Chapter 8.5, Problem 21SE

A random sample of 10 items is drawn from a population whose standard deviation is unknown. The sample mean is $\bar{x}$ = 270 and the sample standard deviation is s = 20. Use Appendix D to find the values of Student’s t.

1. a. Construct an interval estimate of μ with 95 percent confidence.
2. b. Repeat part a assuming that n = 20.
3. c. Repeat part a assuming that n = 40.
4. d. Describe how the confidence interval changes as n increases.

To determine

Using the value of student’s t from Appendix D, construct a 95% confidence interval estimate for $\mu$.

Answer to Problem 21SE

The 95% confidence interval estimate for $\mu$ is (255.694, 284.306).

Explanation of Solution

Calculation:

The given information is that a random sample of 10 items is drawn from a population whose standard deviation is unknown. The sample mean is 270 and sample standard deviation is 20.

Since, the population standard deviation is unknown; the sampling distribution is t-distribution.

Confidence interval:

The confidence interval estimate for $\mu$ is,

$\overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)$

where $\overline{x}$ is the sample mean $t_{\frac{\alpha }{2}}$ is the student’s t value and  $\frac{s}{\sqrt{n}}$ is the estimated standard error of the mean.

The degrees of freedom for the test is 9$\left(=10-1\right)$ and value of student’s t for a 95% confidence interval is obtained.

Procedure the value of Student’s t using Appendix D:

• Go through the row corresponding to the degrees of freedom 9 in Appendix D of student’s t critical values.
• Go through the row corresponding to 9 and column corresponding to the confidence level 0.95.
• Obtain the value corresponding to (9, 0.95) from the table.

Thus, the value $t_{\frac{\alpha }{2}}$ is 2.262.

Substitute $\overline{x}=270,t_{\frac{\alpha }{2}}=2.262,s=20\text{ and }n=10$. The confidence interval estimate for $\mu$ is,

$\begin{array}{c}\overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=\left[270-2.262\left(\frac{20}{\sqrt{10}}\right),270+2.262\left(\frac{20}{\sqrt{10}}\right)\right]\\ =\left[270-2.262\left(6.3245\right),270+2.262\left(6.3245\right)\right]\\ =\left[270-14.306,270+14.306\right]\\ =\left(255.694,284.306\right)\end{array}$

Thus, the 95% confidence interval estimate for $\mu$ is (255.694, 284.306).

To determine

Using the value of student’s t from Appendix D, construct a 95% confidence interval estimate for $\mu$.

Answer to Problem 21SE

The 95% confidence interval estimate for $\mu$ is (260.64, 279.36).

Explanation of Solution

Calculation:

The given information is that a random sample of 20 items is drawn from a population whose standard deviation is unknown. The sample mean is 270 and sample standard deviation is 20.

Since, the population standard deviation is unknown; the sampling distribution is t-distribution.

The degrees of freedom for the test is 19$\left(=20-1\right)$ and value of student’s t for a 95% confidence interval is obtained.

Procedure the value of Student’s t using Appendix D:

• Go through the row corresponding to the degrees of freedom 19 in Appendix D of student’s t critical values.
• Go through the row corresponding to 19 and column corresponding to the confidence level 0.95.
• Obtain the value corresponding to (19, 0.95) from the table.

Thus, the value $t_{\frac{\alpha }{2}}$ is 2.093.

Substitute $\overline{x}=270,t_{\frac{\alpha }{2}}=2.093,s=20\text{ and }n=20$. The confidence interval estimate for $\mu$ is,

$\begin{array}{c}\overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=\left[270-2.093\left(\frac{20}{\sqrt{20}}\right),270+2.093\left(\frac{20}{\sqrt{20}}\right)\right]\\ =\left[270-2.093\left(4.4721\right),270+.093\left(4.4721\right)\right]\\ =\left[270-9.3601,270+9.3601\right]\\ =\left(260.64,279.36\right)\end{array}$

Thus, the 95% confidence interval estimate for $\mu$ is (260.64, 279.36).

To determine

Using the value of student’s t from Appendix D, construct a 95% confidence interval estimate for $\mu$.

Answer to Problem 21SE

The 95% confidence interval estimate for $\mu$ is (263.603, 276.397).

Explanation of Solution

Calculation:

The given information is that a random sample of 40 items is drawn from a population whose standard deviation is unknown. The sample mean is 270 and sample standard deviation is 20.

Since, the population standard deviation is unknown; the sampling distribution is t-distribution.

The degrees of freedom for the test is 39$\left(=40-1\right)$ and value of student’s t for a 95% confidence interval is obtained.

Procedure the value of Student’s t using Appendix D:

• Go through the row corresponding to the degrees of freedom 39 in Appendix D of student’s t critical values.
• Go through the row corresponding to 39 and column corresponding to the confidence level 0.95.
• Obtain the value corresponding to (39, 0.95) from the table.

Thus, the value $t_{\frac{\alpha }{2}}$ is 2.023.

Substitute $\overline{x}=270,t_{\frac{\alpha }{2}}=2.023,s=20\text{ and }n=40$. The confidence interval estimate for $\mu$ is,

$\begin{array}{c}\overline{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=\left[270-2.023\left(\frac{20}{\sqrt{40}}\right),270+2.093\left(\frac{20}{\sqrt{40}}\right)\right]\\ =\left[270-2.023\left(3.1622\right),270+2.023\left(3.1622\right)\right]\\ =\left[270-6.3971,270+6.3971\right]\\ =\left(260.603,276.39\right)\end{array}$

Thus, the 95% confidence interval estimate for $\mu$ is (260.603, 276.39).

To determine

Explain how the intervals changes as the sample size increases.

Explanation of Solution

As the sample size increases the width of the confidence interval decreases. From the results obtained in parts (a), (b) and (c), the 95% confidence interval for a sample of size 10 is wider than confidence interval for a sample of size 20 which is again wider l for a sample of size 40. A greater confidence implies a greater margin of error or there will be a loss of precision. Since the confidence interval becomes narrower, precise values can be obtained.

Thus, as the confidence level decreases the sample size increases.