Question

Prom totals 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.

Short answer

There is $90\%$confidence that between$462$and$618$seniors in Tanya's school go to prom.

Step-by-step solution

Given Information

It is given that $c=90\%=0.90$

$x=36,n=50$

For$1-\alpha =0.90$, using probability$z_{a/2}=1.645$

Construct and interpret confidence interval

Conditions are:

Random: As samples are selected randomly, it is satisfied.

Independent: Sample of $50$seniors is less than $10\%$of all population of all seniors. This condition is met.

Normal: There are $36$success and $50-36=14$failures which are greater than $10$.

All conditions are fulfilled.

Sample proportion is calculated as: $\hat{p}=\frac{x}{n}$

$=\frac{36}{50}=0.72$

Margin of Error: $E=z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=1.645\times \sqrt{\frac{0.72(1-0.72)}{50}}=0.1045$

Confidence Interval: $\hat{p}-E=0.72-0.1045=0.6155$

$\hat{p}+E=0.72+0.1045=0.8245$

The confidence interval is$(0.6155,0.8245)$.

As population consists of $750$seniors,

$750(0.6155)=461.625\approx 462$

$750(0.8245)=618.375\approx 618$

There is $90\%$confidence that between$462-618$seniors that plan to go to prom.