7.2 A Single Population Mean using the Student t Distribution
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
This is the confidence interval you will actually build on real data, where the population standard deviation is unknown and the sample's own s stands in for it. You pick the t-score from the degrees of freedom, compute the error bound, and read the interval as a claim about the population mean.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
Knowing when s replaces sigma is the technique-selection call this section drills, and invT plus TInterval on the calculator do the arithmetic. You still have to say what the interval means in a plain sentence about the population, which is the communicating half.
Learning Objectives
By the end of this section, you will be able to:
- explain why replacing the unknown \(\sigma\) with the sample standard deviation \(s\) forces a different distribution than the normal;
- state the properties of the Student's t-distribution and describe how the degrees of freedom change its shape;
- calculate the error bound and construct a confidence interval for a population mean when \(\sigma\) is unknown;
- interpret a t-based confidence interval in a plain sentence about the population.
In practice, we almost never know the population standard deviation. Section 7.1 asked you to assume \(\sigma\) was handed to you, and that assumption did a lot of quiet work: it is what let us say the sample mean followed a normal distribution and reach for a z-score. Real data does not come with \(\sigma\) attached. All you have is the spread of the sample in front of you, \(s\).
For a long time statisticians just swapped \(s\) in for \(\sigma\) and carried on as before. With a big sample that works well enough — a large sample gives a good estimate of the population spread, so the answer lands close. With a small sample it does not. A small sample can easily produce an \(s\) that is well off the true \(\sigma\), and treating that shaky estimate as if it were the real thing produces confidence intervals that are too narrow. The interval claims 95% confidence and delivers less.
William S. Gosset (1876–1937) hit this problem head-on. He worked at the Guinness brewery in Dublin, Ireland, testing hops and barley, and his experiments produced very few samples — a handful of batches, not thousands. When he replaced \(\sigma\) with \(s\) and computed a confidence interval the way Section 7.1 does, the results did not hold up. He worked out that the real distribution of the standardized sample mean depends on how big the sample is, and he derived that distribution. Guinness would not let its employees publish under their own names, so he published under the pen name "Student." That is why we call it the Student's t-distribution today, and why nobody named Student was ever involved.
Up until the mid-1970s some statisticians used the normal approximation for large samples and reserved the t-distribution for samples of at most 30. With graphing calculators and computers doing the arithmetic, that shortcut has no reason to exist. The practice now is simple: whenever \(s\) is standing in for \(\sigma\), use the Student's t-distribution, whatever the sample size.
7.2.1 Properties of the Student's t-Distribution
If you draw a simple random sample of size \(n\) from a population that has an approximately normal distribution with mean \(\mu\) and unknown population standard deviation \(\sigma\), and calculate the t-score
$$ t = \frac{\overline{x} - \mu}{\left(\frac{s}{\sqrt{n}}\right)}, $$then the t-scores follow a Student's t-distribution with \(n - 1\) degrees of freedom.
Definition 7.2.1 - The Student's t-curve is shorter in the middle and fatter in the tails than the standard normal, and it climbs toward the normal as the degrees of freedom grow.
The degrees of freedom, written \(df\), is the number \(n - 1\). It comes from the calculation of the sample standard deviation \(s\), which requires the \(n\) deviations \(x - \overline{x}\). Because those deviations must sum to zero, the last one is determined once the other \(n - 1\) are known. Only \(n - 1\) of them can vary freely, and that count is the degrees of freedom.
The normal distribution is one fixed shape. The t-distribution is a whole family — one curve for \(n = 5\), a different one for \(n = 30\). The degrees of freedom pick which member of the family you are standing in.
The t-score reads exactly like the z-score you already know. It measures how far the sample mean \(\overline{x}\) sits from the population mean \(\mu\), in units of the standard error \(\frac{s}{\sqrt{n}}\). The one difference is what happens underneath: the denominator is now an estimate rather than a known constant, and that extra uncertainty has to show up somewhere. It shows up as a distribution with fatter tails.
That family behaves in a few predictable ways, and knowing them is usually enough to sanity-check an answer before you trust it.
- The graph of the Student's t-distribution looks a lot like the standard normal curve — one hump, tails running off both sides.
- Its mean is zero and it is symmetric about zero, exactly like the standard normal.
- It has more probability out in its tails than the standard normal does, because the t-distribution is more spread out. So its graph is thicker in the tails and shorter in the center than the standard normal curve.
- Its exact shape depends on the degrees of freedom. As \(df\) increases, the t-curve gets closer and closer to the standard normal curve. With a few hundred degrees of freedom the two are nearly indistinguishable.
- The population of individual observations is assumed to be normally distributed, with unknown mean \(\mu\) and unknown standard deviation \(\sigma\). The size of that underlying population is generally not relevant unless it is very small. If the population is bell shaped, the assumption is met and needs no further discussion. Random sampling is also assumed, but that is a completely separate assumption from normality — a bell-shaped population sampled badly is still a bad sample.
The thicker tails are the whole point, so it is worth being concrete about what they cost you. To capture the middle 95% of a standard normal curve you go out 1.96 standard errors from the center. To capture the middle 95% of a t-curve with 14 degrees of freedom you have to go out 2.14 standard errors — further, because the curve has pushed more of its probability into the tails and you need a wider net to fence in the same 95%. That is the price of not knowing \(\sigma\), and the price falls as the sample grows: at 30 degrees of freedom the number is 2.04, at 100 it is 1.98, and it keeps creeping toward 1.96 without ever quite getting there.
Calculators and computers handle any Student's t-probability you need. The TI-83, 83+, and 84+ have a tcdf function that finds the probability between two t-values; the syntax is tcdf(lower bound, upper bound, degrees of freedom). For confidence intervals, though, we need to run that backwards — we know the probability and want the t-value. On the TI-84+ that is the invT command on the DISTR menu, and it works like invNorm: invT(area to the left, degrees of freedom) returns the t-score with that much area below it. The TI-83 and 83+ do not have invT. (The TI-89 has an inverse-t command; the TI-86 has neither, so with that calculator you need a printed table.)
A probability table for the Student's t-distribution also works. The table gives t-scores indexed by confidence level (column) and degrees of freedom (row), and you read off the value where the row and column meet. Watch the column headings when you use one — some tables label the columns by confidence level and others label them by the area left over in one tail or in both tails combined, and those are different numbers for the same column. A Student's t table (see Appendix H Tables) gives t-scores from the degrees of freedom and the right-tailed probability. Any printed table is limited to the few confidence levels somebody chose to print; a calculator is not.
Definition 7.2.2 - Four deviations can be anything at all; the fifth is forced, because they have to add to zero.
Denise Halloway is the quality inspector for the bike-share co-op she runs with her wife. She tests a random sample of 25 rechargeable batteries and records how long each one holds a charge. The population standard deviation is unknown.
a. How many degrees of freedom does the t-distribution for this sample have?
b. For those degrees of freedom, the t-score with 0.025 area to its right is 2.064, while the matching z-score is 1.96. What does that difference do to the width of a 95% confidence interval, and why?
