
−2.14 −1.01 −1.41 12.03 −2.56 4.33 2.35

Consider the 39 monthly returns as a random sample from the population of possible monthly returns the brokerage firm would generate during this time period. Are these returns compatible with a population mean of μ=1.38, the average S&P 500 return during that period? Our hypotheses are

H0:  μ=1.38Ha:  μ≠1.38

Figure 7.5 gives a histogram and summary statistics for these data using JMP. There are no outliers, and the distribution shows no strong skewness. Thus, we are reasonably confident that the distribution of x¯ is approximately Normal, and we proceed with our inference based on Normal theory. Minitab, SPSS, and Excel output are shown in Figure 7.6.

J M P output of a histogram and summary statistics.

Figure 7.5 Histogram and summary statistics of monthly rates of return for a stock portfolio using JMP, Example 7.4.

Minitab, S P S S, and Excel outputs of statistics.

Figure 7.6 Minitab, SPSS, and Excel outputs, Examples 7.4 and 7.5.

Here is one way to report the conclusion: the mean monthly return on investment for this client’s account was x¯=−1.1%. This differs significantly from 1.38%, the performance of the S&P 500 index for the same period (t=−2.58, df=38, P=0.014).

The hypothesis test in Example 7.4 leads us to conclude that the mean return on the client’s account differs from that of the S&P 500 index. Now let’s assess the return on the client’s account with a confidence interval.

Example 7.5 Estimating the mean monthly return.

Data set icon for stock.

The mean monthly return on the client’s portfolio was x¯=−1.1%, and the standard deviation was s=5.99%. Figures 7.5 and 7.6 give output for a 95% confidence interval for the population mean μ. Note that Excel gives the margin of error rather than the actual confidence interval next to the label “Confidence Level(95.0%).” We see that the 95% confidence interval is (−3.04, 0.84), or (from Excel) −1.0997±1.9420.

Because the S&P 500 return, 1.38%, falls outside this interval, we know that μ differs significantly from 1.38% at the α=0.05 level. The confidence interval suggests that the broker’s management of this account had a long-term mean somewhere between a loss of 3.04% and a gain of 0.84% per month. However, we are interested not in the actual mean but in the difference between the performance of the broker’s portfolio and that of the diversified S&P 500 index.

Example 7.6 Estimating the difference from a standard.

Data set icon for stock.

The difference between the mean of the investor’s account and the S&P 500 is x¯−μ=−1.10−1.38=−2.48%. In Example 7.5, we found that the 95% confidence interval for the investor’s account was (−3.04, 0.84). To obtain the corresponding interval for the difference, subtract 1.38 from each of the endpoints. The resulting interval is (−3.04−1.38, 0.84−1.38), or (−4.42, −0.54). This interval is presented in the SPSS output of Figure 7.6. We conclude with 95% confidence that the underperformance was between −4.42% and −0.54%. This estimate helps to set the compensation owed the investor.

The assumption that these 39 monthly returns represent an SRS from the population of monthly returns from the brokerage during this time period is certainly debatable. One could also argue that we should not consider the S&P 500 monthly average return as a constant standard. Its return would also vary from month to month. If those monthly returns were available, an alternative analysis would be to compare the average difference between each monthly return for this account and for the S&P 500. This type of comparison is discussed next.

Check-in

  1. 7.7 Using software to obtain a confidence interval. In Example 7.1 (page 387), we calculated the 95% confidence interval for average delivery time of a delivery robot. Use software to compute this interval and verify that you obtain the same interval. Data set icon for botdel.

  2. 7.8 Using software to perform a significance test. In Example 7.2 (page 389), we tested whether the average delivery time for the robot differs from 15 minutes at the 0.05 significance level. Use software to perform this test and obtain the exact P-value. Data set icon for botdel.

Matched pairs t procedures

Data set icon for Vtm.

The robot delivery problem of Example 7.1 concerns only a single population. We know that comparative studies are usually preferred to single-sample investigations because of the protection they offer against confounding. For that reason, inference about a parameter of a single distribution is less common than comparative inference.

One common comparative design, however, makes use of single-sample procedures. In a matched pairs study, subjects are matched in pairs, and their outcomes are compared within each matched pair. For example, an experiment to compare the interest in two smartphone packages might use pairs of subjects who are the same age, sex, and income level. The experimenter could toss a coin to assign the two packages to the two subjects in each pair. The idea is that matched subjects are more similar than unmatched subjects, so comparing outcomes within each pair is more efficient (i.e., reduces the standard deviation of the estimated difference of treatment means).

Matched pairs are also common when randomization is not possible. For example, one situation calling for matched pairs is when observations are taken on the same subjects under two different conditions. Here is an example.

Example 7.7 The effect of altering a software parameter.

Data set icon for GEparts.

The MeasureMind® 3D MultiSensor metrology software is used by various companies to measure complex machine parts. As part of a technical review of the software, researchers at GE Healthcare discovered that unchecking one software option reduced measurement time by 10%. This time reduction would help the company’s productivity, provided that the option had no impact on the measurement outcome. To investigate this, the researchers measured 51 parts using the software both with and without this option checked.3 The experimenters tossed a fair coin to decide which measurement (with or without the option) to take first.

TABLE 7.2 gives the measurements (in microns) for the first 20 parts. For analysis, we subtract the measurement with the option on from the measurement with the option off. These differences form a single sample and appear in the “Diff” columns for each part.

Table 7.2 Parts measurements using optical software

Part OptionOn OptionOff Diff Part OptionOn OptionOff Diff
 1 118.63 119.01 0.38 11 119.03 118.66 −0.37
 2 117.34 118.51 1.17 12 118.74 118.88 0.14
 3 119.30 119.50 0.20 13 117.96 118.23 0.27
 4 119.46 118.65 −0.81 14 118.40 118.96 0.56
 5 118.12 118.06 −0.06 15 118.06 118.28 0.22
 6 117.78 118.04 0.26 16 118.69 117.46 −1.23
 7 119.29 119.25 −0.04 17 118.20 118.25 0.05
 8 120.26 118.84 −1.42 18 119.54 120.26 0.72
 9 118.42 117.78 −0.64 19 118.28 120.26 1.98
10 119.49 119.66 0.17 20 119.13 119.15 0.02

To assess whether there is a difference between the measurements with and without this option, we test

H0:  μ=0Ha:  μ≠0

Here, μ is the mean difference for the entire population of parts. The null hypothesis says that there is no difference, and Ha says that there is a difference but does not specify a direction.

