Chapter 7.2, Problem 23P

(a)

To determine

Find the 90% confidence interval for $\mu$ using method 1.

Find the 95% confidence interval for $\mu$ using method 1.

Find the 99% confidence interval for $\mu$ using method 1.

Answer to Problem 23P

The 90% confidence interval for $\mu$ using method 1is $43.58<\mu <46.82$.

The 95% confidence interval for $\mu$ using method 1is $43.26<\mu <47.14$.

The 99% confidence interval for $\mu$ using method 1is $42.58<\mu <47.82$.

Explanation of Solution

Calculation:

For confidence level 90%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘8’ numbered key.
• In Input, select Stats.
• Enter $\overline{x}$ as 45.2, $s_x$ as 5.3, and n as 31.
• Enter C-Level as 0.90.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(43.58,46.82\right)$.

Hence, the 90% confidence interval for $\mu$ using method 1 is $43.58<\mu <46.82$.

For confidence level 95%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘8’ numbered key.
• In Input, select Stats.
• Enter $\overline{x}$ as 45.2, $s_x$ as 5.3, and n as 31.
• Enter C-Level as 0.95.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(43.26,47.14\right)$.

Hence, the 95% confidence interval for $\mu$ using method 1is $43.26<\mu <47.14$.

For confidence level 99%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘8’ numbered key.
• In Input, select Stats.
• Enter $\overline{x}$ as 45.2, $s_x$ as 5.3, and n as 31.
• Enter C-Level as 0.99.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(42.58,47.82\right)$.

Hence, the 99% confidence interval for $\mu$ using method 1is $42.58<\mu <47.82$.

(b)

To determine

Find the 90% confidence interval for $\mu$ using method 2.

Find the 95% confidence interval for $\mu$ using method 2.

Find the 99% confidence interval for $\mu$ using method 2.

Answer to Problem 23P

The 90% confidence interval for $\mu$ using method 2 is $43.63<\mu <46.77$.

The 95% confidence interval for $\mu$ using method 2 is $43.33<\mu <47.07$.

The 99% confidence interval for $\mu$ using method 2 is $42.75<\mu <47.65$.

Explanation of Solution

Calculation:

For confidence level 90%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘7’ numbered key.
• In Input, select Stats.
• Enter $\sigma$ as 5.3, $\overline{x}$ as 45.2, and n as 31.
• Enter C-Level as 0.90.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(43.63,46.77\right)$.

Hence, the 90% confidence interval for $\mu$ using method 2 is $43.63<\mu <46.77$.

For confidence level 95%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘7’ numbered key.
• In Input, select Stats.
• Enter $\sigma$ as 5.3, $\overline{x}$ as 45.2, and n as 31.
• Enter C-Level as 0.95.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(43.33,47.07\right)$.

Hence, the 95% confidence interval for $\mu$ using method 2 is $43.33<\mu <47.07$.

For confidence level 99%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘7’ numbered key.
• In Input, select Stats.
• Enter $\sigma$ as 5.3, $\overline{x}$ as 45.2, and n as 31.
• Enter C-Level as 0.99.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(42.75,47.65\right)$.

Hence, the 99% confidence interval for $\mu$ using method 2 is $42.75<\mu <47.65$.

(c)

To determine

Compare the confidence intervals of the two methods.

Explain whether the confidence intervals using a Student’s t distribution are more conservative or not.

Explanation of Solution

Calculation:

From part (a), 90% confidence interval using method 1 is $\left(43.58,46.82\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.82-43.58\\ =3.24\end{array}$

The 95% confidence interval using method 1 is $\left(43.26,47.14\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =47.14-43.26\\ =3.88\end{array}$

The 99% confidence interval using method 1 is $\left(42.58,47.82\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =47.82-42.58\\ =5.24\end{array}$

From part (b), the 90% confidence interval using method 2is $\left(43.63,46.77\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.77-43.63\\ =3.14\end{array}$

The 95% confidence interval using method 2is $\left(43.33,47.07\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =47.07-43.33\\ =3.74\end{array}$

The 99% confidence interval using method 2is $\left(42.75,47.65\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =47.65-42.75\\ =4.9\end{array}$

It can be observed that, length of the confidence interval calculated using student’s t distribution is more when compared to standard normal distribution. This shows that, confidence intervals using a Student’s t distribution can be considered as more conservative with respect to length.

(d)

To determine

Find the 90% confidence interval for $\mu$ using method 1 for sample size 81.

Find the 95% confidence interval for $\mu$ using method 1for sample size 81.

Find the 99% confidence interval for $\mu$ using method 1for sample size 81.

Find the 90% confidence interval for $\mu$ using method 2for sample size 81.