Solution — Try It Now 7.2.1
Step 1 — Find the degrees of freedom. The degrees of freedom is always one less than the sample size:
$$ df = n - 1 = 25 - 1 = 24 $$Step 2 — Compare the two multipliers. The error bound is the multiplier times the standard error \(\frac{s}{\sqrt{n}}\). Using 2.064 instead of 1.96 multiplies the error bound by \(\frac{2.064}{1.96} \approx 1.053\), so the interval is about 5% wider on each side.
Step 3 — Say why. We do not know \(\sigma\); we estimated it with \(s\) from only 25 observations, and that estimate could be off. The t-distribution builds that extra uncertainty into its fatter tails, so it demands a bigger multiplier to still deliver 95% coverage. A wider interval is what honesty about the unknown \(\sigma\) costs.
Answer: a. \(df = 24\). b. The interval is roughly 5% wider on each side, because the t-distribution's fatter tails require a larger multiplier to capture the same 95%.
Try it in rāSHio
You do not need a TI-84 to get that 2.064. Open rāSHio, choose Distributions → T, set Degrees of freedom to 24, and read the critical value t* beside the Confidence field with the level set to 0.95. Typing a t-value into the Two-Tailed box instead runs the comparison the other way and shades the area in both tails.
Figure 7.2.1 — The t-distribution panel in rāSHio: Distributions → T, on its Two-Tailed tab. The clip types a t-value and degrees of freedom and reads the tail probability; the critical value t* shown beside the Confidence field is the multiplier this section's error bound needs.
7.2.2 Notation and the Error Bound for a Mean
When the population standard deviation is not known, the error bound for a population mean is
$$ EBM = \left(t_{\frac{\alpha}{2}}\right)\left(\frac{s}{\sqrt{n}}\right), $$where \(t_{\frac{\alpha}{2}}\) is the t-score with area \(\frac{\alpha}{2}\) to its right, using \(df = n - 1\) degrees of freedom, and \(s\) is the sample standard deviation. The confidence interval is then
$$ (\overline{x} - EBM,\ \overline{x} + EBM). $$Section 7.1 built the interval as \(\overline{x} \pm z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\). Swap the known \(\sigma\) for the sample's \(s\), swap \(z\) for \(t\), and you have this section. Nothing else changes.
Using \(T\) as the random variable, the notation for the Student's t-distribution is \(T \sim t_{df}\), where \(df = n - 1\). If a sample has \(n = 20\) items, then \(df = 20 - 1 = 19\), and we write the distribution as \(T \sim t_{19}\). The subscript is doing real work here — it names which member of the t-family you are using, so writing \(t_{19}\) instead of just \(t\) is the difference between a specific curve and a vague gesture at a family of them.
Because the structure carries over, so does the procedure, and it is worth walking through once so the steps are in order before the arithmetic starts. Begin with the confidence level \(CL\) the problem asks for, and turn it into the leftover probability \(\alpha = 1 - CL\). Split that leftover between the two tails, so each tail holds \(\frac{\alpha}{2}\). Now find the t-score that leaves exactly \(\frac{\alpha}{2}\) of the area to its right. On a calculator you feed invT the area to the left, which is \(1 - \frac{\alpha}{2}\), together with the degrees of freedom. Multiply that t-score by the standard error \(\frac{s}{\sqrt{n}}\) to get the error bound. Then subtract and add the error bound to \(\overline{x}\). The only place students reliably lose points is the left-versus-right tail flip, so say out loud which one the tool wants before you type anything: a table indexed by right-tail area wants \(\frac{\alpha}{2}\), and invT wants \(1 - \frac{\alpha}{2}\). Both give the same t-score; they just ask for it differently.
To build the confidence interval directly, without computing the error bound by hand:
1. Press STAT.
2. Arrow over to TESTS.
3. Arrow down to 8:TInterval and press ENTER (or just press 8).
4. Choose Data if you have the raw list, or Stats if you have \(\overline{x}\), \(s\), and \(n\).
5. Enter the values, set C-Level to the confidence level as a decimal, and select Calculate.
Definition 7.2.3 - The error bound is one distance measured off the t-curve, then laid out either side of the sample mean.
You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.
8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5
Solution — Try It Now 7.2.2
Step 1 — Summarize the sample. With \(n = 12\) values, one-variable statistics give
$$ \overline{x} = 8.9833, \qquad s = 1.2904, \qquad n = 12, \qquad df = 12 - 1 = 11 $$Step 2 — Find the t-score. For \(CL = 0.95\), \(\alpha = 1 - 0.95 = 0.05\) and \(\frac{\alpha}{2} = 0.025\). The area to the left is \(1 - 0.025 = 0.975\), so invT(0.975, 11) gives
Step 3 — Compute the error bound.
$$ EBM = (2.201)\left(\frac{1.2904}{\sqrt{12}}\right) = (2.201)(0.3725) = 0.8199 $$Step 4 — Build the interval.
$$ \overline{x} - EBM = 8.9833 - 0.8199 = 8.1634 $$ $$ \overline{x} + EBM = 8.9833 + 0.8199 = 9.8032 $$On a TI-83/83+/84/84+: enter the 12 values into a list, then STAT → TESTS → 8:TInterval, choose Data, name the list, Freq: 1, C-Level: 0.95, Calculate.
Answer: the 95% confidence interval is (8.1634, 9.8032). We estimate with 95% confidence that the true mean number of hours slept per night for this population is between 8.16 and 9.80 hours.
Rey Salazar (they/them) runs a study of acupuncture to determine how effective it is in relieving pain. They measure sensory rates for 15 subjects with the results given. Use their sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which the data was taken.
8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9
Solution — Example 7.2.1
Step 1 — Summarize the sample. To find the confidence interval you need the sample mean \(\overline{x}\) and the error bound. Running one-variable statistics on the 15 values gives
$$ \overline{x} = 8.2267, \qquad s = 1.6722, \qquad n = 15 $$ $$ df = 15 - 1 = 14 $$Step 2 — Turn the confidence level into a tail area. With \(CL = 0.95\),
$$ \alpha = 1 - CL = 1 - 0.95 = 0.05, \qquad \frac{\alpha}{2} = 0.025 $$so the t-score we want is \(t_{\frac{\alpha}{2}} = t_{0.025}\).
Step 3 — Find the t-score. The area to the right of \(t_{0.025}\) is 0.025, so the area to its left is \(1 - 0.025 = 0.975\). Using invT(.975, 14) on the TI-84+ calculator,
Step 4 — Compute the error bound.
$$ EBM = \left(t_{\frac{\alpha}{2}}\right)\left(\frac{s}{\sqrt{n}}\right) = (2.14)\left(\frac{1.6722}{\sqrt{15}}\right) = 0.924 $$Step 5 — Build the interval.
$$ \overline{x} - EBM = 8.2267 - 0.9240 = 7.30 $$ $$ \overline{x} + EBM = 8.2267 + 0.9240 = 9.15 $$On a TI-83/83+/84/84+: press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or just press 8). Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. There should be a 1 after Freq. Arrow down to C-Level and enter 0.95. Arrow down to Calculate and press ENTER. The calculator returns (7.3006, 9.1527), which differs from the hand computation only by rounding.
