Question

Normal body temperature (8.2) If "normal" body temperature really is $98.6F$, 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?

Short 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-by-step solution

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

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

localid="1650429756908" $=65\ge 10$

Also

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

$=130\times 0.5$

localid="1650429761719" $=65\ge 10$

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 localid="1650429765277" $1300$persons will have a body temperature of less than. As a result, the localid="1650429768820" $10\%$requirement is met. The critical valuelocalid="1650429773704" $z=1.96$has a localid="1650429778950" $95\%$confidence interval. The localid="1650429785799" $p$localid="1650429791900" $95\%$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}}$

$=0.5\pm 0.086$

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