A Single Population Mean Using the Student's t-Distribution

Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95 percent confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.
The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators.

• 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

• The first solution is step-by-step (Solution A).
• The second solution uses the TI-83+ and TI-84 calculators (Solution B).

To find the confidence interval, you need the sample mean, $\overline{x}$, and the EBM.

$\begin{array}{c}\overline{x}=\frac{8.6 + 9.4 + 7.9 + 6.8 + 8.3 + 7.3 + 9.2 + 9.6 + 8.7 + 11.4 + 10.3 + 5.4 + 8.1+ 5.5+ 6.9}{15}=8.2267;\\ \begin{array}{c}\\ s=\frac{{(8.6-\overline{x})}^2+ {(9.4 - \overline{x})}^2+\cdots +{(5.5 - \overline{x})}^2+ {(6.9 - \overline{x})}^2}{14}=1.6722;\end{array}\\ \begin{array}{c}\\ n=15\end{array}\end{array}$

df = 15 – 1 = 14 CL, so α = 1 – CL = 1 – 0.95 = 0.05

$\frac{\alpha }{2}$ = 0.025; $t_{\frac{\alpha }{2}}=t_{0.025}$

The area to the right of t_0.025 is 0.025, and the area to the left of t_0.025 is 1 – 0.025 = 0.975.

$t_{\frac{\alpha }{2}}=t_{0.025}=2.14$ using invT(.975,14) on the TI-84+ calculator.

$EBM=\left(t_{\frac{\alpha }{2}}\right)\left(\frac{s}{\sqrt{n}}\right)$ $EBM=\left(2.14\right)\left(\frac{1.6722}{\sqrt{15}}\right)=0.924$ $\overline{x}–EBM\text{ = 8}\text{.2267 – 0}\text{.9240 = 7}\text{.3}$ $\overline{x}+EBM\text{= 8}\text{.2267 + 0}\text{.9240 = 9}\text{.15}$

The 95 percent confidence interval is (7.30, 9.15).

We estimate with 95 percent confidence that the true population mean sensory rate is between 7.30 and 9.15.

A group of researchers 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. In October 2008, the scientists 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. eip-222 shows how many of the targeted chemicals were found in each infant’s cord blood.

Use this sample data to construct a 90 percent confidence interval for the mean number of targeted industrial chemicals to be found in an infant’s blood.

Solution A

From the sample data, you can calculate $\begin{array}{c}\begin{array}{c}\\ \overline{x}=\frac{79 + 145 + \cdots + 139+ 99}{20}=127.45\\ \end{array}\\ s=\frac{{(79-\overline{x})}^2+ {(145 - \overline{x})}^2+\cdots +{(139 - \overline{x})}^2+ {(99 - \overline{x})}^2}{19}=25.965\text{.}\end{array}$ There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19.

You are asked to calculate a 90 percent confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10.

• $\frac{\alpha }{2}=\mathrm{0.05,}{t}_{\frac{\alpha }{2}}=t_{0.05}$

By definition, the area to the right of t_0.05 is 0.05, and so the area to the left of t_0.05 is 1 – 0.05 = 0.95.

Use a table, calculator, or computer to find that t_0.05 = 1.729.

$\begin{array}{c}EBM=t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=1.729\left(\frac{25.965}{\sqrt{20}}\right) \approx 10.038\\ \overline{x}\text{ – }EBM\text{ = 127}\text{.45 – 10}\text{.038 = 117}\text{.412 }\\ \overline{x}\text{ + }EBM\text{ = 127}\text{.45 + 10}\text{.038 = 137}\text{.488}\end{array}$

We estimate with 90 percent 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.

Solution B

Identify the following:

1. $\overline{x}$ =_______
2. $s_x$ =_______
3. n =_______
4. n – 1 =_______

Define the random variables X and $\overline{X}$ in words.

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.

Which distribution should you use for this problem?

Construct a 95 percent confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.

CI: (1.3808, 1.6192)

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 1.5 on the horizontal axis. A central region is shaded between points 1.38 and 1.62.

EBM = 0.12

Explain in complete sentences what the confidence interval means.

Identify the following:

1. $\overline{x}$ =_______
2. $s_x$ =_______
3. n =_______
4. n – 1 =_______

1. $\overline{x}$ = 151
2. $s_x$ = 32
3. n = 108
4. n – 1 = 107

Define the random variable X in words.

Define the random variable $\overline{X}$ in words.

$\overline{X}$ is the mean number of hours spent watching television per month from a sample of 108 Americans.

Which distribution should you use for this problem?

