Chapter 8.2, Problem 47E

To determine

To use your interval from exercise $39$ to construct and interpret a $90\%$ confidence interval for the total number of seniors planning to go to the prom.

Answer to Problem 47E

We are then $90\%$ confident that between $462$ seniors and $618$ seniors in Tonya’s school plan to go to prom.

Explanation of Solution

It is given that,

  $\begin{array}{l}c=90\%=0.90\\ x=36\\ n=50\end{array}$

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 $50$ seniors is less than $10\%$ of the population of all seniors.

Normal: It is satisfied because there are $36$ successes and $50-36=14$ 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:

  $\begin{array}{l}\widehat{p}=\frac{x}{n}\\ =\frac{36}{50}\\ =0.72\end{array}$

For the confidence interval $1-\alpha =0.90$ determine $z_{\alpha /2}=z_{0.05}$ using the normal probability table in the appendix as:

  $z_{\alpha /2}=1.645$

Thus, the margin of error is then:

  $\begin{array}{l}E=z_{\alpha /2}\times \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\\ =1.645\times \sqrt{\frac{0.72(1-0.72)}{50}}\\ =0.1045\end{array}$

Thus, the confidence interval is as:

  $\begin{array}{l}\widehat{p}-E=0.72-0.1045\\ =0.6155\\ \widehat{p}+E=0.72+0.1045\\ =0.8245\end{array}$

So, we are $90\%$ confident that the true proportion of seniors in Tonya’s school who plan to go to prom is between $(0.6155,0.8245)$ .

However, we know that the population contains $750$ seniors. Thus, we have that,

  $\begin{array}{l}750(0.6155)=461.625\\ \approx 462\\ 750(0.8245)=618.375\\ \approx 618\end{array}$

Thus, we are then $90\%$ confident that between $462$ seniors and $618$ seniors in Tonya’s school plan to go to prom.