Answer: the 95% confidence interval is (7.30, 9.15). We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
Try it in rāSHio
rāSHio builds this interval without the t-score lookup at all. Choose Stats → T Stats, enter the sample mean 8.2267, the sample standard deviation 1.6722 and the sample size 15, tick Confidence Interval (μ), set the level to 0.95, and press Calculate — it returns the same (7.30, 9.15) the TInterval command does.
Figure 7.2.2 — The t panel in rāSHio: Stats → T Stats. The clip walks this panel running a one-sample hypothesis test on its own demo numbers; the Confidence Interval (μ) checkbox visible beside it is the box that builds the interval this Example computes.
7.2.3 Confidence Intervals from Raw Data
The two problems above both handed you a list of numbers and asked for an interval. That is the ordinary case in practice, and it is worth naming the one extra step it adds: before any of the confidence-interval machinery runs, you have to reduce the list to \(\overline{x}\), \(s\), and \(n\) yourself. Every calculator has a one-variable-statistics routine that does this, and it is the same routine you used back in the descriptive-statistics chapter — nothing new. What is new is being careful about which standard deviation you copy down. A calculator that reports both will show \(S_x\) (the sample standard deviation, dividing by \(n - 1\)) and \(\sigma_x\) (the population version, dividing by \(n\)). This section always wants \(S_x\), because the whole reason we are using a t-distribution is that we are estimating the spread from a sample. Grabbing \(\sigma_x\) by mistake gives an answer that is slightly too small and looks perfectly reasonable, which is exactly what makes it a hard error to catch.
Kayla Brennan asked a random sample of statistics students to estimate the total number of hours they spend watching television in an average week. She recorded their responses in Table 7.2.1. Use her sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.
| 0 | 3 | 1 | 20 | 9 |
| 5 | 10 | 1 | 10 | 4 |
| 14 | 2 | 4 | 4 | 5 |
Solution — Try It Now 7.2.3
Step 1 — Summarize the sample. There are 15 responses, so
$$ \overline{x} = 6.133, \qquad s = 5.514, \qquad n = 15, \qquad df = 15 - 1 = 14 $$Step 2 — Find the t-score. For \(CL = 0.98\), \(\alpha = 1 - 0.98 = 0.02\) and \(\frac{\alpha}{2} = 0.01\). The area to the left is \(1 - 0.01 = 0.99\), so invT(0.99, 14) gives
Step 3 — Compute the error bound.
$$ EBM = (2.624)\left(\frac{5.514}{\sqrt{15}}\right) = (2.624)(1.4238) = 3.736 $$Step 4 — Build the interval.
$$ \overline{x} - EBM = 6.133 - 3.736 = 2.397 $$ $$ \overline{x} + EBM = 6.133 + 3.736 = 9.869 $$Answer: the 98% confidence interval is (2.397, 9.869). We estimate with 98% confidence that statistics students watch, on average, between 2.4 and 9.9 hours of television per week. The interval is wide because the sample is small and the responses ranged from 0 to 20 hours — real disagreement in the data shows up as a wide interval, not as a wrong one.
The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. The scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the "in utero/newborn" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table 7.2.2 shows how many of the targeted chemicals were found in each infant's cord blood.
| 79 | 145 | 147 | 160 | 116 | 100 | 159 | 151 | 156 | 126 |
| 137 | 83 | 156 | 94 | 121 | 144 | 123 | 114 | 139 | 99 |
Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an infant's blood.
Solution — Example 7.2.2
Step 1 — Summarize the sample. From the 20 values in Table 7.2.2,
$$ \overline{x} = 127.45, \qquad s = 25.965, \qquad n = 20, \qquad df = 20 - 1 = 19 $$Step 2 — Turn the confidence level into a tail area. You are asked for a 90% confidence interval, so \(CL = 0.90\) and
$$ \alpha = 1 - CL = 1 - 0.90 = 0.10, \qquad \frac{\alpha}{2} = 0.05 $$and the t-score we want is \(t_{\frac{\alpha}{2}} = t_{0.05}\).
Step 3 — Find the t-score. By definition the area to the right of \(t_{0.05}\) is 0.05, so the area to its left is \(1 - 0.05 = 0.95\). Using a table, calculator, or computer,
$$ t_{0.05} = 1.729 $$Step 4 — Compute the error bound.
$$ EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right) = 1.729\left(\frac{25.965}{\sqrt{20}}\right) \approx 10.038 $$Step 5 — Build the interval.
$$ \overline{x} - EBM = 127.45 - 10.038 = 117.412 $$ $$ \overline{x} + EBM = 127.45 + 10.038 = 137.488 $$On a TI-83/83+/84/84+: enter the data as a list. Press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or just press 8). Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. Arrow down to Freq and enter 1. Arrow down to C-Level and enter 0.90. Arrow down to Calculate and press ENTER. The calculator returns (117.41, 137.49).
Answer: the 90% confidence interval is (117.412, 137.488). We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.
Try it in rāSHio
Those twenty counts have to become \(\overline{x}\), \(s\) and \(n\) before any of the interval machinery runs. Paste Table 7.2.1's values into rāSHio with File → Delimited List…, then choose Stats → Summary Statistics and read 127.45 and 25.965 straight off the panel — taking the sample standard deviation, the one computed with \(n - 1\), not the population figure printed beside it.
Figure 7.2.3 — Reducing a raw list to its summary statistics in rāSHio: Stats → Summary Statistics. The clip walks the panel on its own demo data; run it on Table 7.2.1's twenty counts and it returns the \(\overline{x} = 127.45\) and \(s = 25.965\) this Example starts from.
Problem Set 7.2
Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. Dr. Wei Chen chairs the investigation committee, and his team randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.
Problem 1. Identify the following:
a) \(\overline{x} =\) ____
b) \(s_x =\) ____
c) \(n =\) ____
d) \(n - 1 =\) ____
Problem 1 Solution
Step 1 — Read the summary statistics straight off the setup: the committee surveyed 70 patients, the sample mean wait was 1.5 hours, and the sample standard deviation was 0.5 hours.
Step 2 — Compute the degrees of freedom: it is always one less than the sample size, \(n - 1 = 70 - 1 = 69\).
Answer: a) \(\overline{x} = 1.5\) hours; b) \(s_x = 0.5\) hours; c) \(n = 70\); d) \(n - 1 = 69\).
Problem 2. Define the random variables \(X\) and \(\overline{X}\) in words.
Problem 2 Solution
Step 1 — Identify what one observation is: each patient contributes one wait time, so \(X\) describes a single patient.
Step 2 — Identify what the sample mean describes: \(\overline{X}\) averages the 70 individual wait times, so it describes the whole sample at once.