The 51 differences have

x¯=0.0504 and s=0.6943

Figure 7.7 shows a histogram and estimated Normal density curve of the differences. The data are reasonably symmetric, with no outliers, so based on the guidelines we discuss next (page 398), we can comfortably use the one-sample t procedures. Remember to always check assumptions before proceeding with statistical inference.

A histogram and normal density curve.

Figure 7.7 Histogram of differences in measurements, Example 7.7.

The one-sample t statistic is

t=x¯−0s/n=0.05040.6943/51=0.52

The P-value is found from the t(50) distribution. Remember that the degrees of freedom are 1 less than the sample size.

Table D shows that 0.52 lies to the left of the first column entry. This means the P-value is greater than 2(0.25)=0.50. Software gives the exact value P=0.6054. There is little evidence to suggest that this option has an impact on the measurements. When reporting results, it is usual to omit the details of routine statistical procedures; our test would be reported in the form “The difference in measurements was not statistically significant (t=0.52, df=50, P=0.61).”

This result, however, does not fully address the goal of this study. caution A lack of statistical significance does not prove the null hypothesis is true. If that were the case, we would simply design poor experiments whenever we wanted to prove the null hypothesis. The more appropriate method of inference in this setting is to consider equivalence testing. With this approach, we try to prove that the mean difference is within some acceptable region around 0. We can actually perform this test using a confidence interval.

Example 7.8 Are the two means equivalent?

Data set icon for GEparts.

Suppose the GE Healthcare researchers state that a mean difference less than 0.20 micron is not important. To see if the data support a mean difference within 0.00±0.20 micron, we construct a 90% confidence interval for the mean difference.

The standard error is

SEx¯=sn=0.694351=0.0972

so the margin of error is

m=t*×SEx¯=(1.676)(0.0972)=0.1629

where the critical value t*=1.676 comes from Table D using 50 degrees of freedom. The confidence interval is

x¯±m=0.0504±0.1629=(−0.112, 0.2133)
df=50
t* 1.676 2.009
C 90% 95%

This interval is not entirely within the 0.00±0.20 micron region that the researchers state is not important. Thus, we do not conclude at the 5% significance level that the two means are equivalent. Because the observed mean difference is close to zero and well within the “equivalence region,” the company may want to consider a larger study to improve precision.

One can also use statistical software to perform an equivalence test. Figure 7.8 shows the Minitab output for Example 7.8. It is common to visually present the test using the confidence interval and the user-specified upper and lower equivalence limits.

A Minitab output.

Figure 7.8 Minitab output of the equivalence test, Example 7.8.

Check-in

  1. 7.9 Female wolf spider mate preferences. As part of a study on factors affecting mate choice, researchers exposed 18 premature female wolf spiders twice a day until maturity to iPod videos of three courting males with average-size tufts. Once mature, each female spider was exposed to two videos, one involving a male with large tufts and the other involving a male with small tufts. The number of receptivity displays by the female toward each male was recorded.4 Explain why a paired t-test is appropriate in this setting.

  2. 7.10 Cold plasma technology. Cold plasma (CP) technology has been shown to be an effective tool for shelf-life extension of food.5 For it to become widely used, however, CP’s effects on food quality must be understood. Consider the following study, which compared the taste of fruit juice with and without CP treatment. Each juice was rated, by a set of taste experts, using a 0 to 100 scale, with 100 being the highest rating. For each expert, a coin was tossed to see which juice was tasted first. Data set icon for juice.

    Expert
    1 2 3 4 5 6 7 8 9 10
    With CP: 93 65 37 78 89 49 88 55 63 62
    Without CP: 92 67 47 79 84 52 80 52 67 69

    Is there a difference in taste? State the appropriate hypotheses and carry out a matched pairs t test using α=0.05.

  3. 7.11 Are the means equivalent? Refer to the previous Check-in question. Other research has suggested that a difference of ±4 on this taste scale is indistinguishable. Use the data to get a 90% confidence interval for the mean difference in taste scores between CP-treated and non-CP-treated fruit juice. Is there enough evidence to claim the two means are equivalent? Data set icon for juice.

Robustness of the t procedures

The matched pairs t procedures and the test of equivalence use one-sample t confidence intervals and significance tests for differences. They are, therefore, based on an assumption that the population of differences has a Normal distribution. For the histogram of the 51 differences in Example 7.7 shown in Figure 7.7 (page 395), the data appear to be slightly skewed. Does this slight non-Normality suggest that we should not use the t procedures for these data?

All inference procedures are based on some conditions, such as Normality. Procedures that are not strongly affected by violations of a condition are called robust. Robust procedures are very useful in statistical practice because they can be used over a wide range of conditions with good performance.

The condition that the population is Normal rules out outliers, so the presence of outliers shows that this assumption is not valid. The t procedures are not robust against outliers because x¯ and s are not resistant to outliers.

Fortunately, the t procedures are quite robust against non-Normality of the population, particularly when the sample size is large. This is true because:

  1. The central limit theorem says that the sampling distribution of the sample mean x¯ from a large sample is close to Normal. Normality of the individual observations is of little concern when the sample is large.

  1. As the sample size n grows, the sample standard deviation s will be an accurate estimate of σ whether or not the population has a Normal distribution. This fact is closely related to the law of large numbers.

