6.1 Estimating with Confidence

The SAT and ACT are two standardized tests used to assess readiness for college study. Until recently, one of these two tests was required by most colleges for admission. Even though many universities are now considering test-optional admissions, the tests are still used by states and universities for comparison and benchmarking purposes.

The SAT consists of two sections, one for mathematics (SATM) and one for critical reading and writing (SATW), and an optional essay. Possible scores on each section range from 200 to 800, for a total range of 400 to 1600. Since 1995, section scores have been recentered so that the mean is approximately 500 with a standard deviation of 100 in a large “standardized group.” This scale has been maintained so that scores have a constant interpretation over time.

Example 6.3 Estimating the mean SATM score for seniors in California.

Suppose that you want to estimate the mean SATM score for the 489,650 high school seniors in California.2 You know better than to trust data from the students who choose to take the SAT. Only about 56% of California students typically take the SAT.3 These self-selected students are planning to attend college and are not representative of all California seniors. At considerable effort and expense, you give the test to an SRS of 500 California high school seniors. The mean score for your sample is x¯=505. What can you say about the mean score μ in the population of all 489,650 seniors?

The sample mean x¯ is the natural estimator of the unknown population mean μ. We know that x¯ is an unbiased estimator of μ. More importantly, the law of large numbers says that the sample mean must approach the population mean as the size of the sample grows. The value x¯=505, therefore, appears to be a reasonable estimate of the mean score μ that all 489,650 students would achieve if they took the test.

But how reliable is this estimate? A second sample of 500 students would surely not give a sample mean of 505 again. Unbiasedness says only that there is no systematic tendency to underestimate or overestimate the truth. Could we plausibly get a sample mean of 493 and a sample mean of 517 in repeated samples? An estimate without an indication of its variability is of little value.

Statistical confidence

The unbiasedness of an estimator concerns the center of its sampling distribution, but questions about variation are answered by looking at its spread. The central limit theorem says that if the entire population of SATM scores has mean μ and standard deviation σ, then in repeated SRSs of size 500, the sample mean x¯ is approximately N(μ,σ/500). Let us suppose that we know that the standard deviation σ of SATM scores in our California population is σ=100. (We will see in the next chapter how to proceed when σ is not known. For now, we are more interested in statistical reasoning than in details of realistic methods.) This means that in repeated sampling the sample mean x¯ has an approximately Normal distribution centered at the unknown population mean μ and a standard deviation of

σx¯=100500=4.5

Now we are ready to proceed. Consider this line of thought, which is illustrated in Figure 6.2:

A normal distribution curve.

Figure 6.2 Distribution of the sample mean, Example 6.3.

We have simply restated a fact about the sampling distribution of x¯. The language of statistical inference uses this fact about what would happen in the long run to express our confidence in the results of any one sample. Our sample gave x¯=505. We say that we are 95% confident that the unknown mean score for all California seniors lies between

x¯9=5059=496

and

x¯+9=505+9=514

Be sure you understand the grounds for our confidence. There are only two possibilities for our SRS:

  1. The interval between 496 and 514 contains the true μ.
  2. The interval between 496 and 514 does not contain the true μ.

We cannot know whether our sample is one of the 95% for which the interval x¯±9 contains μ or one of the unlucky 5% for which it does not contain μ. The statement that we are 95% confident is shorthand for saying, “We arrived at these numbers by a method that gives correct results 95% of the time.”

Check-in
  1. 6.1 How much do you spend on lunch? The average amount you spend on a lunch during the week is not known. Based on past experience, you are willing to assume that the standard deviation is $1.75. If you take a random sample of 25 lunches, what is the value of the standard deviation of x¯?

  2. 6.2 Applying the 68–95–99.7 rule. In the setting of the previous Check-in question, the 68–95–99.7 rule says that the probability is about 0.95 that x¯ is within $______ of the population mean μ. Fill in the blank.

  3. 6.3 Constructing a 95% confidence interval. In the setting of the previous two Check-in questions, about 95% of all samples will capture the true mean in the interval x¯ plus or minus $______. Fill in the blank.

Confidence intervals

In the setting of Example 6.3, the interval of numbers between the values x¯±9 is called a 95% confidence interval for μ. Like most confidence intervals we will discuss, this one has the form

estimate ± margin of error

The estimate (x¯=505 in this case) is our guess for the value of the unknown parameter. The margin of error (9 here) reflects how accurate we believe our guess is, based on the variability of the estimate, and how confident we are that the procedure will produce an interval that will contain the true population mean μ.