Answer: \(X\) is the number of hours a patient waits in the emergency room before being called back to be examined. \(\overline{X}\) is the mean wait time of 70 patients in the emergency room.
Problem 3. Which distribution should you use for this problem?
Problem 3 Solution
Step 1 — Ask whether \(\sigma\) is known: the setup gives a sample standard deviation of 0.5 hours. The population standard deviation is never stated, so \(\sigma\) is unknown.
Step 2 — Pick the matching distribution: whenever \(s\) stands in for \(\sigma\), the standardized sample mean follows a Student's t-distribution with \(df = n - 1 = 69\).
Answer: use the Student's t-distribution, \(\overline{X} \sim t_{69}\).
Problem 4. Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.
Problem 4 Solution
Step 1 — Split the confidence level into tails: with \(CL = 0.95\), \(\alpha = 1 - 0.95 = 0.05\) and \(\frac{\alpha}{2} = 0.025\).
Step 2 — Find the t-score: the area to the left is \(1 - 0.025 = 0.975\), so invT(0.975, 69) gives \(t_{0.025} = 1.995\).
Step 3 — Compute the error bound:
$$ EBM = (1.995)\left(\frac{0.5}{\sqrt{70}}\right) = (1.995)(0.0598) = 0.12 $$Step 4 — Build the interval:
$$ 1.5 - 0.12 = 1.3808, \qquad 1.5 + 0.12 = 1.6192 $$Step 5 — Sketch the graph: draw a t-curve centered at 1.5, shade the middle 95%, mark the boundaries at 1.3808 and 1.6192, and label 0.025 of area in each unshaded tail.
Answer: the 95% confidence interval is (1.3808, 1.6192), with \(EBM = 0.12\) hours.
Problem 5. Explain in complete sentences what the confidence interval means.
Problem 5 Solution
Step 1 — State what the interval claims: the interval is a range of plausible values for the population mean wait time, not for any one patient.
Step 2 — Attach the confidence level correctly: the 95% describes the method. If the committee repeated this survey many times and built an interval each time, about 95% of those intervals would contain the true mean wait time.
Answer: we estimate with 95% confidence that the true mean wait time for all emergency room patients at this hospital is between 1.3808 hours and 1.6192 hours — roughly 1 hour 23 minutes to 1 hour 37 minutes.
Use the following information to answer the next six exercises. One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.
Problem 6. Identify the following:
a) \(\overline{x} =\) ____
b) \(s_x =\) ____
c) \(n =\) ____
d) \(n - 1 =\) ____
Problem 6 Solution
Step 1 — Read the summary statistics off the setup: 108 Americans watched an average of 151 hours per month with a standard deviation of 32 hours.
Step 2 — Compute the degrees of freedom: \(n - 1 = 108 - 1 = 107\).
Answer: a) \(\overline{x} = 151\); b) \(s_x = 32\); c) \(n = 108\); d) \(n - 1 = 107\).
Problem 7. Define the random variable \(X\) in words.
Problem 7 Solution
Step 1 — Identify the single observation: one person in the survey reports one monthly viewing total.
Answer: \(X\) is the number of hours one American spends watching television each month.
Problem 8. Define the random variable \(\overline{X}\) in words.
Problem 8 Solution
Step 1 — Identify what gets averaged: \(\overline{X}\) is the average of the 108 individual monthly totals.
Answer: \(\overline{X}\) is the mean number of hours spent watching television per month from a sample of 108 Americans.
Problem 9. Which distribution should you use for this problem?
Problem 9 Solution
Step 1 — Check what spread we were given: the 32 hours is a sample standard deviation, so \(\sigma\) is unknown.
Step 2 — Pick the distribution: with \(s\) estimating \(\sigma\), use Student's t with \(df = 107\).
Answer: use the Student's t-distribution, \(\overline{X} \sim t_{107}\).
Problem 10. Construct a 99% confidence interval for the population mean hours spent watching television per month.
a) State the confidence interval.
b) Sketch the graph.
c) Calculate the error bound.
Problem 10 Solution
Step 1 — Split the confidence level into tails: with \(CL = 0.99\), \(\alpha = 0.01\) and \(\frac{\alpha}{2} = 0.005\).
Step 2 — Find the t-score: the area to the left is \(1 - 0.005 = 0.995\), so invT(0.995, 107) gives \(t_{0.005} = 2.623\).
Step 3 — Compute the error bound:
$$ EBM = (2.623)\left(\frac{32}{\sqrt{108}}\right) = (2.623)(3.0792) = 8.08 $$Step 4 — Build the interval:
$$ 151 - 8.08 = 142.92, \qquad 151 + 8.08 = 159.08 $$Step 5 — Sketch the graph: a t-curve centered at 151, middle 99% shaded, boundaries labelled 142.92 and 159.08, with 0.005 of area in each tail.
Answer: a) the 99% confidence interval is (142.92, 159.08); b) see the sketch described above; c) \(EBM = 8.08\) hours.
Problem 11. Why would the error bound change if the confidence level were lowered to 95%?
Problem 11 Solution
Step 1 — See which piece of the formula moves: \(EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right)\). Lowering the confidence level does not touch \(s\) or \(n\), only the t-score.
Step 2 — Track the t-score: at 95% confidence, \(\frac{\alpha}{2} = 0.025\) instead of 0.005, and \(t_{0.025} = 1.982\) instead of 2.623. A smaller multiplier gives a smaller error bound: \(EBM = (1.982)(3.0792) = 6.10\).
Answer: the error bound would shrink from about 8.08 to about 6.10 hours. Asking for less confidence means you need to cast a narrower net, so the interval gets tighter.
Use the following information to answer the next 13 exercises. The data in Table 7.2.3 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let \(X =\) the number of colors on a national flag.
| \(X\) | Freq. |
|---|---|
| 1 | 1 |
| 2 | 7 |
| 3 | 18 |
| 4 | 7 |
| 5 | 6 |
Problem 12. Calculate the following:
a) \(\overline{x} =\) ____
b) \(s_x =\) ____
c) \(n =\) ____
Problem 12 Solution
Step 1 — Total the observations using the frequencies: the survey covers \(1 + 7 + 18 + 7 + 6 = 39\) flags, and the colors total
$$ 1(1) + 2(7) + 3(18) + 4(7) + 5(6) = 1 + 14 + 54 + 28 + 30 = 127 $$Step 2 — Compute the mean:
$$ \overline{x} = \frac{127}{39} = 3.26 $$Step 3 — Compute the sample standard deviation: enter the values in L1 and the frequencies in L2, then run 1-Var Stats L1, L2 to read \(S_x = 1.02\).
Answer: a) \(\overline{x} = 3.26\); b) \(s_x = 1.02\); c) \(n = 39\).
Problem 13. Define the random variable \(\overline{X}\) in words.
Problem 13 Solution
Step 1 — Identify what is being averaged: each of the 39 sampled flags contributes its own color count, and \(\overline{X}\) averages those counts.
Answer: \(\overline{X}\) is the mean number of colors on the flags in a sample of 39 national flags.