Applet To convince yourself of this fact, use the Distribution of the One-Sample t Statistic applet to study the sampling distribution of the one-sample t statistic. From one of three population distributions, 10,000 SRSs of a user-specified sample size n are generated, and a histogram of the t statistics is constructed. You can then compare this estimated sampling distribution with the t(n−1) distribution. When the population distribution is Normal, the sampling distribution of the t statistic is always t distributed. For the other two population distributions, you should see that as n increases, the histogram of t statistics looks more like the t(n−1) distribution.

To assess whether the t procedures can be used in practice, a Normal quantile plot, stemplot, or boxplot can be used to check for skewness and outliers. For most purposes, the one-sample t procedures can be safely used when n≥15 unless an outlier or clearly marked skewness is present. In fact, the condition that the data are an SRS from the population of interest is the more crucial assumption, except in the cases of small samples. Here are practical guidelines, based on the sample size and plots of the data, for inference on a single mean:6

For the robot delivery study in Example 7.1, there are only n=8 observations. We are comfortable that the t procedures give approximately correct results because the data are close to Normal. On the other hand, the measurement study in Example 7.7 (page 394) has n=51 observations. With that large a sample, our main concern is the presence of outliers.

Check-in

  1. 7.12 t procedures for average business start time? Consider the data from Example 1.19 (page 26). Would you be comfortable using the t procedures for inference on the average business start time? Explain your answer. Data set icon for tts24.

  2. 7.13 t procedures for a larger sample? Now consider the data for all 187 countries as an SRS from the population of potential start times. Would you be comfortable applying the t procedures in this case? Explain your answer. Data set icon for tts.

Inference for non-normal populations

So what do you do when the populations are clearly non-Normal and you are not comfortable assuming that the sample size is large enough to rely on the robustness of the t procedures? Our recommendation is to consult an expert. There are three general strategies, but each quickly moves beyond the basic practice of statistics:

  1. In some cases, a distribution other than a Normal distribution describes the data well. There are many non-Normal models for data, and inference procedures for these models are available.
  2. Because skewness is the chief barrier to the use of t procedures on data without outliers, you can attempt to transform skewed data so that the distribution is symmetric and as close to Normal as possible. Confidence levels and P-values from the t procedures applied to the transformed data will be quite accurate for even moderate sample sizes. Methods are generally available for transforming the results back to the original scale.
  3. Use a distribution-free inference procedure. Such procedures do not assume that the population distribution has any specific form, such as Normal. Distribution-free procedures are often called nonparametric procedures. Chapter 15 discusses several of these procedures.

We emphasize procedures based on Normal distributions because they are the most common in practice, because their robustness makes them widely useful, and (most importantly) because we are first of all concerned with understanding the principles of inference. Nonetheless, it is helpful to illustrate by example a couple of these approaches—in particular, the use of a transformation and the use of a simple distribution-free procedure.

Transforming data

When the distribution of a variable is skewed, it often happens that a transformation results in a variable whose distribution is symmetric and even close to Normal. The most common transformation is the logarithm, or log. The logarithm tends to pull in the right tail of a distribution. For example, the data 2, 3, 4, 20 show an outlier in the right tail. Their natural logarithms 0.30, 0.48, 0.60, 1.30 are much less skewed. Taking logarithms is a possible remedy for right-skewness. Instead of analyzing values of the original variable X, we compute their logarithms and analyze the values of X.

Example 7.9 Length of songs in a Spotify library.

Data set icon for songs.

TABLE 7.3 presents data on the length (in seconds) of songs found in a “Liked Songs” Spotify library. The library contained a total of 8173 songs, and 50 songs were randomly selected using the “shuffle play” option.7 We would like to give a confidence interval for the average song length μ for this library.

Table 7.3 Length (in seconds) of songs sampled from a Spotify library

 240 316 259  46 871 411 1366
 233 520 239 259 535 213 492
 315 696 181 357 130 373 245
 305 188 398 140 252 331 47
 309 245 69 293 160 245 184
 326 612 474 171 498 484 271
 207 169 171 180 269 297 266
1847

A Normal quantile plot of these data (Figure 7.9) shows that the distribution is non-negative and skewed to the right. Even for this clearly skewed distribution, the guidelines suggest that the sample mean of the 50 observations will nonetheless have an approximately Normal sampling distribution. Thus, the t procedures could be used for approximate inference. For more exact inference, we will transform the data so that the distribution is more nearly Normal. Figure 7.10 is a Normal quantile plot of the natural logarithms of the time measurements. The transformed data are very close to Normal, so t procedures will give quite exact results.

A normal quantile plot.

Figure 7.9 Normal quantile plot of audio file length, Example 7.9. This sort of pattern occurs when a distribution is skewed to the right.

A normal quantile plot.

Figure 7.10 Normal quantile plot of the logarithms of the audio file lengths, Example 7.9. This distribution appears approximately Normal.

The application of the t procedures to the transformed data is straightforward. Call the original length values from Table 7.3 the variable X. The transformed data are values of Xnew=logX. In most software packages, it is an easy task to transform data in this way and then analyze the new variable.

Example 7.10 Software output of audio length data.

Data set icon for songs.

Analysis of the natural log of the length values in Minitab produces the following output:

For comparison, the 95% t confidence interval for the original mean μ is found from the original data as follows:

The advantage of analyzing transformed data is that use of procedures based on the Normal distributions is better justified, and the results are more exact. The disadvantage is that a confidence interval for the mean μ in the original scale (in our example, seconds) cannot be easily recovered from the confidence interval for the mean of the logs. One approach based on the lognormal distribution8 results in an interval of (285.5, 435.5), which is narrower and slightly asymmetric compared with the t interval.

Use of a distribution-free procedure

Perhaps the most straightforward way to cope with non-Normal data is to use a distribution-free procedure. As the name indicates, these procedures do not require that the data follow any specific distribution, such as Normal. This gain in generality, however, is not free. First, if the data are close to Normal, distribution-free procedures are generally less powerful than the t test. Second, they don’t quite answer the same question. One-sample distribution-free procedures ask about the population median, as this is a natural measure of population center for a skewed distribution.

