Q.108 - The Practice of Statistics for AP

Question

Normal body temperature (8.2) If "normal" body temperature really is $98.6 F$, we would expect the proportion $p$ of all healthy 18- to 40 -year-olds who have body temperatures less than this value to be $0.5$. Construct and interpret a $95\%$ confidence interval for $p$. What conclusion would you draw?

Step-by-Step Solution

Answer

We are $95\%$confident that the interval from $0.414$ to$0.586$ contains the true proportion of people who has body temperature is than.

Step 1: Given Information

Given in the question that,

$n=130$

$\hat{p}=0.5$

we have to Construct and interpret a $95\%$ confidence interval for $p$.

Step-2 Explanation

We have to use a one-sample $z$ interval for $p$ if the conditions are satisfied.

1) We have selected a random sample of $130$ people at random

2) We know that

$n\overbrace{p}=130\times 0.5$

      $=65\ge 10$

Also

$n\left(1-\overbrace{p}\right)=130\times \left(1-0.5\right)$

              $=130\times 0.5$

              $=65\ge 10$

Since $n\hat{p},n\left(1-\overbrace{p}\right)\ge 10$, we should be safe to do calculations

3) To verify independence, we must examine the $10\%$condition

We estimate that at least $1300$ persons will have a body temperature of less than. As a result, the $10\%$ requirement is met. The critical value$z=1.96$ has a $95\%$ confidence interval. The $p$$95\%$ confidence interval is calculated as follows: $\hat{p}\pm z\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}$$=0.5\pm 1.96\times \sqrt{\frac{0.5\left(1-0.5\right)}{130}}$

                                   $=0.5\pm 0.086$

                                   $=\left(0.414,0.586\right)$