Student’s t-probabilities

In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for [latex]\sigma[/latex] and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.

William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing [latex]\sigma[/latex] with [latex]s[/latex] did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name "Student."

Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and only used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for [latex]\sigma[/latex].

If you draw a simple random sample of size n from a population that has an approximately a normal distribution with mean μ and unknown population standard deviation [latex]\sigma[/latex] and calculate the t-score [latex]t = \frac{\overline{x}–\mu }{\left(\frac{s}{\sqrt{n}}\right)}[/latex], then the t-scores follow a Student's t-distribution with n – 1 degrees of freedom (df). The t-score has the same interpretation as the z-score. It measures how far [latex]\overline{x}[/latex] is from its mean [latex]\mu[/latex]. For each sample size [latex]n[/latex], there is a different Student's t-distribution.

The degrees of freedom, [latex]n – 1[/latex], come from the calculation of the sample standard deviation s. Previously, we used n deviations [latex]\left(x–\overline{x}\text{values}\right)[/latex] to calculate s. Because the sum of the deviations is zero, we can find the last deviation once we know the other [latex]n – 1[/latex] deviations. The other [latex]n – 1[/latex] deviations can change or vary freely. We call the number n – 1 the degrees of freedom.

Properties of the Student's t-Distribution

A probability table for the Student's t-distribution can also be used to calculate Student's t-probabilities. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row). When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only the corresponding area in one or both tails. A Student's t table gives t-scores given the degrees of freedom and the right-tailed probability.

There is one in this book found in the Back Matter - Statistics Tables,

If the population standard deviation is not known and the population has an approximate normal distribution, the error bound for a population mean is:

• [latex]\text{EBM}=\left({t}_{\frac{\alpha }{2}}\right)\left(\frac{s}{\sqrt{n}}\right)[/latex],
• [latex]{t}_{\frac{\sigma }{2}}[/latex] is the t-score with area to the right equal to [latex]\frac{\alpha }{2}[/latex],
• Use [latex]df = n – 1[/latex] degrees of freedom, and s = sample standard deviation.

The format for the confidence interval is:

[latex]\left(\overline{x} - \text{EBM}, \overline{x} + \text{EBM}\right)[/latex].

Section 7.2 Review

In many cases, the researcher does not know the population standard deviation, [latex]\sigma[/latex], of the measure being studied. In these cases, it is common to use the sample standard deviation, s, as an estimate of [latex]\sigma[/latex]. The normal distribution creates accurate confidence intervals when [latex]\sigma[/latex] is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula: [latex]t = \frac{\overline{x}–\mu }{\left(\frac{s}{\sqrt{n}}\right)}[/latex]

The t-score follows the Student’s t-distribution with [latex]n - 1[/latex] degrees of freedom. The confidence interval under this distribution is calculated with [latex]\text{EBM} = \left({t}_{\frac{\alpha }{2}}\right)\frac{s}{\sqrt{n}}[/latex] where [latex]{t}_{\frac{\alpha }{2}}[/latex] is the t-score with area to the right equal to [latex]\frac{\alpha }{2}[/latex], [latex]s[/latex] is the sample standard deviation, and [latex]n[/latex] is the sample size.

Formula Review

• [latex]s =[/latex]the standard deviation of sample values.
• [latex]t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}[/latex] is the formula for the t-score which measures how far away a measure is from the population mean in the Student’s t-distribution
• [latex]df = n - 1[/latex]; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample
• [latex]T \sim t_{df}[/latex] the random variable, [latex]T[/latex], has a Student’s t-distribution with [latex]df[/latex] degrees of freedom
• [latex]\text{EBM}={t}_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}}[/latex] = the error bound for the population mean when the population standard deviation is unknown
• [latex]{t}_{\frac{\alpha }{2}}[/latex] is the t-score in the Student’s t-distribution with area to the right equal to [latex]\frac{\alpha }{2}[/latex]
• The general form for a confidence interval for a single mean, population standard deviation unknown, Student's t is given by
  • [latex](\text{lower bound, upper bound}) = (\text{point estimate} - \text{EBM}, \text{point estimate} + \text{EBM})
  • = \left(\overline{x}–\frac{ts}{\sqrt{n}},\overline{x} + \frac{ts}{\sqrt{n}}\right)[/latex]

Section 7.2 Practice

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. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.

1. Identify the following:
   • [latex]\overline{x} =\underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} =\underline{\hspace{2cm}}[/latex]
   • [latex]n =\underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 =\underline{\hspace{2cm}}[/latex]
2. Define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex] in words.
3. Which distribution should you use for this 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.
5. Explain in complete sentences what the confidence interval means.

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.

1. Identify the following:
   • [latex]\overline{x}=\underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} =\underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 =\underline{\hspace{2cm}}[/latex]
2. Define the random variable [latex]X[/latex] in words.
3. Define the random variable [latex]\overline{X}[/latex] in words.
4. Which distribution should you use for this problem?
5. Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound.
6. Why would the error bound change if the confidence level were lowered to 95%?