The simplest distribution-free procedure, and one of the most useful, is the sign test. The test gets its name from the fact that we look only at the signs of the differences, not their actual values. The following example illustrates this test.

Example 7.11 The effect of altering a software parameter.

Data set icon for GEparts.

Example 7.7 (page 394) describes an experiment to compare the measurements obtained from two software algorithms. In that example, we used the matched pairs t test on these data, despite some skewness, which makes the P-value only roughly correct. The sign test is based on the following simple observation: of the 51 parts measured, 29 had a larger measurement with the option off, and 22 had a larger measurement with the option on.

To perform a significance test based on these counts, let p be the probability that a randomly chosen part would have a larger measurement with the option turned off. The null hypothesis of “no effect” says that the two measurements on each part are just repeat measurements. Thus, measurement with the option off is equally likely to be larger or smaller than the measurement with the option on. Therefore, we want to test

H0:  p=1/2Ha:  p≠1/2

The 51 parts are independent trials, so the number that had larger measurements with the option off has the binomial distribution B(51, 1/2) if H0 is true. The P-value for the observed count 29 is, therefore, 2P(X≥29), where X has the B(51, 1/2) distribution. You can compute this probability with software or the Normal approximation to the binomial:

2P(X≥29)=2P(Z≥29−51(0.5)51(0.5)(1−0.5))=2P(Z≥29−25.512.75)=2P(Z≥0.98)=2(0.1635)=0.3270

As in Example 7.7, there is not strong evidence that the two measurements are different.

There are several varieties of sign test, all based on counts and the binomial distribution. The sign test for matched pairs is the most useful. The null hypothesis of “no effect” is then always Ha:  p=1/2. The alternative can be one-sided in either direction or two-sided, depending on the type of change we are considering.

The matched pairs t test in Example 7.7 tested the hypothesis that the mean of the distribution of differences is 0. The sign test in Example 7.11 is, in fact, testing the hypothesis that the median of the differences is 0. If p is the probability that a difference is positive, then p=1/2 when the median is 0. This is true because the median of the distribution is the point with probability 1/2 lying to its right. As Figure 7.11 illustrates, p>1/2 when the median is greater than 0, again because the probability to the right of the median is always 1/2. The sign test of H0:  p=1/2 against Ha:  p>1/2 is a test of

H0:  population median=0Ha:  population median>0
A right-skewed distribution curve has 0 marked just right of the peak and the median marked just right of 0. The area under the curve to the right of 0 is shaded representing p equals the probability that the difference is positive.

Figure 7.11 Why the sign test tests the median difference: when the median is greater than 0, the probability p of a positive difference is greater than 1/2 and vice versa.

The sign test in Example 7.11 makes no use of the actual scores; it just counts how many parts had a larger measurement with the option off. Any parts that did not have different measurements would be ignored altogether. Because the sign test uses so little of the available information, it is much less powerful than the t test when the population is close to Normal. Chapter 15 describes other distribution-free procedures that are more powerful than the sign test.

Check-in

  1. 7.14 Sign test for the cold-plasma study. Check-in question 7.10 (page 397) gives data on the taste of fruit juice with and without CP treatment. Is there evidence that the median difference in taste is nonzero? State the hypotheses, carry out the sign test, and report your conclusion.

Example 7.12 A bootstrap confidence interval.

Data set icon for botdel.

Consider the eight delivery time measurements (in minutes) in Example 7.1:

13.7 26.3 20.0 45.3 8.5 43.6 10.1 17.3

We defended the use of the one-sided t confidence interval for an earlier analysis. Let’s now compare those results with the confidence interval constructed using the bootstrap.

This can be done using JMP by right-clicking on the numerical mean value in the summary statistics produced from the distribution platform and selecting bootstrap. We decide to collect the x¯’s from 1000 resamples of size n=8. Figure 7.12 shows the histogram of these 1000 x¯’s. One resample was

13.7 26.3 45.3 45.3 8.5 43.6 10.1 17.3

with x¯=26.2625. The middle 95% of the 1000 x¯’s runs from 14.4 to 33.8. We repeat the procedure and get the interval (14.5, 32.9). The two bootstrap intervals are relatively close to each other and are more narrow than the one-sample t confidence interval (11.1, 35.1).

A histogram of a sample mean shows a normal distribution between 10 and 40 on a horizontal axis, with class sizes of 2.5. The highest point is from 20 to less than 22.5, and the center is at approximately 22.5 to less than 25.

Figure 7.12 Histogram of x¯’s based on resamples of size n=8 using JMP. The middle 95% of these x¯’s represents a 95% bootstrap confidence interval.

Using the middle 95% of the resampled x¯’s is just one of many methods to form the 95% confidence interval from the resamples. While conceptually straightforward, the resulting intervals tend to be too narrow when the sample size is small. Other methods do a better job with coverage but require more computation. As a result, the bootstrap is practical only when you can use a computer to take and store a large number of samples quickly. More details about the bootstrap can be found in Chapter 16.

Section 7.1 SUMMARY

  • Significance tests and confidence intervals for the mean μ of a Normal population are based on the sample mean x¯ of an SRS. Because of the central limit theorem, the resulting procedures are approximately correct for other population distributions when the sample is large.

  • The standard error of x¯ is

    SEx¯=sn

  • The standardized sample mean, or one-sample z statistic,

    Z=x¯-μσ/n

    has the N(0, 1) distribution. If the standard deviation σ/n of x¯ is replaced by the standard error s/n, the one-sample t statistic

    t=x¯-μs/n

    has the t distribution with n−1 degrees of freedom.

  • There is a t distribution for every positive degrees of freedom k. All are symmetric distributions similar in shape to Normal distributions. The t(k) distribution approaches the N(0, 1) distribution as k increases.

  • The level C margin of error of x¯ is

    m=t*×SEx¯=t*sn

    where t* is the value for the t(n−1) density curve with area C between −t* and t*.