Figure 6.3 illustrates the behavior of 95% confidence intervals in repeated sampling from a Normal distribution with mean μ. The center of each interval (marked by a dot) is at x¯ and varies from sample to sample. The sampling distribution of x¯ (also Normal) appears at the top of the figure to show the long-term pattern of this variation.

A normal distribution curve and a plot of 25 confidence intervals.

Figure 6.3 Twenty-five samples from the same population gave these 95% confidence intervals. In the long run, 95% of all samples give an interval that covers μ.

The 95% confidence intervals, x¯± margin of error, from 25 SRSs appear below the sampling distribution. The arrows on either side of the dot (x¯) span the confidence interval. All except 1 of the 25 intervals contain the true value of μ. In those intervals that contain μ, sometimes μ is near the middle of the interval and sometimes it is closer to one of the ends. This again reflects the variation of x¯. In practice, we don’t know the value of μ, but we have a method such that, in a very large number of samples, 95% of the confidence intervals will contain μ.

We can construct confidence intervals for many different parameters based on a variety of designs for data collection. We will learn the details of a number of these in later chapters. Two important things about a confidence interval are common to all settings:

  1. It is an interval of the form (a, b), where a and b are numbers computed from the sample data.
  2. It has a property called a confidence level that gives the probability of producing an interval that contains the unknown parameter.

Users can choose the confidence level, but 95% is the standard for most situations. Occasionally, 90% or 99% is used. We use C to stand for the confidence level in decimal form. For example, a 95% confidence level corresponds to C=0.95.

Applet With the Confidence Intervals applet, you can construct diagrams similar to the one displayed in Figure 6.3. The only differences are that the applet displays both the Normal population distribution and the Normal sampling distribution of x¯ at the top and that at most 20 confidence intervals are displayed, with the data that make up the last SRS displayed below its confidence interval. To use the applet, choose the confidence level C, the sample size n, and how many SRSs (between 1 and 100) to generate. The number (and percent) of intervals that contain μ is displayed.

The spread of the data for the latest SRS reflects the spread of the population distribution. This spread is assumed known, and it does not change with sample size. What does change, as you vary n, is the spread of the sampling distribution of x¯; this is expressed through the margin of error. As you increase n, you’ll find that the span of the confidence interval gets smaller. More on this property soon.

Check-in
  1. Applet 6.4 Generating a single confidence interval. Set the Confidence Intervals applet to a 95% confidence level and n=20 and use the applet to choose an SRS and display its confidence interval.

    1. Is the spread in the data, shown as dots below the confidence interval, larger than the span of the confidence interval? Explain why this would typically be the case.

    2. For the same data set, you can compare the span of the confidence interval for different values of C by sliding the confidence level to a new value. For the SRS you generated in part (a), what happens to the span of the interval when you move C to 99%? What about 90%? Describe the relationship you find between the confidence level C and the span of the confidence interval.

  2. Applet 6.5 80% confidence intervals. The idea of an 80% confidence interval is that the interval captures the true parameter value in 80% of all samples. That’s not high enough confidence for practical use, but 80% hits and 20% misses make it easy to see how a confidence interval behaves in repeated samples from the same population.

    1. Set the confidence level in the Confidence Intervals applet to 80% and n=20. Choose 20 SRSs and display their confidence intervals. How many of the 20 intervals contain the true mean μ? What proportion contain the true mean?

    2. We can’t determine whether a new SRS will result in an interval that contains μ or not. The confidence level only tells us what percent will contain μ in the long run. Choose another 20 SRSs and record the number of hits. What is the proportion of hits among the 40 SRSs? Keep sampling and record the proportion of hits among 100, 200, 300, 400, and 500 SRSs. As the number of samples increases, we expect the percent of captures to get closer to the confidence level, 80%. Do you find this pattern in your results?

Confidence interval for a population mean

Data set icon for Vtm.

We now construct a level C confidence interval for the mean μ of a population when the data are an SRS of size n. The construction is based on the sampling distribution of the sample mean x¯. This distribution is exactly N(μ,σ/n) when the population has the N(μ,σ) distribution. The central limit theorem says that this same sampling distribution is approximately correct for large samples. For now, we will assume we are in one of these two situations. We discuss what we mean by “large sample” after we briefly study these intervals.

Our construction of a 95% confidence interval for the mean SATM score began by noting that any Normal distribution has probability about 0.95 within ±2 standard deviations of its mean. To construct a level C confidence interval, we first catch the central C area under a Normal curve. That is, we must find the number z* such that any Normal distribution has probability C within ±z* standard deviations of its mean.