Problem 14. What is \(\overline{x}\) estimating?
Problem 14 Solution
Step 1 — Separate statistic from parameter: \(\overline{x} = 3.26\) is a number computed from the sample, so it is a statistic. The unknown it stands in for is the population's own mean.
Answer: \(\overline{x}\) is estimating \(\mu\), the true mean number of colors on all national flags.
Problem 15. Is \(\sigma_x\) known?
Problem 15 Solution
Step 1 — Look for a stated population spread: the problem gives only a survey of 39 flags. Nothing tells us the spread of the whole population of flags.
Answer: no, \(\sigma_x\) is not known. Only the sample standard deviation \(s_x = 1.02\) is available.
Problem 16. As a result of your answer to Problem 7.2.15, state the exact distribution to use when calculating the confidence interval.
Construct a 95% confidence interval for the true mean number of colors on national flags.
Problem 16 Solution
Step 1 — Apply the rule: because \(\sigma_x\) is unknown and we are using \(s_x\) in its place, the standardized sample mean follows a Student's t-distribution.
Step 2 — Get the degrees of freedom: \(df = n - 1 = 39 - 1 = 38\).
Answer: \(\overline{X} \sim t_{38}\).
Problem 17. How much area is in both tails (combined)?
Problem 17 Solution
Step 1 — Turn the confidence level into leftover area: a 95% confidence interval leaves \(\alpha = 1 - 0.95 = 0.05\) outside the interval, split between the two tails.
Answer: 0.05 of the area is in the two tails combined.
Problem 18. How much area is in each tail?
Problem 18 Solution
Step 1 — Split the leftover evenly: the t-distribution is symmetric, so each tail holds half of \(\alpha\).
$$ \frac{\alpha}{2} = \frac{0.05}{2} = 0.025 $$Answer: 0.025 in each tail.
Problem 19. Calculate the following:
a) lower limit
b) upper limit
c) error bound
Problem 19 Solution
Step 1 — Find the t-score: with \(df = 38\) and 0.025 in the right tail, invT(0.975, 38) gives \(t_{0.025} = 2.024\).
Step 2 — Compute the error bound:
$$ EBM = (2.024)\left(\frac{1.02}{\sqrt{39}}\right) = (2.024)(0.1633) = 0.33 $$Step 3 — Build the limits:
$$ 3.26 - 0.33 = 2.93, \qquad 3.26 + 0.33 = 3.59 $$Answer: a) lower limit 2.93; b) upper limit 3.59; c) error bound 0.33.
Problem 20. The 95% confidence interval is ____.
Problem 20 Solution
Step 1 — Assemble the limits from the previous problem: the interval runs from \(\overline{x} - EBM\) to \(\overline{x} + EBM\).
Answer: the 95% confidence interval is (2.93, 3.59).
Problem 21. Fill in the blanks on the graph with the areas, the upper and lower limits of the confidence interval, and the sample mean.
Exercise Figure - Fill in the confidence level, the two tail areas, and the three values on the axis.
Problem 21 Solution
Step 1 — Label the confidence level: the shaded middle region carries the confidence level, so C.L. = 0.95.
Step 2 — Label the tails: the unshaded area outside the interval splits evenly, so each α/2 blank gets 0.025.
Step 3 — Label the three marks under the axis: left to right they are the lower limit, the sample mean, and the upper limit — 2.93, 3.26, and 3.59.
Answer: C.L. = 0.95; \(\frac{\alpha}{2} = 0.025\) on the left and \(\frac{\alpha}{2} = 0.025\) on the right; lower limit 2.93, \(\overline{x} = 3.26\), upper limit 3.59.
Problem 22. In one complete sentence, explain what the interval means.
Problem 22 Solution
Step 1 — Name the population parameter, not the sample: the sentence has to be about all national flags, not about the 39 that were surveyed.
Answer: we are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.
Problem 23. Using the same \(\overline{x}\), \(s_x\), and level of confidence, suppose that \(n\) were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
Problem 23 Solution
Step 1 — See which pieces change: raising \(n\) from 39 to 69 shrinks the standard error \(\frac{s_x}{\sqrt{n}}\) and also nudges the t-score down, because \(df\) rises from 38 to 68.
Step 2 — Recompute: \(t_{0.025} = 1.996\) at \(df = 68\), and \(\frac{1.02}{\sqrt{69}} = 0.1228\), so
$$ EBM = (1.996)(0.1228) = 0.245 $$Answer: the error bound would become \(EBM = 0.245\), which is smaller. As sample sizes increase, variability decreases and we need less interval length to capture the true mean.
Problem 24. Using the same \(\overline{x}\), \(s_x\), and \(n = 39\), how would the error bound change if the confidence level were reduced to 90%? Why?
Problem 24 Solution
Step 1 — See which piece changes: \(\overline{x}\), \(s_x\), and \(n\) all stay put, so only the t-score moves.
Step 2 — Recompute the t-score: at 90% confidence, \(\frac{\alpha}{2} = 0.05\), and invT(0.95, 38) gives \(t_{0.05} = 1.686\) instead of 2.024.
Step 3 — Recompute the error bound:
$$ EBM = (1.686)(0.1633) = 0.275 $$Answer: the error bound would drop from 0.33 to about 0.275. A lower confidence level asks the interval to succeed less often, so it does not have to reach as far from the sample mean.
Problem 25. In six packages of "The Flintstones® Real Fruit Snacks" there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces.
a) Define the random variables \(X\) and \(P'\) in words.
b) Which distribution should you use for this problem? Explain your choice.
c) Calculate \(p'\).
d) Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e) Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?
Problem 25 Solution
Step 1 — Notice what is actually being estimated: this problem asks for a proportion, not a mean, so the Student's t-distribution of this section does not apply. Confidence intervals for a proportion use the normal distribution — see Section 7.3.
Step 2 — Define the variables: \(X\) is the number of Bam-Bam snack pieces among the snack pieces examined, and \(P'\) is the proportion of Bam-Bam pieces in the sample.
Step 3 — Compute the sample proportion:
$$ p' = \frac{5}{68} = 0.0735, \qquad q' = 1 - 0.0735 = 0.9265 $$Step 4 — Find the z-score: with \(CL = 0.96\), \(\alpha = 0.04\) and \(\frac{\alpha}{2} = 0.02\), so invNorm(0.98, 0, 1) gives \(z_{0.02} = 2.054\).
Step 5 — Compute the error bound:
$$ EBP = (2.054)\sqrt{\frac{(0.0735)(0.9265)}{68}} = (2.054)(0.0317) = 0.065 $$Step 6 — Build the interval:
$$ 0.0735 - 0.065 = 0.0085, \qquad 0.0735 + 0.065 = 0.1385 $$Step 7 — Sketch the graph: a normal curve centered at 0.0735 with the middle 96% shaded between 0.0085 and 0.1385, and 0.02 of area in each tail.