  • A level C confidence interval for μ of a Normal population is

    x¯±m

  • Significance tests for H0:   μ=μ0 are based on the one-sample t statistic. P-values and critical values are computed from the t(n−1) distribution.

  • A matched pairs analysis is needed when subjects or experimental units are matched in pairs or when there are two measurements on each individual or experimental unit and the question of interest concerns the difference between the two measurements.

  • The one-sample t procedures are used to analyze matched pairs data by first taking the differences within the matched pairs to produce a single sample of differences.

  • One-sample equivalence testing assesses whether a population mean μ is practically different from a hypothesized mean μ0. This test requires a threshold δ, which represents the largest difference between μ and μ0 such that the means are considered equivalent.

  • The t procedures are relatively robust against non-Normal populations. The t procedures are useful for non-Normal data when n≥15 unless the data show outliers or strong skewness.

  • The sign test is a distribution-free procedure because it uses probability calculations that are correct for a wide range of population distributions.

  • The sign test for “no treatment effect” in matched pairs counts the number of positive differences. The P-value is computed from the B(n, 1/2) distribution, where n is the number of nonzero differences. The sign test is less powerful than the t test in cases where use of the t test is justified.

Section 7.1 EXERCISES

  1. 7.1 What’s wrong? For each of the following statements, explain what is wrong and why.

    1. The degrees of freedom for the one-sample t statistic is n−2.

    2. The margin of error for the population mean μ is m=t*s.

    3. The P-value in a one-sample equivalence test quantifies the evidence provided by the data that μ=μ0.

    4. A matched pairs analysis is needed whenever you have a study with two treatment conditions.

  2. 7.2 What’s wrong? For each of the following statements, explain what is wrong and why.

    1. As the degrees of freedom k decrease, the t distribution density curve gets closer to the N(0, 1) curve.

    2. The standard error of the sample mean is s2/n.

    3. A researcher wants to test H0:   x¯=30 versus the one-sided alternative Ha:   x¯<30.

    4. The 95% margin of error for the mean μ of a Normal population with unknown σ is the same for all SRS of size n.

  3. 7.3 Finding the critical value t*. What critical value t* from Table D should be used to calculate the margin of error for a confidence interval for the mean of the population in each of the following situations?

    1. A 95% confidence interval based on n=18 observations.

    2. A 95% confidence interval from an SRS of 26 observations.

    3. A 90% confidence interval from a sample of size 26.

    4. These cases illustrate how the size of the margin of error depends upon the confidence level and the sample size. Summarize these relationships.

  4. 7.4 Distribution of the t statistic. Assume a sample size of n=20. Draw a picture of the distribution of the t statistic under the null hypothesis. Use Table D and your picture to illustrate the values of the test statistic that would lead to rejection of the null hypothesis at the 1% level for a two-sided alternative.

  5. 7.5 More on the distribution of the t statistic. Repeat the previous exercise for the two situations where the alternative is one-sided.

  6. 7.6 A one-sample t test. The one-sample t statistic for testing

    H0:  μ=8Ha:  μ>8

    from a sample of n=22 observations has the value t=2.24.

    1. What are the degrees of freedom for this statistic?

    2. Give the two critical values t* from Table D that bracket t.

    3. Between what two values does the P-value of the test fall?

    4. Is the value t=2.24 significant at the 5% level? Is it significant at the 1% level?

    5. If you have software available, find the exact P-value.

  7. 7.7 Another one-sample t test. The one-sample t statistic for testing

    H0:  μ=40Ha:  μ≠40

    from a sample of n=13 observations has the value t=2.78.

    1. What are the degrees of freedom for t?

    2. Locate the two critical values t* from Table D that bracket t.

    3. Between what two values does the P-value of the test fall?

    4. Is the value t=2.78 statistically significant at the 5% level? At the 1% level?

    5. If you have software available, find the exact P-value.

  8. 7.8 A final one-sample t test. The one-sample t statistic for testing

    H0:  μ=20Ha:  μ<20

    based on n=9 observations has the value t=−1.85.

    1. What are the degrees of freedom for this statistic?

    2. Between what two values does the P-value of the test fall?

    3. If you have software available, find the exact P-value.

  9. 7.9 Margin of error. For each of the following situations, compute the 95% margin of error for the population mean μ.

    1. An SRS of n=25 gives x¯=23 and SEx¯=0.5.

    2. An SRS of n=16 gives x¯=32 and s=4.

    3. An SRS of n=25 gives x¯=44 and s2=25.

  10. 7.10 Business bankruptcies in Canada. Business bankruptcies in Canada are monitored by the Office of the Superintendent of Bankruptcy Canada (OSB).10 Included in each report are the assets and liabilities the company declared at the time of the bankruptcy filing. A study is based on a random sample of 75 bankruptcy reports filed during the year. The average debt (liabilities minus assets) is $5.32 million, with a standard deviation of $5.87 million.

    1. Construct a 95% one-sample t confidence interval for the average debt of Canadian companies filing bankruptcy reports this year.

    2. For this sample of companies, debt is positive, and the sample standard deviation is larger than the sample mean. This suggests the debt distribution is right skewed. Provide a defense for using the t confidence interval in this case.

  11. 7.11 Fuel economy. Although the Environmental Protection Agency (EPA) establishes the tests to determine the fuel economy of new cars, it often does not perform them. Instead, the test protocols are given to the car companies, and the companies perform the tests themselves. To keep the industry honest, the EPA runs some audits each year. Recently, the EPA announced that Audi and Volkswagen must lower their fuel economy estimates for some models.11 Here are some city miles per gallon (mpg) values for one of the models the EPA investigated: Data set icon for mileage.

    18.0 15.7 15.8 18.0 18.5 19.8 20.2 20.4
    16.9 18.3 19.8 17.2 16.7 17.7 19.5 18.0

    Give a 95% confidence interval for μ, the mean city mpg for this model.