Because all Normal distributions have the same standardized form, we can obtain everything we need from the standard Normal curve. Figure 6.4 shows how C and z* are related. Values of z* for many choices of C appear in the row labeled z* at the bottom of Table D at the back of the book. Here are the most important entries from that row:

z* 1.645 1.960 2.576
C 90% 95% 99%

Notice that for 95% confidence, the value 2 obtained from the 68–95–99.7 rule is replaced with the more precise 1.96.

A normal distribution curve.

Figure 6.4 To construct a level C confidence interval, we must find the number z* such that the probability between z* and z* under the standard Normal curve is C.

As Figure 6.4 reminds us, any Normal curve has probability C between the point z* standard deviations below the mean and the point z* standard deviations above the mean. The sample mean x¯ has the Normal distribution with mean μ and standard deviation σ/n, so there is probability C that x¯ lies between

μz*σnandμ+z*σn

This is exactly the same as saying that the unknown population mean μ lies between

x¯z*σnandx¯+z*σn

That is, there is probability C that the interval x¯±z*σ/n contains μ. This is our confidence interval. The estimate of the unknown μ is x¯, and the margin of error is z*σ/n.

Since 2008, Sallie Mae, a major provider of education loans and savings programs, has conducted an annual study titled “How America Pays for College.” In the 2019 survey, 2000 randomly selected individuals (1000 parents of undergraduate students and 1000 undergraduate students) were surveyed online.4

Many of the survey questions focus on the composition of funding sources used to pay for college, so the undergraduates in the survey are often responding for their family. For example, each participant is asked to report how much of the parents’ current income is used to pay for college. Do you think it is wise to combine responses across the parents and undergraduates? Are you fully aware of how much money your parents are spending and borrowing for college? The authors consider this one population and report overall averages and percents in their report. We will also consider this a sample from one population, but this is certainly debatable.

Example 6.4 Average college savings fund contribution.

One Sallie Mae survey question asked how much money from a college savings fund, such as a 529 plan, is used to pay for college. Of the 2000 who were surveyed, the average amount is $1577. This population distribution is highly skewed to the right, with many zeros and likely a few very large amounts. Nevertheless, because the sample size is quite large, we can rely on the central limit theorem to assure us that the confidence interval based on the Normal distribution will be a good approximation.

Let’s compute an approximate 95% confidence interval for the true mean amount contributed from a college savings fund among all undergraduates. We’ll assume that the standard deviation for the population of college savings fund contributions is $2076. For 95% confidence, we see from Table D that z*=1.960. The margin of error for the 95% confidence interval for μ is, therefore,

m=z*σn=1.96020762000=90.98

We have computed the margin of error with more digits than we really need. Our mean is rounded to the nearest $1, so we will do the same for the margin of error. Keeping additional digits would provide no additional useful information. Therefore, we will use m=91. The approximate 95% confidence interval is

x¯±m=1577±91=(1486,1668)

We are 95% confident that the mean amount contributed from a college savings fund among all undergraduates is between $1486 and $1668.

Suppose that the researchers who designed this study had used a different sample size. How would this affect the confidence interval? We can answer this question by changing the sample size in our calculations and assuming that the sample mean is the same.

Example 6.5 How sample size affects the confidence interval.

As in Example 6.4, the sample mean of the college savings fund contribution is $1577, and the population standard deviation is $2076. Suppose that the sample size is only 500, but it is still large enough for us to rely on the central limit theorem. In this case, the margin of error for 95% confidence is

m=z*σn=1.9602076500=181.97

and the approximate 95% confidence interval is

x¯±m=1577±182=(1395,1759)

Notice that the margin of error for this example is twice as large as the margin of error that we computed in Example 6.4. The only change that we made was to assume that the sample size is 500 rather than 2000. This sample size is one-fourth of the original 2000. Thus, we double the margin of error when we reduce the sample size to one-fourth of the original value. Figure 6.5 illustrates the effect in terms of the intervals.

A plot of two confidence intervals.

Figure 6.5 Confidence intervals for n=500 and n=2000, Examples 6.4 and 6.5. A sample size four times as large results in a confidence interval that is one-half as wide.

Check-in
  1. 6.6 Average amount paid for college. Refer to Example 6.4. The average annual amount the n=2000 families paid for college was $26,226.5 If the population standard deviation is $8721, give the 95% confidence interval for μ, the average annual amount a family pays for a college undergraduate.

