Chapter 8.2, Problem 48E

To determine

To use your interval from exercise $40$ to construct and interpret a $99\%$ confidence interval for the total number of students at the school that the student body president can identify by name and then use your interval to evaluate the president’s claim.

Answer to Problem 48E

We are then $99\%$ confident that the number of students that the student body present knows is between $597$ students and $1059$ students.

Explanation of Solution

It is given that,

  $\begin{array}{l}c=99\%=0.99\\ x=46\\ n=100\end{array}$

Conditions to be met are:

The three conditions for calculating a confidence interval for the population proportion p are:

Random: It is satisfied because the samples are randomly selected.

Independent: It is satisfied because the sample of $100$ student chips is less than $10\%$ of the population of all $1800$ students.

Normal: It is satisfied because there are $46$ successes and $100-46=54$ failures, which are about at least ten.

Thus, all the three conditions are satisfied.

The calculation is as:

The sample proportion is the number of successes divided by the sample size as:

  $\begin{array}{l}\widehat{p}=\frac{x}{n}\\ =\frac{46}{100}\\ =0.46\end{array}$

For the confidence interval $1-\alpha =0.99$ determine $z_{\alpha /2}=z_{0.005}$ using the normal probability table in the appendix as:

  $z_{\alpha /2}=2.575$

Thus, the margin of error is then:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =2.575\times \sqrt{\frac{0.46(1-0.46)}{100}}\\ =0.1283\end{array}$

Thus, the confidence interval is as:

  $\begin{array}{l}\widehat{p}-E=0.46-0.1283\\ =0.3317\\ \widehat{p}+E=0.46+0.1283\\ =0.5883\end{array}$

So, we are $99\%$ confident that the true proportion of all students that the student body president knows is between $(0.3317,0.5883)$ .

However, we know that the population contains $1800$ students. Thus, we have,

  $\begin{array}{l}1800(0.3317)=597.06\\ \approx 597\\ 1800(0.5883)=1058.94\\ \approx 1059\end{array}$

Thus, we are then $99\%$ confident that the number of students that the student body present knows is between $597$ students and $1059$ students.