  12. 7.12 Testing the sticker information. Refer to the previous exercise. The vehicle sticker information for this model stated a city average of 19 mpg. Are these mpg values consistent with the vehicle sticker? Perform a significance test using the 0.05 significance level. Be sure to specify the hypotheses, the test statistic, the P-value, and your conclusion. Data set icon for mileage.

  13. 7.13 Uber driver earnings. On its blog back in 2014, Uber posted a scatterplot using a sample of several thousand drivers in New York City. The plot showed each driver’s average net earnings per hour versus the number of hours worked.12 Here is a sample of earnings (dollars) for 27 drivers working 40 hours a week. Data set icon for uberx.

    26.25 33.51 43.91 31.91 31.78 43.37 36.66 31.69 31.25
    46.86 35.44 40.30 30.93 37.80 42.44 43.80 49.64 36.79
    34.10 37.54 30.93 38.40 37.83 21.73 41.62 26.25 33.51

    1. Do you think it is appropriate to use the t methods of this section to compute a 95% confidence interval for the 2014 average earnings per hour of New York City Uber drivers working 40 hours a week? Generate a plot to support your answer.

    2. Report the 95% confidence interval for μ, the average earnings per hour of New York City Uber drivers working 40 hours a week, as an estimate and margin of error.

    3. Report the 95% confidence interval for the average annual earnings of New York City Uber drivers working 40 hours a week.

    4. According to Uber in 2014, the median wage of an Uber driver working at least 40 hours in New York City is $90,766. Can these data be used to assess this claim? Explain your answer. (Note: Due to an increase in driver supply, these earning rates have dropped substantially since then.)

  14. 7.14 Number of friends on Facebook. To mark Facebook’s 10th birthday, Pew Research surveyed people using Facebook to see what they like and dislike about the site. The survey found that among adult Facebook users, the average number of friends is 338. This distribution takes only integer values, so it is certainly not Normal. It is also highly skewed to the right, with a median of 200 friends.13 Consider the following SRS of n=30 Facebook users from your large university. Data set icon for facefr.

    107 246 289 177 155  101  80 461 336  78
    463 264 827 180 221 1065  79 691  70 921
    126 672 296  60  11  227  84 787  18  82

    1. Are these data also heavily skewed? Use graphical methods to examine the distribution. Write a short summary of your findings.

    2. Do you think it is appropriate to use the t methods of this section to compute a 95% confidence interval for the mean number of friends that Facebook users at your large university have? Explain why or why not.

    3. Compute the sample mean and standard deviation, the standard error of the mean, and the margin of error for 95% confidence.

    4. Report the 95% confidence interval for μ, the average number of friends for Facebook users at your large university.

  15. 7.15 Beer before wine, and you’ll feel fine? Some researchers recently performed a study to investigate the order of beer and wine on hangover intensity.14 The study was run with each participant undergoing both orders of beer and wine with at least a week between sessions. Each session involved drinking the first beverage type until a reaching a breath alcohol concentration (BrAC) of 0.05% and then the other beverage type until a BrAC of 0.11%. The following day, once their BrAC had returned to 0, the participant filled out the Acute Hangover Scale (AHS).

    1. Explain why this design is a version of a matched-pairs design.

    2. To compare the two orders, the researchers computed the difference in AHS scores (Session 1−Session 2) for each participant. For the 31 participants who did beer/wine first and then wine/beer, the difference in AHS scores was −0.32, with a standard deviation of 7.63. For the 31 participants who did wine/beer first and then beer/wine, the difference in AHS scores was 0.36, with a standard deviation of 6.82. Comparing the mean differences of these two groups is comparable to comparing the two orders. Perform this comparison assuming a common variance. Report the test statistic and its P-value. What do you conclude?

  16. 7.16 Using an app or the Web on a smartphone. The Nielsen Company reports that U.S. residents aged 18 to 34 years spend an average of 2.83 hours per day using an app or the Web on a smartphone.15 You wonder if this it true for students at your large university because you rarely see students not using their smartphones. You collect an SRS of n=75 students and obtain x¯=3.17 hours with s=1.36 hours.

    1. Report the 95% confidence interval for μ, the average number of hours per month that students at your university use the Internet on a computer.

    2. Use this interval to test whether the average time for students at your university is different from the average reported by Nielsen. Use the 5% significance level. Summarize your results.

  17. 7.17 Rudeness and its effect on onlookers. Many believe that an uncivil environment has a negative effect on people. A pair of researchers performed a series of experiments to test whether witnessing rudeness and disrespect affects task performance.16 In one study, 34 participants met in small groups and witnessed the group organizer being rude to a “participant” who showed up late for the group meeting. After the exchange, each participant performed an individual brainstorming task in which he or she was asked to produce as many uses for a brick as possible in five minutes. The mean number of uses was 7.88, with a standard deviation of 2.35.

    1. Suppose that prior research has shown that the average number of uses a person can produce in five minutes under normal conditions is 10. Given that the researchers hypothesize that witnessing this rudeness will decrease performance, state the appropriate null and alternative hypotheses.

    2. Carry out the significance test using a significance level of 0.05. Give the P-value and state your conclusion.

  18. 7.18 Fuel efficiency t test. Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled, and the computer was then reset.17 Here are the mpg values for a random sample of 20 of these records: Data set icon for mpg.

    41.5 50.7 36.6 37.3 34.2 45.0 48.0 43.2 47.7 42.2
    43.2 44.6 48.4 46.4 46.8 39.2 37.3 43.5 44.3 43.3

    1. Describe the distribution using graphical methods. Is it appropriate to analyze these data using methods based on Normal distributions? Explain why or why not.

    2. Find the mean, standard deviation, standard error, and margin of error for 95% confidence.

    3. Report the 95% confidence interval for μ, the mean miles per gallon for this vehicle based on these data.

  19. 7.19 Tree diameter confidence interval. A study of 584 longleaf pine trees in the Wade Tract in Thomas County, Georgia, is described in Example 6.1 (page 329). For each tree in the tract, the researchers measured the diameter at breast height (DBH). This is the diameter of the tree at a height of 4.5 feet, and the units are centimeters (cm). Only trees with DBH greater than 1.5 cm were sampled. Here are the diameters of a random sample of 40 of these trees: Data set icon for pines.