The argument leading to the form of confidence intervals for the population mean μ rested on the fact that the sampling distribution of the statistic x¯ used to estimate μ is Normal. Because many sample estimates have Normal (at least approximately), it is useful to notice that the confidence interval has the form

estimate±z*σestimate

The estimate based on the sample is the center of the confidence interval. The margin of error is z*σestimate. The desired confidence level determines z* from Table D. The standard deviation of the estimate is found from knowledge of the sampling distribution in a particular case. When the estimate is x¯ from an SRS, the standard deviation of the estimate is σestimate=σ/n. We return to this general form numerous times in the following chapters.

How confidence intervals behave

The margin of error z*σ/n for the mean of a Normal population illustrates several important properties that are shared by all confidence intervals in common use. The user chooses the sample size and confidence level, and the margin of error follows from these choices.

Both high confidence and a small margin of error are desirable characteristics of a confidence interval. High confidence says that our method almost always gives correct answers. A small margin of error says that we have pinned down the parameter quite precisely.

Suppose that in planning a study, you calculate the margin of error and decide that it is too large. Here are your choices to reduce it:

For most problems, you would choose a confidence level of 90%, 95%, or 99%, so z* will be 1.645, 1.960, or 2.576, respectively. Figure 6.4 (page 335) shows that z* will be smaller for lower confidence (smaller C). The bottom row of Table D also shows this. If n and σ are unchanged, a smaller z* leads to a smaller margin of error.

Example 6.6 How the confidence level affects the confidence interval.

Suppose that for the college saving fund contribution data in Example 6.4 (page 336), we wanted 99% confidence. Table D tells us that for 99% confidence, z*=2.576. The margin of error for 99% confidence based on 2000 observations is

m=z*σn=2.57620762000=119.58

and the 99% confidence interval is

x¯±m=1577±120=(1457,1697)

Requiring 99%, rather than 95%, confidence has increased the margin of error from 91 to 120. Figure 6.6 compares the two intervals.

A plot of two confidence intervals.

Figure 6.6 Confidence intervals, Examples 6.4 and 6.6. The larger the value of C, the wider the interval.

Similarly, choosing a larger sample size n reduces the margin of error for any fixed confidence level. The square root in the formula implies that we must multiply the number of observations by 4 in order to cut the margin of error in half. Likewise, if we want to reduce the standard deviation of x¯ by a factor of 4, we must take a sample 16 times as large.

The standard deviation σ measures the variation in the population. You can think of the variation among individuals in the population as noise that obscures the average value μ. It is harder to pin down the mean μ of a highly variable population; that is why the margin of error of a confidence interval increases with σ. In practice, we can sometimes reduce σ by carefully controlling the measurement process. We also might change the mean of interest by restricting our attention to only part of a large population. Focusing on a subpopulation (e.g., women in rural counties) will typically result in a smaller σ. The trade-off, however, is less generalizable results.

Check-in
  1. 6.7 Changing the sample size. In the setting of Check-in question 6.6 (page 338), would the margin of error for 95% confidence be doubled or halved if the sample size were raised to n=8000? Verify your answer by performing the calculations.

  2. 6.8 Changing the confidence level. In the setting of Check-in question 6.6 (page 338), would the margin of error for 90% confidence be larger or smaller? Verify your answer by performing the calculations.

Choosing the sample size

A wise user of statistics never plans data collection without, at the same time, planning the inference. You can arrange to have both high confidence and a small margin of error through the margin of error formula for a population mean

m=z*σn

By rearranging this equation, we can find the sample size n for a desired margin of error m. Here is the result.

Recall that the value of σ is known, and the value z* comes from the choice of confidence level C. Notice also that the size of the population (as long as the population is much larger than the sample) does not influence the sample size we need.

A drawback of this formula is that it does not account for collection costs. In practice, taking observations costs time and money. The required sample size may be impossibly expensive. If you face such a situation, you might consider a larger margin of error and/or a lower confidence level to find a workable sample size.

caution This formula also does not adjust for the fact that the actual number of usable observations is often less than what is planned at the beginning of a study. This is particularly true of data collected in surveys but is an important consideration in most studies. Careful study designers often assume a nonresponse rate or dropout rate that specifies what proportion of the originally planned sample will fail to provide data. This information is used to adjust the sample size to be used at the start of the study.

Example 6.7 How many undergraduates should we survey?

Suppose that we are planning a survey similar to the one described in Example 6.4 (page 336). If we want the margin of error for the average amount contributed from a college savings plan to be $50 with 95% confidence, what sample size n do we need?

For 95% confidence, Table D gives z*=1.960. For σ we will use the value from the previous study, $2076. If the margin of error is $50, we have

n=(z*σm)2=(1.96×207650)2=6622.57

Because n=6622 will give a slightly wider interval than desired and n=6623 a slightly narrower interval, we should choose n=6623. It is always safe to round up to the next higher whole number when finding n because doing so will give us a smaller margin of error. We need information from 6623 undergraduates to determine an estimate of the mean college savings fund contribution with the desired margin of error.

