Chapter 8.1, Problem 20P

Confidence Intervals: Values of $\sigma$ A random sample of size 36 is drawn

from an x distribution. The sample mean is 100.

(a) Suppose the x distribution has $\sigma$ = 30. Compute a 90% confidence interval for µ. What is the value of the margin of error?

(b) Suppose the x distribution has $\sigma$ = 20. Compute a 90% confidence interval for µ. What is the value of the margin of error?

(c) Suppose the x distribution has $\sigma$= 10. Compute a 90% confidence interval for What is the value of the margin of error?

(d) Compare the margins of error for pans (a) through (c). As the standard deviation decreases, does the margin of error decrease?

(e)Critical Thinking Compare the lengths of the confidence intervals for pans (a) through (c). As the standard deviation decreases, does the length of a 90% confidence interval decrease?

To determine

To find: A 90% confidence interval for $\mu$ supposing that the x distribution has $\sigma =30.$

Also, find the margin of error.

Answer to Problem 20P

Solution:

The 90% confidence interval for $\mu$ is (91.78, 108.23) and margin of error $E\text{is}\mathrm{8.225.}$

Explanation of Solution

Calculation:

Given that x distribution with sample mean $\overline{x}=100$ and sample size $n=36$. Supposing that the x distribution has $\sigma =30.$

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.90}=1.645\\ \sigma =30\\ n=36\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=1.645\frac{30}{\sqrt{36}}\\ E=8.225\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 100-8.225<\mu <100+8.225\\ 91.78<\mu <108.23\end{array}$

The 90% confidence interval for $\mu$ is (91.78, 108.23).

Margin of error is $E=8.225\text{.}$

To determine

To find: A 90% confidence interval for $\mu$ supposing that the x distribution has $\sigma =20.$

Also, find the margin of error.

Answer to Problem 20P

Solution:

The 90% confidence interval for $\mu$ is (94.52, 105.48) and margin of error $E\text{is}\mathrm{5.483.}$

Explanation of Solution

Calculation:

Given that x distribution with sample mean $\overline{x}=100$ and sample size $n=36$. Supposing that the x distribution has $\sigma =20.$

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.90}=1.645\\ \sigma =20\\ n=36\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=1.645\frac{20}{\sqrt{36}}\\ E=5.483\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 100-5.483<\mu <100+5.483\\ 94.52<\mu <105.48\end{array}$

The 90% confidence interval for $\mu$ is (94.52, 105.48).

Margin of error is $E=\mathrm{5.483.}$

To determine

To find: A 90% confidence interval for $\mu$ supposing that the x distribution has $\sigma =10.$

Also, find the margin of error.

Answer to Problem 20P

Solution:

The 90% confidence interval for $\mu$ is (97.26, 102.74) and margin of error $E\text{is}\mathrm{2.742.}$

Explanation of Solution

Calculation:

Given that x distribution with sample mean $\overline{x}=100$ and sample size $n=36$. Supposing that the x distribution has $\sigma =10.$

We have to find 90% confidence interval,

$\begin{array}{l}c=0.90\\ \text{Using}\text{Table}\text{3}\text{of}\text{Appendix}\\ z_{0.90}=1.645\\ \sigma =10\\ n=36\\ E=z_c\frac{\sigma }{\sqrt{n}}\\ E=1.645\frac{10}{\sqrt{36}}\\ E=2.742\end{array}$

90% confidence interval is

$\begin{array}{l}\overline{x}-E<\mu <\overline{x}+E\\ 100-2.742<\mu <100+2.742\\ 97.26<\mu <102.74\end{array}$

The 90% confidence interval for $\mu$ is (97.26, 102.74).

Margin of error is $E=\mathrm{2.742.}$

To determine

The comparison of the margins of error for parts (a) through (c) and determines whether the margin of error decrease as the standard deviation decreases.

Answer to Problem 20P

Solution:

Yes, the margin of error also decreases as the standard deviation decreases.

Explanation of Solution

For $\sigma =30,$ the margin of error is $E\text{is}\mathrm{8.225.}$

For $\sigma =20,$, the margin of error is $E\text{is}\mathrm{5.483.}$

For $\sigma =10,$, the margin of error is $E=\mathrm{2.742.}$

As standard deviation decreases, the margin of error also decreases.

To determine

The comparison of confidence interval for part (a) through (c). Also determine whether the length of confidence interval decreases, as the standard deviation decreases.

Answer to Problem 20P

Solution:

Yes, the length of confidence interval also decreases, as the standard deviation decreases.

Explanation of Solution

For $\sigma =30,$, the length of confidence interval is $108.23-91.78=\mathrm{16.45.}$

For $\sigma =20,$, the length of confidence interval is $105.83-94.52=\mathrm{11.31.}$

For $\sigma =10,$ the length of confidence interval is $102.42-97.26=\mathrm{5.16.}$

As the standard deviation decreases, the length of confidence interval also decreases.