    10.5 13.3 26.0 18.3 52.2  9.2 26.1 17.6 40.5 31.8
    47.2 11.4  2.7 69.3 44.4 16.9 35.7  5.4 44.2  2.2
     4.3  7.8 38.1  2.2 11.4 51.5  4.9 39.7 32.6 51.8
    43.6  2.3 44.6 31.5 40.3 22.3 43.3 37.5 29.1 27.9

    1. Use a histogram or stemplot and a boxplot to examine the distribution of DBHs. Include a Normal quantile plot if you have the necessary software. Write a careful description of the distribution.

    2. Is it appropriate to use the methods of this section to find a 95% confidence interval for the mean DBH of all trees in the Wade Tract? Explain why or why not.

    3. Report the mean with the margin of error and the confidence interval. Write a short summary describing the meaning of the confidence interval.

    4. Do you think these results would apply to other similar trees in the same area? Give reasons for your answer.

  20. 7.20 Nutritional intake among Canadian high-performance male athletes. Recall Exercise 6.44 (page 366). For one part of the study, n=114 male athletes from eight Canadian sports centers were surveyed. Their average caloric intake was 3077.0 kilocalories per day (kcal/d), with a standard deviation of 987.0. The recommended amount is 3421.7. Is there evidence that Canadian high-performance male athletes are deficient in their caloric intake?

    1. State the appropriate H0 and Ha to test this.

    2. Carry out the test, give the P-value, and state your conclusion.

    3. Construct a 95% confidence interval for the average deficiency in caloric intake.

  21. 7.21 Corn seed prices. The U.S. Department of Agriculture (USDA) uses sample surveys to obtain important economic estimates. One USDA pilot study estimated from a sample of 30 farms the amount a farmer will pay per planted acre of corn. The mean price was reported as $674.96 with a standard error of $17.19. Give a 95% confidence interval for the amount a farmer will pay per planted acre of corn.18

  22. 7.22 Stress levels in parents of children with ADHD. In a study of parents who have children with attention-deficit/hyperactivity disorder (ADHD), parents were asked to rate their overall stress level using the Parental Stress Scale (PSS).19 This scale has 18 items that contain statements regarding both positive and negative aspects of parenthood. Respondents are asked to rate their agreement with each statement using a five-point scale (1=strongly disagree to 5=strongly agree). The scores are summed such that a higher score indicates greater stress. The mean rating for the 50 parents in the study was reported as 52.98, with a standard deviation of 10.34.

    1. Do you think that these data are approximately Normally distributed? Explain why or why not.

    2. Is it appropriate to use the methods of this section to compute a 90% confidence interval? Explain why or why not.

    3. Find the 90% margin of error and the corresponding confidence interval. Write a sentence explaining the interval and the meaning of the 90% confidence level.

    4. To recruit parents for the study, the researchers visited a psychiatric outpatient service in Rohtak, India, and selected 50 consecutive families who met the inclusion and exclusion criteria. To what extent do you think the results can be generalized to all parents with children who have ADHD in India or in other locations around the world?

  23. 7.23 Are the parents feeling extreme stress? Refer to the previous exercise. The researchers considered a score greater than 45 to represent extreme stress. Is there evidence that the average stress level for the parents in this study is above this level? Perform a test of significance using α=0.10 and summarize your results.

  24. 7.24 Potential insurance fraud? Insurance adjusters are concerned about the high estimates they are receiving from Jocko’s Garage. To see if the estimates are unreasonably high, each of 10 damaged cars was taken to Jocko’s and to another garage, and the estimates (in dollars) were recorded. Here are the results: Data set icon for jocko.

    Car 1 2 3 4 5
    Jocko’s 1410 1550 1250 1300 900
    Other 1250 1300 1250 1200 950
    Car 6 7 8 9 10
    Jocko’s 1520 1750 3600 2250 2840
    Other 1575 1600 3380 2125 2600

    1. For each car, subtract the estimate of the other garage from Jocko’s estimate. Find the mean and the standard deviation for this difference.

    2. Test the null hypothesis that there is no difference between the estimates of the two garages. Be sure to specify the null and alternative hypotheses, the test statistic with degrees of freedom, and the P-value. What do you conclude using the 0.05 significance level?

    3. Construct a 95% confidence interval for the difference in estimates.

    4. The insurance company is considering seeking repayment from 1000 claims filed with Jocko’s last year. Using your answer to part (c), what repayment would you recommend the insurance company seek? Explain your answer.

  25. 7.25 Fuel efficiency comparison t test. Refer to Exercise 7.18. In addition to the computer calculating miles per gallon, the driver also recorded the miles per gallon by dividing the miles driven by the number of gallons at fill-up. The driver wants to determine if these calculations are different. Data set icon for mpgdiff.

    Fill-up 1 2 3 4 5 6 7 8 9 10
    Computer 41.5 50.7 36.6 37.3 34.2 45.0 48.0 43.2 47.7 42.2
    Driver 36.5 44.2 37.2 35.6 30.5 40.5 40.0 41.0 42.8 39.2
    Fill-up 11 12 13 14 15 16 17 18 19 20
    Computer 43.2 44.6 48.4 46.4 46.8 39.2 37.3 43.5 44.3 43.3
    Driver 38.8 44.5 45.4 45.3 45.7 34.2 35.2 39.8 44.9 47.5

    1. State the appropriate H0 and Ha.

    2. Carry out the test using a significance level of 0.05. Give the P-value and then interpret the result.

  26. 7.26 Counts of picks in a one-pound bag. A guitar supply company must maintain strict oversight on the number of picks it packages for sale to customers. The company’s current advertisement specifies between 900 and 1000 picks in every bag. An SRS of 36 one-pound bags of picks was collected as part of a quality improvement effort within the company. The number of picks in each bag is shown in the following table. Data set icon for picks.