The data in the table below 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.

Table 1: x and Frequency for 39 national flags from various countries 

X	Freq.
1	1
2	7
3	18
4	7
5	6

1. Calculate the following:
   • [latex]\overline{x}=\underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} =\underline{\hspace{2cm}}[/latex]
   • [latex]n =\underline{\hspace{2cm}}[/latex]
2. Define the random variable [latex]\overline{X}[/latex] in words.
3. What is [latex]\overline{x}[/latex] estimating?
4. Is [latex]{\sigma }_{x}[/latex] known?
5. As a result of your answer to exercise 4, 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.

1. How much area is in both tails (combined)?
2. How much area is in each tail?
3. Calculate the following:
   • lower limit
   • upper limit
   • error bound
4. The 95% confidence interval is [latex]\underline{\hspace{2cm}}[/latex].
5. Fill in the blanks on the graph with the areas, the upper and lower limits of the Confidence Interval and the sample mean.
6. In one complete sentence, explain what the interval means.
7. Using the same [latex]\overline{x}[/latex], [latex]{s}_{x}[/latex], and level of confidence, suppose that [latex]n[/latex] were 69 instead of 39. Would the error bound become larger or smaller? How do you know?
8. Using the same [latex]\overline{x}[/latex], [latex]{s}_{x}[/latex], and [latex]n = 39[/latex], how would the error bound change if the confidence level were reduced to 90%? Why?

The image below is to assist with confidence intervals. Remember, a normal curve has a peak at [latex]\overline{x}[/latex], points [latex]\overline{x}-\text{EBM}[/latex] and [latex]\overline{x}+\text{EBM}[/latex] labeled with each unshaded tail has area a [latex]\frac{\alpha }{2}[/latex].

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.

1. Define the random variables [latex]X[/latex] and [latex]P^{\prime}[/latex] in words.
2. Which distribution should you use for this problem? Explain your choice
3. Calculate [latex]p^{\prime}[/latex].
4. Construct a 96% confidence interval for the population proportion of Bam-Bam snack pieces per bag.
   • State the confidence interval.
   • Sketch the graph.
   • 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. Identify the following:
   • [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 = \underline{\hspace{2cm}}[/latex]
2. Define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex] in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
5. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why?

Solution

1. Identify the following:
   •  8629
   • 6944
   • 35
   • 34
2. [latex]X[/latex] is the number of students enrolled at a single community college. [latex]\overline{X}[/latex] is the mean number of students enrolled in the sample of 35 community colleges.
3. [latex]{t}_{34}[/latex]
4. Construct a 95% confidence interval
   • [latex]\text{CI}: (6244, 11,014)[/latex]
   • Graph below.
   • [latex]\text{EBM} = 2385[/latex]
5. It will become smaller.

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.

1. Identify the following:
   • [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 = \underline{\hspace{2cm}}[/latex]
2. Define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex] in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95% confidence interval for the population mean time wasted.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
5. Explain in a complete sentence what the confidence interval means.

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 the 2011–2012 election cycle. The data below shows the total receipts during this cycle for a random selection of 20 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

[latex]\overline{x}=$251,854.23[/latex]

[latex]s=\$521,130.41[/latex]

Use this sample data to construct a 96% 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.

Solution

[latex]\overline{x}=$251,854.23[/latex]

[latex]s=\$521,130.41[/latex]

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 [latex]n = 30[/latex], and [latex]df = 30 - 1 = 29[/latex]

[latex]\text{CL} = 0.96[/latex], so [latex]\alpha = 1 - \text{CL} = 1 - 0.96 = 0.04[/latex]

[latex]\frac{\alpha }{2}=0.02{t}_{\frac{\alpha }{2}}={t}_{0.02}= 2.150[/latex]

[latex]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[/latex]

[latex]\overline{x}[/latex] - EBM = $251,854.23 - $204,561.66 = $47,292.57

[latex]\overline{x}[/latex] + EBM = $251,854.23+ $204,561.66 = $456,415.89

We estimate with 96% 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.

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 [latex]\overline{X}[/latex] in words.
3. Construct a 95% confidence interval for the population mean cost of a used car.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
4. Explain what a “95% 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% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets.
   • State the confidence interval.
   • Sketch the graph.
   • 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?

Solution

1. Construct a 90% confidence interval
   • [latex]\text{CI}: (7.64 , 9.36)[/latex]
   • Graph below
   • [latex]\text{EBM}: 0.86[/latex]
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 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.

1. Identify the following:
   • [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 = \underline{\hspace{2cm}}[/latex]
2. Define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex] in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 95% confidence interval for the population mean worth of coupons.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
5. 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.

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.

1. Find the 95% Confidence Interval for the true population mean for the amount of soda served.
2. What is the error bound?

References

“America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013).

Data from Microsoft Bookshelf.

Data from http://www.businessweek.com/.

Data from http://www.forbes.com/.

“Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013).

“Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013).

“Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013).