Construct a 99 percent confidence interval for the population mean hours spent watching television per month. State the confidence interval, sketch the graph, and calculate the error bound.

CI: (142.92, 159.08)

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a normal distribution curve. The peak of the curve coincides with the point 151 on the horizontal axis. A central region is shaded between points 142.92 and 159.08.

EBM = 8.08

Why would the error bound change if the confidence level were lowered to 95 percent?

Calculate the following:

1. $\overline{x}$ =______
2. $s_x$ =______
3. n =______

1. 3.26
2. 1.02
3. 39

Define the random variable $\overline{X}$ in words.

What is $\overline{x}$ estimating?

μ

Is ${\sigma }_x$ known?

As a result of your answer to element-704, state the exact distribution to use when calculating the confidence interval.

t₃₈

How much area is in both tails (combined)?

How much area is in each tail?

0.025

Calculate the following:

1. lower limit
2. upper limit
3. error bound

The 95 percent confidence interval is_____.

(2.93, 3.59)

Fill in the blanks on the graph with the areas, the upper and lower limits of the confidence interval, and the sample mean.

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a template of a normal distribution curve with the central region shaded to represent a confidence interval. The residual areas are on either side of the shaded region. Blanks indicate that students should label the confidence level, residual areas, and points that define the confidence interval.

In one complete sentence, explain what the interval means.

We are 95 percent confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.

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?

The error bound would become EBM = 0.245. This error bound decreases, because as sample sizes increase, variability decreases, and we need less interval length to capture the true mean.

Using the same $\overline{x}$, $s_x$, and n = 39, how would the error bound change if the confidence level were reduced to 90 percent? Why?

In six packages of multicolored fruit snacks, there were five red snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96 percent confidence interval for the population proportion of red snack pieces.

1. Define the random variables X and P′ in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Calculate p′.
4. Construct a 96 percent confidence interval for the population proportion of red snack pieces per bag.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
5. Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?

A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414, 1,550, 2,109, 9,350, 21,828, 4,300, 5,944, 5,722, 2,825, 2,044, 5,481, 5,200, 5,853, 2,750, 10,012, 6,357, 27,000, 9,414, 7,681, 3,200, 17,500, 9,200, 7,380, 18,314, 6,557, 13,713, 17,768, 7,493, 2,771, 2,861, 1,263, 7,285, 28,165, 5,080, 11,622. Assume the underlying population is normal.

1. 1. $\overline{x}$ = __________
   2. $s_x$ = __________
   3. n = __________
   4. n – 1 = __________
2. Define the random variables $X$ and $\overline{X}$ in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95 percent confidence interval for the population mean enrollment at community colleges in the United States.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
5. What will happen to the error bound and confidence interval if 500 community colleges are surveyed? Why?

1. 1. 8,629
   2. 6,944
   3. 35
   4. 34
2. $t_{34}$
3. 1. CI: (6244, 11,014)

   2. View original figure (external; image not reproduced in this text edition)

      Original figure description: A graph showing a bell shaped curve of normal distribution with two vertical lines to the left and right of center. The vertical line to the left is labeled 7.64 and the vertical line to the right is labeled 9.36. The area between the vertical lines is shaded purple. The horizontal axis is unlabeled. There are three tick marks on the horizontal axis at the bottom of the vertical lines and at a point equidistant between the lines. The tick mark in the center of the bell curve is labeled 8.5.

   3. EB = 2385
4. It will become smaller.

Suppose that a committee is studying whether there is wasted 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 8 hr, with a sample standard deviation of 4 hr.

1. 1. $\overline{x}$ = __________
   2. $s_x$ = __________
   3. n = __________
   4. n – 1 = __________
2. Define the random variables $X$ and $\overline{X}$ in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95 percent confidence interval for the population mean time wasted.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
5. Explain in a complete sentence what the confidence interval means.

A pharmaceutical company makes a drug used during surgery. It is assumed that the distribution for the length of time the drug lasts is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the antibiotic drug 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.

1. 1. $\overline{x}$ = __________
   2. $s_x$ = __________
   3. n = __________
   4. n – 1 = __________
2. Define the random variable $X$ in words.
3. Define the random variable $\overline{X}$ in words.
4. Which distribution should you use for this problem? Explain your choice.
5. Construct a 95 percent confidence interval for the population mean length of time.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
6. What does it mean to be 95 percent confident in this problem?