If past studies suggested that 70% of those contacted will respond, we would need to start with a sample size of 6623/0.7=9642 undergraduates to expect usable information from 6623 of them. Accounting for nonresponse often results in a much larger initial sample size.

The purpose of these calculations is to determine a sample size that is sufficient to provide useful results, but the determination of what is useful is a matter of judgment that requires subject-matter knowledge. In Section 7.3 (page 433), we revisit this issue of sample size determination under the more realistic setting when σ is unknown. Although the calculations are more complex, the design principles are the same. Furthermore, software now does most of these calculations for us.

Check-in
  1. 6.9 Starting salaries. You are planning a survey of starting salaries for recent computer science majors. In the latest survey by the National Association of Colleges and Employers, the average starting salary was reported to be $68,668.6 If you assume that the standard deviation is $6650, what sample size do you need in order to have a margin of error equal to $500 with 95% confidence?

  2. 6.10 Changes in sample size. Suppose that in the setting of the previous Check-in question you have the resources to contact 1000 recent graduates. If all respond, will your margin of error be larger or smaller than $500? What if only 60% respond? Verify your answers by computing the margin of errors.

Some cautions

We have already seen that small margins of error and high confidence can require large numbers of observations. caution You should also be keenly aware that any formula for inference is correct only in specific circumstances. If the government required statistical procedures to carry warning labels like those on drugs, most inference methods would have long labels. Our formula x¯±z*σ/n for estimating a population mean comes with the following list of warnings for the user:

The most important caution concerning confidence intervals is a consequence of the first of these warnings. The margin of error in a confidence interval covers only random sampling errors. The margin of error is obtained from the sampling distribution and indicates how much error can be expected because of chance variation in randomized data production.

caution Practical difficulties such as undercoverage and nonresponse in a sample survey cause additional errors. These errors can be larger than the random sampling error. This often happens when the sample size is large (so that σ/n is small). Remember this unpleasant fact when reading the results of an opinion poll or other sample survey. The practical conduct of the survey influences the trustworthiness of its results in ways that are not included in the announced margin of error.

Every inference procedure that we will meet has its own list of warnings. Because many of the warnings are similar to those we have mentioned, we will not print the full warning label each time. It is easy to state (from the mathematics of probability) conditions under which a method of inference is exactly correct. These conditions are never fully met in practice.

For example, no population is exactly Normal. Deciding when a statistical procedure should be used in practice often requires judgment assisted by exploratory analysis of the data. Mathematical facts are, therefore, only a part of statistics. The difference between statistics and mathematics can be stated thusly: mathematical theorems are true; statistical methods are often effective when used with skill.

Finally, you should understand what statistical confidence does not say. Based on our SRS in Example 6.3, we are 95% confident that the mean SATM score for the California students lies between 496 and 514. This says that this interval was calculated by a method that gives correct results in 95% of all possible samples. It does not say that the probability is 0.95 that the true mean falls between 496 and 514. No randomness remains after we draw a particular sample and compute the interval. The true mean either is or is not between 496 and 514. The probability calculations of standard statistical inference describe how often the method, not a particular sample, gives correct answers.

Check-in
  1. 6.11 Nonresponse in a survey. In earlier versions of the Sallie Mae survey of Example 6.4 (page 336), participants were asked to report the undergraduate’s outstanding credit card balance. Only about a third reported this amount. Provide a couple of reasons why a survey respondent might not provide an amount. Based on these reasons, do you think the sample mean using just the reported amounts is biased? Is the margin of error based just on the reported amounts a good measure of precision? Explain your answers.

Section 6.1 SUMMARY

  • The purpose of a confidence interval is to estimate an unknown parameter with an indication of how accurate the estimate is and of how confident we are that the result is correct.

  • Any confidence interval has two parts: an interval computed from the data and a confidence level. The interval often has the form

    estimate ± margin of error
  • The confidence level states the probability that the method will give a correct answer. That is, if you use 95% confidence intervals, in the long run 95% of your intervals will contain the true parameter value. When you apply the method once (that is, to a single sample), you do not know if your interval gave a correct answer (which happens 95% of the time) or not (which happens 5% of the time).

  • Other things being equal, the margin of error of a confidence interval decreases as

    • – The confidence level C decreases.
    • – The sample size n increases.
    • – The population standard deviation σ decreases.
  • The level C margin of error of x¯, based on an SRS of size n from a Normal population with known standard deviation σ, is

    m=z*σn

    Here z* is obtained from the row labeled z* at the bottom of Table D. The probability is C that a standard Normal random variable takes a value between z* and z*.