Answer: a) \(X\) = number of Bam-Bam pieces, \(P'\) = sample proportion of Bam-Bam pieces; b) the normal distribution, because this is a proportion and both \(np'\) and \(nq'\) are being treated as large enough; c) \(p' = 0.0735\); d) the 96% confidence interval is (0.0085, 0.1385) with \(EBP = 0.065\); e) no — with only five Bam-Bam pieces in the sample, \(np' = 5\) falls short of the usual rule of thumb of at least 10 successes, so the normal approximation is shaky and the interval is very wide relative to the estimate.
Problem 26. A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal.
a) Identify the following:
i. \(\overline{x} =\) ____
ii. \(s_x =\) ____
iii. \(n =\) ____
iv. \(n - 1 =\) ____
b) Define the random variables \(X\) and \(\overline{X}\) in words.
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e) What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why?
Problem 26 Solution
Step 1 — Summarize the 35 enrollment figures: entering them in a list and running one-variable statistics gives \(\overline{x} = 8{,}629\), \(s_x = 6{,}944\), \(n = 35\), and \(n - 1 = 34\).
Step 2 — Define the variables: \(X\) is the enrollment at one community college; \(\overline{X}\) is the mean enrollment for a sample of 35 community colleges.
Step 3 — Pick the distribution: the population standard deviation is unknown, so use \(t_{34}\).
Step 4 — Find the t-score: for \(CL = 0.95\), invT(0.975, 34) gives \(t_{0.025} = 2.032\).
Step 5 — Compute the error bound:
$$ EBM = (2.032)\left(\frac{6{,}944}{\sqrt{35}}\right) = (2.032)(1{,}173.7) = 2{,}385 $$Step 6 — Build the interval:
$$ 8{,}629 - 2{,}385 = 6{,}244, \qquad 8{,}629 + 2{,}385 = 11{,}014 $$Answer: a) i. 8,629; ii. 6,944; iii. 35; iv. 34. b) \(X\) is one college's enrollment, \(\overline{X}\) is the mean enrollment of the 35 sampled colleges. c) \(t_{34}\), because \(\sigma\) is unknown. d) the 95% confidence interval is (6,244, 11,014) with \(EBM = 2{,}385\); sketch a t-curve centered at 8,629 with those limits marked. e) surveying 500 colleges makes \(\sqrt{n}\) much larger, so the error bound shrinks and the confidence interval gets narrower.
Problem 27. Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours.
a) Identify the following:
i. \(\overline{x} =\) ____
ii. \(s_x =\) ____
iii. \(n =\) ____
iv. \(n - 1 =\) ____
b) Define the random variables \(X\) and \(\overline{X}\) in words.
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 95% confidence interval for the population mean time wasted.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e) Explain in a complete sentence what the confidence interval means.
Problem 27 Solution
Step 1 — Read off the statistics: \(\overline{x} = 8\) hours, \(s_x = 4\) hours, \(n = 81\), \(n - 1 = 80\).
Step 2 — Define the variables: \(X\) is the number of hours one juror waits to be called; \(\overline{X}\) is the mean wait time of the 81 sampled jurors.
Step 3 — Pick the distribution: \(\sigma\) is unknown and \(s\) is standing in for it, so use \(t_{80}\).
Step 4 — Find the t-score: for \(CL = 0.95\), invT(0.975, 80) gives \(t_{0.025} = 1.990\).
Step 5 — Compute the error bound:
$$ EBM = (1.990)\left(\frac{4}{\sqrt{81}}\right) = (1.990)(0.4444) = 0.88 $$Step 6 — Build the interval:
$$ 8 - 0.88 = 7.12, \qquad 8 + 0.88 = 8.88 $$Answer: a) i. 8; ii. 4; iii. 81; iv. 80. b) as defined in Step 2. c) \(t_{80}\), because the population standard deviation is unknown. d) the 95% confidence interval is (7.12, 8.88) with \(EBM = 0.88\) hours. e) we estimate with 95% confidence that the mean time a juror waits to be called at this courthouse is between 7.12 and 8.88 hours.
Problem 28. A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.
a) Identify the following:
i. \(\overline{x} =\) ____
ii. \(s_x =\) ____
iii. \(n =\) ____
iv. \(n - 1 =\) ____
b) Define the random variable \(X\) in words.
c) Define the random variable \(\overline{X}\) in words.
d) Which distribution should you use for this problem? Explain your choice.
e) Construct a 95% confidence interval for the population mean length of time.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
f) What does it mean to be "95% confident" in this problem?
Problem 28 Solution
Step 1 — Summarize the nine effective periods: running one-variable statistics on 2.7, 2.8, 3.0, 2.3, 2.3, 2.2, 2.8, 2.1, 2.4 gives \(\overline{x} = 2.51\), \(s_x = 0.318\), \(n = 9\), \(n - 1 = 8\).
Step 2 — Define the variables: \(X\) is the effective length of time of the tranquilizer for one patient; \(\overline{X}\) is the mean effective length of time for a sample of nine patients.
Step 3 — Pick the distribution: we need a Student's t-distribution, because we do not know the population standard deviation. With \(df = 8\), \(\overline{X} \sim t_8\).
Step 4 — Find the t-score: for \(CL = 0.95\), invT(0.975, 8) gives \(t_{0.025} = 2.306\).
Step 5 — Compute the error bound:
$$ EBM = (2.306)\left(\frac{0.318}{\sqrt{9}}\right) = (2.306)(0.106) = 0.25 $$Step 6 — Build the interval:
$$ 2.51 - 0.25 = 2.27, \qquad 2.51 + 0.25 = 2.76 $$Answer: a) i. 2.51; ii. 0.318; iii. 9; iv. 8. b) and c) as defined in Step 2. d) Student's t with 8 degrees of freedom, because \(\sigma\) is unknown. e) i. the 95% confidence interval is (2.27, 2.76); ii. sketch a t-curve centered at 2.51 with those limits; iii. \(EBM = 0.25\) hours. f) if we were to sample many groups of nine patients, 95% of the intervals we built would contain the true population mean length of time.
Problem 29. Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal.
a) Identify the following:
i. \(\overline{x} =\) ____
ii. \(s_x =\) ____
iii. \(n =\) ____
iv. \(n - 1 =\) ____
b) Define the random variable \(X\) in words.
c) Define the random variable \(\overline{X}\) in words.
d) Which distribution should you use for this problem? Explain your choice.
e) Construct a 99% confidence interval for the population mean length of time using training wheels.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
f) Why would the error bound change if the confidence level were lowered to 90%?
Problem 29 Solution
Step 1 — Read off the statistics: \(\overline{x} = 6\) months, \(s_x = 3\) months, \(n = 14\), \(n - 1 = 13\).
Step 2 — Define the variables: \(X\) is the number of months one child used training wheels; \(\overline{X}\) is the mean number of months for the sample of 14 children.
Step 3 — Pick the distribution: \(\sigma\) is unknown, so use \(t_{13}\).
Step 4 — Find the t-score: for \(CL = 0.99\), \(\frac{\alpha}{2} = 0.005\), and invT(0.995, 13) gives \(t_{0.005} = 3.012\).
