Chapter 23, Problem 13E

(a)

To determine

To check the conditions for creating a t -interval.

Answer to Problem 13E

All conditions are met.

Explanation of Solution

It is given in the question that the researcher described in previous exercise also measured the body temperature of that randomly selected group of adults. Thus, the conditions for creating a t -interval to be checked are:

Random: It is satisfied because given is that the sample contains randomly selected adults.

Independent condition: It is satisfied because it can be assumed that the $52$ sampled adults is less than $10\%$ of the population.

Normal condition: It is satisfied because the sample size of $52\text{ is 30}$ or more and thus the sampling distribution of the sample mean is approximately normal, which is confirmed by the given histogram in the question.

Thus, we have that all the three conditions are met.

(b)

To determine

To find a $98\%$ confidence interval for mean body temperature.

Answer to Problem 13E

The confidence interval is $(98.0576,98.5124)$ .

Explanation of Solution

It is given in the question that the researcher described in previous exercise also measured the body temperature of that randomly selected group of adults. And all conditions are met.

It is given:

  $\begin{array}{l}n=52\\ \overline{x}=98.285\\ s=0.6824\end{array}$

Now we will find the t -value by looking in the row starting with degrees of freedom. The degree of freedom is as:

  $\begin{array}{l}df=n-1\\ =52-1\\ =51\end{array}$

Thus, the t -value with $98\%$ confidence interval be:

  $t_{\alpha /2}=2.403$

The confidence interval is calculated as:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =98.285-2.403\times \frac{0.6824}{\sqrt{52}}\\ =98.0576\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =98.285+2.403\times \frac{0.6824}{\sqrt{52}}\\ =98.5124\end{array}$

Thus the confidence interval is $(98.0576,98.5124)$ .

(c)

To determine

To explain the meaning of that interval.

Explanation of Solution

It is given in the question that the researcher described in previous exercise also measured the body temperature of that randomly selected group of adults. And all conditions are met.

It is given:

  $\begin{array}{l}n=52\\ \overline{x}=98.285\\ s=0.6824\end{array}$

And the confidence interval is $(98.0576,98.5124)$ . Thus, it means that we are $98\%$ confident that the true average body temperature of all adults is between $98.0576\text{F}\text{ and }98.5124\text{F}$ .

(d)

To determine

To explain what “ $98\%$ confidence” means in this context.

Explanation of Solution

It is given in the question that the researcher described in previous exercise also measured the body temperature of that randomly selected group of adults. And all conditions are met.

It is given:

  $\begin{array}{l}n=52\\ \overline{x}=98.285\\ s=0.6824\end{array}$

And the confidence interval is $(98.0576,98.5124)$ . The term “ $98\%$ confidence” means in this context that the confidence interval will contain the true population mean in $98\%$ of all possible samples.

(e)

To determine

To explain do these data suggest otherwise.

Answer to Problem 13E

Yes.

Explanation of Solution

It is given in the question that the researcher described in previous exercise also measured the body temperature of that randomly selected group of adults. And all conditions are met.

It is given:

  $\begin{array}{l}n=52\\ \overline{x}=98.285\\ s=0.6824\end{array}$

And the confidence interval is $(98.0576,98.5124)$ . This data suggest that the confidence interval does not contain $98.6\text{F}$ and thus this data suggest that the body temperature of $98.6\text{F}$ is not “normal”. Therefore, yes, this data suggest otherwise.