Chapter 23, Problem 36E

(a)

To determine

To check the assumptions and conditions for inference.

Explanation of Solution

We have to check the conditions for inference for the following data provided in the question. Thus, now we will check the same as:

Random condition: It is satisfied because the bags were randomly selected.

Independent condition: It is satisfied because since the sample of $14$ brands of vanilla yogurt is less than $10\%$ of the population of all brands of vanilla yogurt.

Normal condition: It is satisfied because the histogram shows no outlier and the normal quantile plot is nearly normal.

Thus, all conditions and assumptions of the inference are met.

(b)

To determine

To construct a $95\%$ confidence interval for the average calorie content of vanilla yogurt.

Answer to Problem 36E

The confidence interval is $(-3959.97,4275.68)$ .

Explanation of Solution

It is given: $n=14$ .

Thus, the mean of the data given in the question is:

  $\begin{array}{l}\overline{x}=\frac{160+200+220+230+\mathrm{....}+170}{14}\\ =157.8571\end{array}$

The standard deviation of the data given is as:

  $\begin{array}{l}s=\sqrt{\frac{{(160-157.8571)}^2+\mathrm{....}+{(170-157.8571)}^2}{14-1}}\\ =7133.10\end{array}$

The degrees of freedom is:

  $\begin{array}{l}df=n-1\\ =14-1\\ =13\end{array}$

Thus, the t -value is:

  $t_{\alpha /2}=2.160$

Therefore the confidence interval is:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =157.8571-2.160\times \frac{7133.10}{\sqrt{14}}\\ =-3959.97\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =157.8571+2.160\times \frac{7133.10}{\sqrt{14}}\\ =4275.68\end{array}$

Thus, the confidence interval is $(-3959.97,4275.68)$ .

(c)

To determine

To explain what does this evidence indicate.

Explanation of Solution

It is given: $n=14$ . And, the confidence interval is $(-3959.97,4275.68)$ . Thus, it is given that a diet guide claims that you will get $120$ calories from a serving of vanilla yogurt. Since the confidence interval does not contain $120$ and only contains values above $120$ , therefore there is sufficient evidence to support the claim that the mean number vanilla yogurt of all brands is greater than $120$ .