Chapter 10.2, Problem 10.39E

(a)

To determine

To construct: and interpret a 95% confidence internal for the mean score in the population of all young women.

Answer to Problem 10.39E

For 95% confidence interval

(271.536, 278.464)

Explanation of Solution

Given:

  $\begin{array}{l}n=1077\\ \overline{x}=275\\ s=58\end{array}$

Formula used:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}$

For intervals

  $\begin{array}{l}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

  $c=95\%=0.95$

For confidence level $1-\alpha =0.95$ , $z_{a/2}=z_{0.025}$ using the normal probability

  $z_{a/2}=1.96$

The margin of error is then:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}=1.96\times \frac{58}{\sqrt{1077}}=3.4640$

The boundaries of the confidence interval then become:

  $\begin{array}{l}\overline{x}-E=275-3.4640=271.536\\ \overline{x}+E=275+3.4640=278.464\end{array}$

It is 95% confident that the mean quantitative score of all young women is between 271.536 and 278.464.

(b)

To determine

To Calculate: the 90% and 99% confidence intervals for mean.

Answer to Problem 10.39E

For 90% confidence interval

(270.1056, 277.8944)

For 99% confidence interval

(270.4403, 279.5597)

Explanation of Solution

Formula used:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}$

For intervals

  $\begin{array}{l}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

90% confidence interval

  $c=90\%=0.90$

For confidence level $1-\alpha =0.90$ , $z_{a/2}=z_{0.05}$ using the normal probability

  $z_{a/2}=1.64$

The margin of error is then:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}=1.64\times \frac{58}{\sqrt{1077}}=2.8944$

The boundaries of the confidence interval then become;

  $\begin{array}{l}\overline{x}-E=275-2.8944=272.1056\\ \overline{x}+E=275+2.8944=277.8944\end{array}$

99% confidence interval

  $c=99\%=0.99$

For confidence level $1-\alpha =0.99$ , $z_{a/2}=z_{0.005}$ using the normal probability

  $z_{a/2}=2.58$

The margin of error is then:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}=2.58\times \frac{58}{\sqrt{1077}}=4.5597$

The boundaries of the confidence interval then become;

  $\begin{array}{l}\overline{x}-E=275-4.5597=270.4403\\ \overline{x}+E=275+4.597=279.5597\end{array}$

(c)

To determine

To Explain: the margins of error for 90%, 95% and 99% confidence and increasing the confidence level affect the margin of error of a confidence interval.

Answer to Problem 10.39E

2.8944, 3.4640, 4.5597, increases

Explanation of Solution

The margins of errors were determined in part (a) and (b):

90% confidence interval: 2.8944

95% confidence interval: 3.4640

99% confidence interval: 4.5597

It is observed that the margin of error increases as the confidence level increases.