Step 5 — Compute the error bound:
$$ EBM = (3.012)\left(\frac{3}{\sqrt{14}}\right) = (3.012)(0.8018) = 2.415 $$Step 6 — Build the interval:
$$ 6 - 2.415 = 3.585, \qquad 6 + 2.415 = 8.415 $$Answer: a) i. 6; ii. 3; iii. 14; iv. 13. b) and c) as defined in Step 2. d) \(t_{13}\), because the population standard deviation is unknown. e) the 99% confidence interval is (3.585, 8.415) with \(EBM = 2.415\) months. f) at 90% confidence the t-score drops from 3.012 to 1.771, so the error bound would shrink to about 1.42 months — less confidence buys a tighter interval.
Problem 30. The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates' campaigns. The FEC has reported financial information for 556 Leadership PACs that operated during a certain election cycle. Table 7.2.4 shows the total receipts during this cycle for a random selection of 30 Leadership PACs.
| $46,500.00 | $0 | $40,966.50 | $105,887.20 | $5,175.00 |
| $29,050.00 | $19,500.00 | $181,557.20 | $31,500.00 | $149,970.80 |
| $2,555,363.20 | $12,025.00 | $409,000.00 | $60,521.70 | $18,000.00 |
| $61,810.20 | $76,530.80 | $119,459.20 | $0 | $63,520.00 |
| $6,500.00 | $502,578.00 | $705,061.10 | $708,258.90 | $135,810.00 |
| $2,000.00 | $2,000.00 | $0 | $1,287,933.80 | $219,148.30 |
The sample statistics are \(\overline{x} = \$251,854.23\) and \(s = \$521,130.41\). Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the specific election cycle. Use the Student's t-distribution.
Problem 30 Solution
Step 1 — Note what we have and what we lack: we are given \(\overline{x} = \$251,854.23\) and \(s = \$521,130.41\). That is the standard deviation of the sample, not of the population, which is why the problem tells us to use the Student's t-distribution.
Step 2 — Get the degrees of freedom: there are 30 measures in the sample, so \(n = 30\) and \(df = 30 - 1 = 29\).
Step 3 — Split the confidence level:
$$ CL = 0.96, \text{ so } \alpha = 1 - CL = 1 - 0.96 = 0.04, \qquad \frac{\alpha}{2} = 0.02 $$Step 4 — Find the t-score: invT(0.98, 29) gives \(t_{0.02} = 2.150\).
Step 5 — Compute the error bound:
$$ EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right) = 2.150\left(\frac{521{,}130.41}{\sqrt{30}}\right) \approx \$204,561.66 $$Step 6 — Build the interval:
$$ \overline{x} - EBM = \$251,854.23 - \$204,561.66 = \$47,292.57 $$ $$ \overline{x} + EBM = \$251,854.23 + \$204,561.66 = \$456,415.89 $$Note on the graphing calculator: entering the raw receipts as a list and running STAT → TESTS → 8:TInterval with Freq: 1 and C-Level: 0.96 would display an interval of about ($47,262, $456,447). The small difference from the hand computation is rounding.
Answer: we estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the specific election cycle lies between $47,292.57 and $456,415.89. The interval is enormous because one PAC raised over $2.5 million while several raised nothing — that spread drives \(s\) far above the typical value.
Problem 31. Forbes magazine published data on the best small firms in a certain year. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. Table 7.2.5 shows the ages of the corporate CEOs for a random sample of these firms. Use this sample data to construct a 90% confidence interval for the mean age of CEOs for these top small firms. Use the Student's t-distribution.
| 48 | 58 | 51 | 61 | 56 |
| 59 | 74 | 63 | 53 | 50 |
| 59 | 60 | 60 | 57 | 46 |
| 55 | 63 | 57 | 47 | 55 |
| 57 | 43 | 61 | 62 | 49 |
| 67 | 67 | 55 | 55 | 49 |
Problem 31 Solution
Step 1 — Summarize the 30 ages: entering the ages from Table 7.2.5 into a list and running one-variable statistics gives
$$ \overline{x} = 56.57, \qquad s_x = 6.91, \qquad n = 30, \qquad df = 29 $$Step 2 — Split the confidence level: with \(CL = 0.90\), \(\alpha = 0.10\) and \(\frac{\alpha}{2} = 0.05\).
Step 3 — Find the t-score: invT(0.95, 29) gives \(t_{0.05} = 1.699\).
Step 4 — Compute the error bound:
$$ EBM = (1.699)\left(\frac{6.91}{\sqrt{30}}\right) = (1.699)(1.261) = 2.14 $$Step 5 — Build the interval:
$$ 56.57 - 2.14 = 54.43, \qquad 56.57 + 2.14 = 58.71 $$Answer: the 90% confidence interval is about (54.43, 58.71). We estimate with 90% confidence that the mean age of CEOs at these top small firms is between roughly 54 and 59 years.
Problem 32. Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.
a) Identify the following:
i. \(\overline{x} =\) ____
ii. \(s_x =\) ____
iii. \(n =\) ____
iv. \(n - 1 =\) ____
b) Define the random variables \(X\) and \(\overline{X}\) in words.
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 92% confidence interval for the population mean number of unoccupied seats per flight.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
Problem 32 Solution
Step 1 — Read off the statistics: \(\overline{x} = 11.6\) seats, \(s_x = 4.1\) seats, \(n = 225\), \(n - 1 = 224\).
Step 2 — Define the variables: \(X\) is the number of unoccupied seats on a single flight; \(\overline{X}\) is the mean number of unoccupied seats from a sample of 225 flights.
Step 3 — Pick the distribution: we will use a Student's t-distribution, because we do not know the population standard deviation. With \(df = 224\), \(\overline{X} \sim t_{224}\).
Step 4 — Find the t-score: for \(CL = 0.92\), \(\alpha = 0.08\) and \(\frac{\alpha}{2} = 0.04\), so invT(0.96, 224) gives \(t_{0.04} = 1.755\).
Step 5 — Compute the error bound:
$$ EBM = (1.755)\left(\frac{4.1}{\sqrt{225}}\right) = (1.755)(0.2733) = 0.48 $$Step 6 — Build the interval:
$$ 11.6 - 0.48 = 11.12, \qquad 11.6 + 0.48 = 12.08 $$Answer: a) i. 11.6; ii. 4.1; iii. 225; iv. 224. b) as defined in Step 2. c) Student's t with 224 degrees of freedom, because \(\sigma\) is unknown. d) i. the 92% confidence interval is (11.12, 12.08); ii. sketch a t-curve centered at 11.6 with those limits and 0.04 of area in each tail; iii. \(EBM = 0.48\) seats.
Problem 33. In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal.
a) Which distribution should you use for this problem? Explain your choice.
b) Define the random variable \(\overline{X}\) in words.
c) Construct a 95% confidence interval for the population mean cost of a used car.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
d) Explain what a "95% confidence interval" means for this study.