1. 1. $\overline{x}$ = 2.51
   2. $s_x$ = 0.318
   3. n = 9
   4. n - 1 = 8
2. The effective length of time for a tranquilizer
3. The mean effective length of time of tranquilizers from a sample of nine patients
4. We need to use a Student’s t-distribution, because we do not know the population standard deviation.
5. 1. CI: (2.27, 2.76)
   2. Check student's solution.
   3. EBM: 0.25
6. If we were to sample many groups of nine patients, 95 percent of the samples would contain the true population mean length of time.

Suppose that 14 children who were learning to ride two-wheel bikes were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months, with a sample standard deviation of three months. Assume that the underlying population distribution is normal.

1. 1. $\overline{x}$ = __________
   2. $s_x$ = __________
   3. n = __________
   4. n – 1 = __________
2. Define the random variable $X$ in words.
3. Define the random variable $\overline{X}$ in words.
4. Which distribution should you use for this problem? Explain your choice.
5. Construct a 99 percent confidence interval for the population mean length of time using training wheels.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
6. Why would the error bound change if the confidence level were lowered to 90 percent?

The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees during 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 the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs.

$\overline{x}=\$251,854.23$

$s=\$521,130.41$

Use the sample data to construct a 96 percent confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's t-distribution.

$\overline{x}=\$251,854.23\text{;}$

$s=\$521,130.41\text{.}$

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29.

CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04.

$\frac{\alpha }{2}=0.02t_{\frac{\alpha }{2}}=t_{0.02}$ = 2.150.

$EBM=t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=2.150\left(\frac{521,130.41}{\sqrt{30}}\right) ~ \$204,561.66\text{.}$

$\overline{x}$ - EBM = $251,854.23 - $204,561.66 = $47,292.57.

$\overline{x}$ + EBM = $251,854.23 + $204,561.66 = $456,415.89.

We estimate with 96 percent confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

Alternate Solution

The difference between solutions arises from rounding differences.

A major business magazine published data on the best small firms in 2012. These were firms that have 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. eip-884 shows the ages of the corporate CEOs for a random sample of these firms.

Use the sample data to construct a 90 percent confidence interval for the mean age of CEOs for these top small firms. Use the Student's t-distribution.

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected, and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats, and the sample standard deviation is 4.1 seats.

1. 1. $\overline{x}$ = __________
   2. $s_x$ = __________
   3. n = __________
   4. n – 1 = __________
2. Define the random variables $X$ and $\overline{X}$ in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 92 percent confidence interval for the population mean number of unoccupied seats per flight.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.

1. 1. $\overline{x}$ =
   2. $s_x$ =
   3. n =
   4. n - 1 =
2. 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.
3. We will use a Student’s t-distribution, because we do not know the population standard deviation.
4. 1. CI: (11.12 , 12.08)
   2. Check student's solution.
   3. EBM: 0.48

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.

1. Which distribution should you use for this problem? Explain your choice.
2. Define the random variable $\overline{X}$ in words.
3. Construct a 95 percent confidence interval for the population mean cost of a used car.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
4. Explain what a 95 percent confidence interval means for this study.

Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8, 8, 10, 7, 9, 9. Assume the underlying distribution is approximately normal.

1. Construct a 90 percent confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
2. 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?
3. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.
4. Calculate the mean.
5. Is the mean within the interval you calculated in Part a? Did you expect it to be? Why or why not?

1. 1. CI: (7.64, 9.36)

   2. View original figure (external; image not reproduced in this text edition)

      Original figure description: A bell curve showing a normal distribution with a shaded region from 7.64 to 9.36, centered at 8.5, often representing a confidence interval or probability.

   3. EBM: 0.86
2. The sample should have been increased.
3. Answers will vary.
4. Answers will vary.
5. Answers will vary.

A survey of the mean number of cents off given by coupons was conducted by randomly surveying one coupon per page from the coupons section of a local newspaper. 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.

1. 1. $\overline{x}$ = __________
   2. $s_x$ = __________
   3. n = __________
   4. n – 1 = __________
2. Define the random variables $X$ and $\overline{X}$ in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95 percent confidence interval for the population mean worth of coupons.
   1. State the confidence interval.
   2. Sketch the graph.
   3. Calculate the error bound.
5. If many random samples were collected with 14 samples as the size, which percentage of the confidence intervals constructed should contain the population mean worth of coupons? Explain why.

Find the 95 percent confidence interval for the true population mean for the amount of soda served.

1. (12.42, 14.18)
2. (12.32, 14.29)
3. (12.50, 14.10)
4. Impossible to determine

b

Which of the following is the error bound?

1. 0.87
2. 1.98
3. 0.99
4. 1.74