    924 925 967 909 959 937 970 936 952
    919 965 921 913 886 956 962 916 945
    957 912 961 950 923 935 969 916 952
    917 977 940 924 957 920 986 895 923

    1. Create (i) a histogram or stemplot, (ii) a boxplot, and (iii) a Normal quantile plot of these counts. Write a careful description of the distribution. Make sure to note any outliers and comment on the skewness and Normality of the data.

    2. Based on your observations in part (a), is it appropriate to analyze these data using the t procedures? Briefly explain your response.

    3. Find the mean, the standard deviation, and the standard error of the mean for this sample.

    4. Calculate the 90% confidence interval for the mean number of picks in a one-pound bag.

  27. 7.27 Significance test for the average number of picks. Refer to the previous exercise. Data set icon for picks.

    1. Do these data provide evidence that the average number of picks in a one-pound bag is greater than 925? Using a significance level of 5%, state your hypotheses, the P-value, and your conclusions.

    2. Do these data provide evidence that the average number of picks in a one-pound bag is greater than 935? Using a significance level of 5%, state your hypotheses, the P-value, and your conclusion.

    3. Explain the relationship between your conclusions in parts (a) and (b) and the 90% confidence interval calculated in the previous problem.

  28. 7.28 A customer satisfaction survey. Many organizations are doing surveys to determine the satisfaction of their customers. Attitudes toward various aspects of campus life were the subject of one such study conducted at Purdue University. Each item was rated on a 1 to 5 scale, with 5 being the highest rating. The average response of 1378 first-year students to “Feeling welcomed at Purdue” was 3.89, with a standard deviation of 1.19. Assuming that the respondents are an SRS, give a 95% confidence interval for the mean of all first-year students.

  29. 7.29 Comparing operators of a DXA machine. Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body bone mineral content (TBBMC). A highly skilled operator is required to take the measurements. Recently, a new DXA machine was purchased by a research lab, and two operators were trained to take the measurements. TBBMC for eight subjects was measured by both operators.20 The units are grams (g). A comparison of the means for the two operators provides a check on the training they received and allows us to determine if one of the operators is producing measurements that are consistently higher than the other. Here are the data: Data set icon for tbbmc.

    Subject
    Operator 1 2 3 4 5 6 7 8
    1 1.328 1.342 1.075 1.228 0.939 1.004 1.178 1.286
    2 1.323 1.322 1.073 1.233 0.934 1.019 1.184 1.304

    1. Take the difference between the TBBMC recorded for Operator 1 and the TBBMC for Operator 2. Describe the distribution of these differences. Is it appropriate to analyze these data using the t methods? Explain why or why not.

    2. Use a significance test to examine the null hypothesis that the two operators have the same mean. Be sure to give the test statistic with its degrees of freedom, the P-value, and your conclusion.

    3. The sample here is rather small, so we may not have much power to detect differences of interest. Use a 95% confidence interval to provide a range of differences that are compatible with these data.

    4. The eight subjects used for this comparison were not a random sample. In fact, they were friends of the researchers whose ages and weights were similar to those of the types of people who would be measured with this DXA machine. Comment on the appropriateness of this procedure for selecting a sample and discuss any consequences regarding the interpretation of the significance-testing and confidence interval results.

  30. 7.30 Equivalence of paper and computer-based questionnaires. Computers are commonly used to complete questionnaires because of the increased efficiency of data collection and reduction in coding errors. Studies, however, have shown that questionnaire format can influence responses, especially for items of a sensitive nature.21 Consider the data for the small study below, comparing paper and computer survey formats of a self-report measure of mental health. Each participant completed both forms on adjacent days with the order determined by a flip of a coin. Data set icon for equiv.

    Subject Paper Computer Diff Subject Paper Computer Diff
    1 5 2     3 11 6 5     1
    2 4 3     1 12 5 5     0
    3 4 4     0 13 3 7 −4
    4 7 8 −1 14 3 6 −3
    5 4 5 −1 15 4 4     0
    6 6 7 −1 16 2 3 −1
    7 4 3     1 17 7 10 −3
    8 6 8 −2 18 8 7     1
    9 6 5     1 19 4 6 −2
    10 2 3 −1 20 6 8 −2

    1. Explain to someone unfamiliar with statistics why this experiment is a matched pairs design.

    2. The measure involves 10 items and produces a whole number score ranging between 0 and 20. Do you think it is appropriate to use the t procedures on the difference in survey scores? Explain your answer.

    3. Perform an equivalence test at the 0.05 level using the limits ±0.5 and state your conclusion.

  31. 7.31 Sign test for potential insurance fraud. The differences in the repair estimates in Exercise 7.24 can also be analyzed using a sign test. Set up the appropriate null and alternative hypotheses, carry out the test, and summarize the results. How do these results compare with those that you obtained in Exercise 7.24? Data set icon for jocko.

  32. 7.32 Sign test for the comparison of operators. The differences in the TBBMC measures in Exercise 7.29 can also be analyzed using a sign test. Set up the appropriate null and alternative hypotheses, carry out the test, and summarize the results. How do these results compare with those that you obtained in Exercise 7.29? Data set icon for tbbmc.

  33. 7.33 Sign test for fuel efficiency comparison. Use the sign test to assess whether the computer calculates a higher mpg than the driver in Exercise 7.25. State the hypotheses, give the P-value using the binomial table (Table C), and report your conclusion. Data set icon for mpgdiff.

  34. 7.34 Insulation study. A manufacturer of electric motors tests insulation at a high temperature (250°C) and records the number of hours until the insulation fails.22 The data for five specimens are

    446 326 372 377 310

    The small sample size makes judgment from the data difficult, but engineering experience suggests that the logarithm of the failure time will have a Normal distribution. Take the logarithms of the five observations and use t procedures to give a 90% confidence interval for the mean of the log failure time for insulation of this type. Data set icon for insulat.