Problem 33 Solution
Step 1 — Pick the distribution: the $3,156 is a sample standard deviation, so \(\sigma\) is unknown and we use Student's t with \(df = 84 - 1 = 83\).
Step 2 — Define the variable: \(\overline{X}\) is the mean cost of a used car for a sample of 84 used car sales.
Step 3 — Find the t-score: for \(CL = 0.95\), invT(0.975, 83) gives \(t_{0.025} = 1.989\).
Step 4 — Compute the error bound:
$$ EBM = (1.989)\left(\frac{3{,}156}{\sqrt{84}}\right) = (1.989)(344.35) = 684.91 $$Step 5 — Build the interval:
$$ 6{,}425 - 684.91 = 5{,}740.09, \qquad 6{,}425 + 684.91 = 7{,}109.91 $$Answer: a) Student's t with 83 degrees of freedom, because the population standard deviation is unknown. b) as defined in Step 2. c) the 95% confidence interval is about ($5,740, $7,110) with \(EBM \approx \$685\); sketch a t-curve centered at $6,425 with those limits. d) it means the method succeeds 95% of the time: if many samples of 84 used car sales were taken and an interval built from each, about 95% of those intervals would contain the true mean cost.
Problem 34. Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal.
a) Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
b) If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done?
c) Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.
d) Calculate the mean.
e) Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not?
Problem 34 Solution
Step 1 — Summarize the six fat measurements: for 8, 8, 10, 7, 9, 9,
$$ \overline{x} = \frac{51}{6} = 8.5, \qquad s_x = 1.049, \qquad n = 6, \qquad df = 5 $$Step 2 — Find the t-score: with \(CL = 0.90\), \(\frac{\alpha}{2} = 0.05\), and invT(0.95, 5) gives \(t_{0.05} = 2.015\).
Step 3 — Compute the error bound:
$$ EBM = (2.015)\left(\frac{1.049}{\sqrt{6}}\right) = (2.015)(0.4282) = 0.86 $$Step 4 — Build the interval:
$$ 8.5 - 0.86 = 7.64, \qquad 8.5 + 0.86 = 9.36 $$Answer: a) i. the 90% confidence interval is (7.64, 9.36); ii. sketch a t-curve centered at 8.5 with those limits and 0.05 of area in each tail; iii. \(EBM = 0.86\) grams. b) increase the sample size — sampling more than six brands shrinks \(\frac{s}{\sqrt{n}}\) and also lowers the t-score, so the error bound falls without giving up any confidence. c), d), e) these parts depend on the brands you record at your own store, so answers vary. Compare your mean against (7.64, 9.36): landing inside is the common outcome, since a 90% interval is built to capture the population mean about nine times in ten, but a store whose brands skew toward low-fat or premium products can easily produce a mean outside it.
Problem 35. A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.
a) Identify the following:
i. \(\overline{x} =\) ____
ii. \(s_x =\) ____
iii. \(n =\) ____
iv. \(n - 1 =\) ____
b) Define the random variables \(X\) and \(\overline{X}\) in words.
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 95% confidence interval for the population mean worth of coupons.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
e) If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why.
Problem 35 Solution
Step 1 — Put every value in the same units: work in cents, so the $1.50 coupon becomes 150¢. The 14 values are 20, 75, 50, 65, 30, 55, 40, 40, 30, 55, 150, 40, 65, 40.
Step 2 — Summarize the sample:
$$ \overline{x} = \frac{755}{14} = 53.93, \qquad s_x = 31.63, \qquad n = 14, \qquad n - 1 = 13 $$Step 3 — Define the variables: \(X\) is the value in cents of one coupon; \(\overline{X}\) is the mean value of the 14 sampled coupons.
Step 4 — Pick the distribution: \(\sigma\) is unknown, so use \(t_{13}\).
Step 5 — Find the t-score: for \(CL = 0.95\), invT(0.975, 13) gives \(t_{0.025} = 2.160\).
Step 6 — Compute the error bound:
$$ EBM = (2.160)\left(\frac{31.63}{\sqrt{14}}\right) = (2.160)(8.455) = 18.26 $$Step 7 — Build the interval:
$$ 53.93 - 18.26 = 35.67, \qquad 53.93 + 18.26 = 72.19 $$Answer: a) i. 53.93¢; ii. 31.63¢; iii. 14; iv. 13. b) as defined in Step 3. c) \(t_{13}\), because the population standard deviation is unknown. d) the 95% confidence interval is about (35.67¢, 72.19¢) with \(EBM = 18.26\)¢; the interval is wide because one $1.50 coupon sits far above the rest and inflates \(s\). e) about 95% of them. The confidence level is a statement about how often the procedure captures \(\mu\), so repeating it with fresh samples of size 14 produces intervals that contain the true mean roughly 95 times in 100.
Use the following information to answer the next two exercises. A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed.
Problem 36. Find the 95% confidence interval for the true population mean for the amount of soda served.
a) (12.42, 14.18)
b) (12.32, 14.29)
c) (12.50, 14.10)
d) Impossible to determine
Problem 36 Solution
Step 1 — Collect the statistics: \(\overline{x} = 13.30\) oz, \(s_x = 1.55\) oz, \(n = 12\), \(df = 11\).
Step 2 — Find the t-score: for \(CL = 0.95\), invT(0.975, 11) gives \(t_{0.025} = 2.201\).
Step 3 — Compute the error bound:
$$ EBM = (2.201)\left(\frac{1.55}{\sqrt{12}}\right) = (2.201)(0.4474) = 0.985 $$Step 4 — Build the interval:
$$ 13.30 - 0.985 = 12.32, \qquad 13.30 + 0.985 = 14.29 $$Answer: b. (12.32, 14.29)
Problem 37. What is the error bound?
a) 0.87
b) 1.98
c) 0.99
d) 1.74
Problem 37 Solution
Step 1 — Reuse the error bound from the previous problem: \(EBM = t_{0.025}\left(\frac{s}{\sqrt{n}}\right) = (2.201)(0.4474) = 0.985\).
Step 2 — Sanity-check it against the interval: the interval (12.32, 14.29) has width 1.97, and half of that is 0.985 — the same number, as it must be.
Answer: c. 0.99 (the error bound is 0.985, which rounds to 0.99).
Key Terms
Student's t-distribution — the distribution followed by the t-score \(\frac{\overline{x} - \mu}{s/\sqrt{n}}\) when a sample of size \(n\) is drawn from an approximately normal population with unknown \(\sigma\); symmetric about zero with thicker tails than the standard normal.
degrees of freedom (df) — the count \(n - 1\), the number of deviations from the sample mean that are free to vary; it selects which t-curve applies.
t-score — the standardized distance \(\frac{\overline{x} - \mu}{s/\sqrt{n}}\) of a sample mean from the population mean, measured in estimated standard errors.
error bound for a population mean (EBM) — the half-width \(\left(t_{\frac{\alpha}{2}}\right)\left(\frac{s}{\sqrt{n}}\right)\) added to and subtracted from \(\overline{x}\) to form the confidence interval.