Chapter 10.2, Problem 45E

(a)

To determine

To write a few sentences comparing the distribution.

Explanation of Solution

A dot plot of the data is shown in the question along with the summary statistics. Thus, on comparing the two distribution we see that:

Shape: the two distributions are slightly skewed to the left because most of the dots lie near the right of the dot plot.

Center: The center for the red wine distribution is higher than the center for the white wine because most of the dots in the dot plot of red wine lie to the right of most of the dots in the dot plot of white wine and the mean and the median are both greater for the red wine.

Spread: The distribution of white wine is little bit more variable than the distribution of red wine because the dot plot of the white wine is little wider than the dot plot of the red wine and the standard deviation of the white wine is little higher than the standard deviation of red wine.

Unusual features: There appear to be no outliers in either distribution because there are no dots that appear to be unusually far from the other dots in the dot plot.

(b)

To determine

To construct and interpret $90\%$ confidence interval for the difference of true mean percent change in polyphenol levels for a healthy men like the ones in the study when drinking red wine versus white wine.

Answer to Problem 45E

There is $90\%$ confidence that the mean percent change in polyphenol levels for healthy men when drinking red wine is between $(2.6977,\text{7}\text{.8363)}$ lower than the mean percent change in polyphenol levels for healthy men when drinking white wine.

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=5.500\\ {\overline{x}}_2=0.233\\ n_1=9\\ n_2=9\\ s_1=2.517\\ s_2=3.292\\ c=0.90\end{array}$

Calculating confidence interval:

Now we will be calculating the t -value for this we need to find out the degree of freedom. Thus, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (9-1,9-1)\\ =8\end{array}$

Then the t -value will be as:

  $t_{\alpha /2}=1.860$

Thus the confidence interval be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(5.500-0.233)-1.860\times \sqrt{\frac{{2.517}^2}{9}+\frac{{3.292}^2}{9}}\\ =5.267-2.5693\\ =2.6977\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(5.500-0.233)+1.860\times \sqrt{\frac{{2.517}^2}{9}+\frac{{3.292}^2}{9}}\\ =5.267+2.5693\\ =7.8363\end{array}$

Thus, we conclude that there is $90\%$ confidence that the mean percent change in polyphenol levels for healthy men when drinking red wine is between $(2.6977,\text{7}\text{.8363)}$ lower than the mean percent change in polyphenol levels for healthy men when drinking white wine.