Find the 95% confidence interval for $\mu$ using method 2for sample size 81.

Find the 99% confidence interval for $\mu$ using method 2for sample size 81.

Compare the confidence intervals of the two methods.

Explain whether the confidence intervals using a Student’s t distribution are more conservative or not.

Answer to Problem 23P

The 90% confidence interval for $\mu$ using method 1for sample size 81 is $44.22<\mu <46.18$.

The 95% confidence interval for $\mu$ using method 1for sample size 81 is $44.03<\mu <46.37$.

The 99% confidence interval for $\mu$ using method 1for sample size 81 is $43.65<\mu <46.75$.

The 90% confidence interval for $\mu$ using method 2for sample size 81 is $44.23<\mu <46.17$.

The 95% confidence interval for $\mu$ using method 2for sample size 81 is $44.05<\mu <46.35$.

The 99% confidence interval for $\mu$ using method 2for sample size 81 is $43.68<\mu <46.72$.

Explanation of Solution

Calculation:

For confidence level 90%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘8’ numbered key.
• In Input, select Stats.
• Enter $\overline{x}$ as 45.2, $s_x$ as 5.3, and n as 81.
• Enter C-Level as 0.90.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(44.22,46.18\right)$.

Hence, the 90% confidence interval for $\mu$ using method 1for sample size 81 is $44.22<\mu <46.18$.

For confidence level 95%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘8’ numbered key.
• In Input, select Stats.
• Enter $\overline{x}$ as 45.2, $s_x$ as 5.3, and n as 81.
• Enter C-Level as 0.95.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(44.03,46.37\right)$.

Hence, the 95% confidence interval for $\mu$ using method 1for sample size 81 is $44.03<\mu <46.37$.

For confidence level 99%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘8’ numbered key.
• In Input, select Stats.
• Enter $\overline{x}$ as 45.2, $s_x$ as 5.3, and n as 81.
• Enter C-Level as 0.99.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(43.65,46.75\right)$.

Hence, the 99% confidence interval for $\mu$ using method 1for sample size 81 is $43.65<\mu <46.75$.

For confidence level 90%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘7’ numbered key.
• In Input, select Stats.
• Enter $\sigma$ as 5.3, $\overline{x}$ as 45.2, and n as 81.
• Enter C-Level as 0.90.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(44.23,46.17\right)$.

Hence, the 90% confidence interval for $\mu$ using method 2for sample size 81 is $44.23<\mu <46.17$.

For confidence level 95%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘7’ numbered key.
• In Input, select Stats.
• Enter $\sigma$ as 5.3, $\overline{x}$ as 45.2, and n as 81.
• Enter C-Level as 0.95.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(44.05,46.35\right)$.

Hence, the 95% confidence interval for $\mu$ using method 2for sample size 81 is $44.05<\mu <46.35$.

For confidence level 99%:

Use Ti 83 calculator tofind the confidence interval as follows:

• Select STAT> take the arrow to the TEST menu and then enter ‘7’ numbered key.
• In Input, select Stats.
• Enter $\sigma$ as 5.3, $\overline{x}$ as 45.2, and n as 81.
• Enter C-Level as 0.99.
• Click Enter.

Output using Ti 83 calculator is given below:

From the Ti 83 calculator output, the confidence interval is $\left(43.68,46.72\right)$.

Hence, the 99% confidence interval for $\mu$ using method 2for sample size 81 is $43.68<\mu <46.72$.

The 90% confidence interval using method 1for sample size 81 is $\left(44.22,46.18\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.18-44.22\\ =1.96\end{array}$

The 95% confidence interval using method 1for sample size 81 is $\left(44.03,46.37\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.37-44.03\\ =2.34\end{array}$

The 99% confidence interval using method 1for sample size 81 is $\left(43.65,46.75\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.75-43.65\\ =3.1\end{array}$

The 90% confidence interval using method 2for sample size 81 is $\left(44.23,46.17\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.17-44.23\\ =1.94\end{array}$

The 95% confidence interval using method 2for sample size 81 is $\left(44.05,46.35\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.35-44.05\\ =2.3\end{array}$

The 99% confidence interval using method 2for sample size 81 is $\left(43.68,46.72\right)$. The length of interval is,

$\begin{array}{c}\text{length}=\text{Upper}\text{limit}-\text{lower}\text{limit}\\ =46.72-43.68\\ =3.04\end{array}$

It can be observed that, length of the confidence interval calculated using student’s t distribution is more when compared to standard normal distribution. But the difference between the lengths of the confidence intervals is less for method 1 and 2 for sample size $n=81$ when compared to sample size $n=31$. As the sample size increases, the confidence intervals for student’s t distribution and standard normal distribution might be approximately closer to each other.