  • The level C confidence interval for μ, the mean of a population with known standard deviation σ, is

    x¯±m

    If the population is not Normal and n is large, the confidence level of this interval is approximately correct.

  • The sample size n required to obtain a confidence interval of specified margin of error m for a population mean is

    n=(z*σm)2

    where z* is the critical point for the desired level of confidence.

  • A specific confidence interval formula is correct only under specific conditions. The most important conditions concern the method used to produce the data. Other factors such as the form of the population distribution may also be important. These conditions should be investigated prior to doing any calculations.

Section 6.1 EXERCISES

  1. 6.1 Student loan debt. The average student loan debt among college graduates is reported using the 95% confidence interval ($23,923, $34,447).

    1. Describe what this interval tells us about average student loan debt.

    2. What is the estimated average student loan debt among college graduates?

    3. What is the margin of error?

  2. 6.2 Margin of error and the confidence interval. A study of stress on the campus of your university using the College Undergraduate Stress Scale (CUSS) reported an average stress level of 1123 (a higher score indicating more stress) with a margin of error of 43 for 95% confidence. The study was based on a random sample of 400 undergraduates.

    1. Give the 95% confidence interval.

    2. If you wanted 99% confidence for the same study, would your margin of error be greater than, equal to, or less than 43? Explain your answer.

  3. 6.3 Changing the sample size. Consider the setting of the previous exercise. Suppose that the sample mean is again 1123, and the population standard deviation is 441. Make a diagram similar to Figure 6.5 (page 338) that illustrates the effect of sample size on the width of a 95% interval. Use the following sample sizes: 100, 144, 225, 324, and 400. Summarize what the diagram shows.

  4. 6.4 Changing the confidence level. Consider the setting of the previous two exercises. Suppose that the sample mean is still 1123, the sample size is 400, and the population standard deviation is 441. Make a diagram similar to Figure 6.6 (page 339) that illustrates the effect of the confidence level on the width of the interval. Use 80%, 90%, 95%, and 99%. Summarize what the diagram shows.

  5. 6.5 Confidence interval mistakes and misunderstandings. Suppose that 500 randomly selected alumni of the University of Okoboji were asked to rate the university’s academic advising services on a scale of 1–10. The sample mean x¯ was found to be 8.6. Assume that the population standard deviation is known to be σ=2.2.

    1. Ima Bitlost computes the 95% confidence interval for the average satisfaction score as 8.6±1.96(2.2). What is her mistake?

    2. After correcting her mistake in part (a), Ima states, “I am 95% confident that the sample mean falls between 8.4 and 8.8.” What is wrong with this statement?

    3. Ima quickly realizes her mistake in part (b) and instead states, “The probability that the true mean is between 8.4 and 8.8 is 0.95.” What misinterpretation is she making now?

    4. Finally, in her defense for using the Normal distribution to determine the confidence interval, Ima says, “Because the sample size is quite large, the population of alumni ratings will be approximately Normal.” Explain to Ima her misunderstanding and correct this statement.

  6. 6.6 More confidence interval mistakes and misunderstandings. Suppose that 100 randomly selected subscribers of Stingray Karaoke on YouTube asked how much time they typically spend on the site during the week.7 The sample mean x¯ is found to be 2.4 hours. Assume that the population standard deviation is known to be σ=1.9.

    1. Cary Oakey computes the 95% confidence interval for the average time on the site as 2.4±1.96(1.9/100). What is his mistake?

    2. Cary corrects this mistake and then states that “95% of the members spend between 2.03 and 2.77 hours a week on the site.” What is wrong with his interpretation of this interval?

    3. To reduce the margin of error in half, Cary says that the sample size needs to be doubled to 200. What is wrong with this statement?

  7. 6.7 The state of stress in the United States. Since 2007, the American Psychological Association has supported an annual nationwide survey to examine stress across the United States.8 This year, a total of 3602 adults were asked to indicate their average stress level (on a 10-point scale) during the past month.

    1. For the 343 Gen Z adults (18- to 22-year-olds), the mean score was 5.5. Assuming that the population standard deviation is 2.9, give the margin of error and find the 95% confidence interval for the mean Gen Z stress level.

    2. In addition, all survey respondents were asked to state what they regarded to be a “healthy” stress level. The mean score was 3.8. Assuming a population standard deviation of 2.2, give the margin of error and find the 95% confidence interval for the mean.

    3. Compare the two intervals. Do you think we can conclude that Gen Z adults have an average stress level that is above what is considered healthy? Explain your answer.

