Chapter 10.2, Problem 10.44E

(a)

To determine

To Explain: a significance test to get whether these data proved convincing evidence that “normal” body temperature is not 98.6 degree F.

Answer to Problem 10.44E

There is enough evidence to help the claim that the mean body temperature is not 98.6F.

Explanation of Solution

Given:

  $\begin{array}{c}n=130\\ \overline{x}=98.25\\ s=0.73\end{array}$

Formula used:

  $z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$

Calculation:

Mean is not 98.6

The claim is either the null hypothesis or the alternative hypothesis..

  $\begin{array}{c}{H}_0:\mu =98.6\\ H_1:\mu \ne 98.6\end{array}$

The z-value is the sample mean decreased by the population mean, divided by the standard deviation:

  $\begin{array}{c}z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}\\ =\frac{98.25-98.6}{0.73/\sqrt{130}}\\ =-5.47\end{array}$

The P-value is the probability of obtain a value more extreme or equal to the standardized test statistic z

  $P=P\left(Z<-5.47\text{or Z>5}\text{.47}\right)=2P\left(Z<-5.47\right)=2\left(0\right)=0$

If the P-value is lesser than the significance level $\alpha$ , then the null hypothesis is rejected.

  $P<0.05\Rightarrow {\text{Reject H}}_0$

There is enough evidence to help the claim that the mean body temperature is not 98.6F.

(b)

To determine

To Calculate: a 95% confidence interval for the mean body temperature in the population of healthy adults and additional information about the confidence intervals provide.

Answer to Problem 10.44E

It is observed that 95% confident that the mean body temperature of all healthy adults is between ${98.4745}^{\circ }F$ and ${98.7255}^{{}^{\circ }}F$ .

Explanation of Solution

Formula used:

  $E=z_{a/2}\times \frac{s}{\sqrt{n}}$

For confidence interval:

  $\begin{array}{c}\overline{x}-E\\ \overline{x}+E\end{array}$

Calculation:

  $c=95\%=0.095$

For confidence level $1-\alpha =0.99$

  $z_{a/2}=z_{0.005}$ using the normal probability table

  $z_{a/2}=1.96$

The margin of error is

  $\begin{array}{c}E=z_{a/2}\times \frac{s}{\sqrt{n}}\\ =1.96\times \frac{0.73}{\sqrt{130}}\\ =0.1255\end{array}$

Confidence interval is

  $\begin{array}{c}\overline{x}-E=98.6-0.1255=98.4745\\ \overline{x}-E=98.6+0.1255=98.7255\end{array}$

It is observed that 95% confident that the mean body temperature of all healthy adults is between ${98.4745}^{\circ }F$ and ${98.7255}^{{}^{\circ }}F$ .