  8. 6.8 Inference based on integer values. Refer to Exercise 6.7. The data for this study are integer values between 1 and 10. Explain why the confidence interval for the mean μ based on the Normal distribution should be a good approximation.

  9. 6.9 Mean TRAP in young women. For many important processes that occur in the body, direct measurement of characteristics of the process is not possible. In many cases, however, we can measure a biomarker, a biochemical substance that is relatively easy to measure and is associated with the process of interest. Bone turnover is the net effect of two processes: the breaking down of old bone, called resorption, and the building of new bone, called formation. One biochemical measure of bone resorption is tartrate-resistant acid phosphatase (TRAP), which can be measured in blood. In a study of bone turnover in young women, serum TRAP was measured in 31 subjects.9 The mean was 13.2 units per liter (U/l). Assume that the standard deviation is known to be 6.5 U/l. Give the margin of error and find a 95% confidence interval for the mean TRAP amount in young women represented by this sample.

  10. 6.10 Mean OC in young women. Refer to the previous exercise. A biomarker for bone formation measured in the same study was osteocalcin (OC), measured in the blood. For the 31 subjects in the study, the mean was 33.4 nanograms per milliliter (ng/ml). Assume that the standard deviation is known to be 19.6 ng/ml. Report the 95% confidence interval.

  11. 6.11 Populations sampled and margins of error. Consider the following two scenarios. (A) Take a simple random sample (SRS) of 250 first-year students at your college or university. (B) Take an SRS of 250 students at your college or university. For each of these samples, you will record the amount spent on textbooks used for classes during the fall semester. Which sample should have the smaller margin of error? Explain your answer.

  12. NAEP 6.12 Average starting salary. The National Association of Colleges and Employers (NACE) Summer Salary Survey shows that the current class of college graduates received an average starting-salary offer of $50,994.10 Your institution collected an SRS(n=200) of its recent graduates and obtained a 95% confidence interval of ($51,756, $53,420). What can we conclude about the difference between the average starting salary of recent graduates at your institution and the overall NACE average? Write a short summary.

  13. 6.13 Consumption of sweet snacks. A study reported that the U.S. per capita consumption of sweet snacks among healthy-weight children aged 12 to 19 years is 251.2 kilocalories per day (kcal/d).11 This was based on 24-hour dietary recall records of n=2265 adolescents.

    1. Suppose that the population distribution is heavily skewed, with a standard deviation equal to 540 kcal/d. What is the margin of error for a 95% confidence interval of the per capita consumption of sweet snacks?

    2. A future study is being planned, and the goal is to have the margin of error no more than 15 kcal/d. Based on your answer to part (a), will this study require an examination of more or fewer 24-hour dietary recall records? Explain your answer without calculations.

    3. Based on your answer in part (b), will more or fewer recall records be required if the confidence level is increased to 99%? Again explain your answer without calculations.

    4. Compute the sample sizes necessary for the studies described in parts (b) and (c).

  14. 6.14 Total sleep time of college students. In Example 5.4 (page 282), the total sleep time per night among college students was approximately Normally distributed with mean μ=7.13 hours and standard deviation σ=1.67 hours. Consider an SRS of size n=175 from this population.

    1. What is the standard deviation of the sample mean in hours? In minutes?

    2. Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean.

    3. What is the probability that your sample average will be below 7.0 hours?

  15. 6.15 Determining sample size. Refer to the previous exercise. You really want to use a sample size such that about 95% of the averages fall within ±5 minutes of the population mean μ.

    1. Based on your answer to part (b) in Exercise 6.14, should the sample size be larger or smaller than 175? Explain your answer.

    2. What standard deviation of x¯ do you need such that 95% of all samples will have a mean within 5 minutes of μ?

    3. Using the standard deviation you calculated in part (b), determine the number of students you need to sample in order to achieve this.

  16. NAEP 6.16 Inference based on skewed data. The mean OC for the 31 subjects in Exercise 6.10 was 33.4 ng/ml. In our calculations, we assumed that the standard deviation was known to be 19.6 ng/ml. Use the 68–95–99.7 rule from Chapter 1 (page 51) to find the approximate bounds on the values of OC that would include these percents of the population. If the assumed standard deviation is correct, this distribution may be highly skewed. Why? (Hint: The measured values for a variable such as this are all positive.) Do you think that this skewness will invalidate the use of the Normal confidence interval in this case? Explain your answer.

  17. 6.17 Average hours per week listening to audio. An iHeartMedia-sponsored survey of 6016 consumers who listen at least once a week to an audio platform reported an average of 17.2 hours a week listening to audio, such as broadcast radio, streaming music services, and podcasts.12 Assume that the standard deviation is 6.8 hours.

    1. Give a 95% confidence interval for the mean time spent per week listening to audio.

    2. Is it true that 95% of the 6016 consumers reported weekly times that lie in the interval you found in part (a)? Explain your answer.

    3. It is likely this population distribution is very skewed with many small listening times and several very large times. Explain why the confidence interval based on the Normal distribution should nevertheless be a good approximation in this case.

  18. 6.18 Average minutes per week listening to audio. Refer to the previous exercise.

    1. Give the sample mean and sample standard deviation in minutes.

    2. Calculate the 95% confidence interval in minutes from your answers to part (a).

    3. Explain how you could have directly calculated this interval from the 95% interval that you calculated in the previous exercise.

  19. 6.19 Outlook on life. Since 2008, the Gallup-Sharecare Well-Being Index tracks how people feel about their daily lives. In 2017, 56.3% of the respondents were classified as “thriving.” This classification is based on how a respondent rates his or her current and future lives. This is the highest percent of respondents in this category since the index started. Material provided with the results noted:

    Results are based on telephone interviews. . . with a random sample of 160,498 adults, living in all 50 U.S. states and the District of Columbia. For results based on the total sample of national adults, the margin of sampling error is ±1 percentage points at the 95% confidence level.13

    The poll uses a complex multistage sample design, but the sample percent has approximately a Normal sampling distribution.

    1. The announced poll result was 56.3%±1%. Can we be certain that the true population percent falls in this interval? Explain your answer.

    2. Explain to someone who knows no statistics what the announced result 56.3%±1% means.

    3. This confidence interval has the same form we have met earlier:

      estimate±z*σestimate

      What is the standard deviation σestimate of the estimated percent?

    4. Does the announced margin of error include errors due to practical problems such as nonresponse? Explain your answer.

  20. Applet 6.20 How many “hits”? The Confidence Intervals applet lets you simulate large numbers of confidence intervals quickly. Select 95% confidence and then sample 50 intervals. Record the number of intervals that cover the true value. Repeat this process until you have 30 counts of hits. Make a stemplot or histogram of the results and find the mean. Describe the results. If you repeated this experiment very many times, what would you expect the average number of hits to be?

  21. 6.21 Required sample size for specified margin of error. A new bone study is being planned that will measure the biomarker TRAP described in Exercise 6.9. Using the value of σ given there, 6.5 U/l, find the sample size required to provide an estimate of the mean TRAP with a margin of error of 1.5 U/l for 95% confidence.

  22. NAEP 6.22 Adjusting required sample size for dropouts. Refer to the previous exercise. In similar previous studies, about 20% of the subjects drop out before the study is completed. Adjust your sample size requirement so that you will have enough subjects at the end of the study to meet the margin of error criterion.

  23. 6.23 Radio poll. A National Public Radio (NPR) station invites listeners to enter a dispute about a proposed “pay as you throw” waste collection program. The station asks listeners to call in and state how much each 10-gallon bag of trash should cost. A total of 179 listeners call in. The station calculates the 95% confidence interval for the average fee to be $0.53 to $1.39. Is this result trustworthy? Explain your answer.

  24. 6.24 Accuracy of a laboratory scale. To assess the accuracy of a laboratory scale, a standard weight known to weigh 10 grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean. (This mean is 10 grams if the scale has no bias.) The standard deviation of the scale readings is known to be 0.0002 gram.

    1. The weight is measured six times. The mean result is 10.0023 grams. Give a 99% confidence interval for the mean of repeated measurements of the weight.

    2. Based on the interval in part (a), do you think the scale is accurate? Explain your answer.

    3. How many measurements must be averaged to get a margin of error of ±0.0001 with 99% confidence?

  25. NAEP 6.25 More than one confidence interval. As we prepare to take a sample and compute a 95% confidence interval, we know that the probability that the interval we compute will cover the parameter is 0.95. That’s the meaning of 95% confidence. If we plan to use several such intervals, however, our confidence that all of them will give correct results is less than 95%. Suppose that we plan to take independent samples each month for five months and report a 95% confidence interval for each set of data.

    1. What is the probability that all five intervals will cover the true means? This probability (expressed as a percent) is our overall confidence level for the five simultaneous statements.

    2. Suppose we instead considered individual 99% confidence intervals. Now, what is the overall confidence level for the five simultaneous statements?

    3. Based on the results of parts (a) and (b), how could you keep the overall confidence level near 95% if you were considering 10 simultaneous intervals?