43. Is red wine better than white wine? Observational studies suggest that moderate use of alcohol by adults reduces heart attacks and that red wine may have special benefits. One reason may be that red wine contains polyphenols, substances that do good things to cholesterol in the blood and so may reduce the risk of heart attacks. In an experiment, healthy men were assigned at random to drink half a bottle of either red or white wine each day for two weeks. The level of polyphenols in their blood was measured before and after the two-week period. Here are the percent
changes in level for the subjects in both groups:

(a) A Fathom dotplot of the data is shown below. Use the graph to answer these questions:

• Are the centers of the two groups similar or different? Explain.
• Are the spreads of the two groups similar or different? Explain.

(b) Construct and interpret a $90\%$ confidence interval for the difference in mean percent change in polyphenol levels for the red wine and white wine treatments.
(c) Does the interval in part (b) suggest that red wine is more effective than white wine? Explain.

A Fathom dotplot of the data is shown as:

To find the centers and spreads of the two groups similar or different.

The center is roughly halfway between the dot on the far left and the dot on the far right.

The upper dotplot's center is more to the right than the lower dotplot's, hence, the centers appear to be different.
The spread is the distance between the farthest left and farthest right dots.And the spread is about the same for both dotplots and therefore the spread appears to be similar.
Hence the Centers are different Spreads are similar.

A $90\%$ confidence interval for the difference in mean percent change
in polyphenol levels for the red wine and white wine treatments.

Let, the total of all quantities is divided by the value count to get the mean.
${\overline{x}}_1=5.5$
${\overline{x}}_2=0.2333$
The amount of square deviations divided by$n-1$ from the norm is termed variance. The standard deviation is the square root of variance..
$s_1=2.5169$
$s_2=3.2920$
Then the degrees of freedom is:
$$\begin{gathered} df=\min \left(n_1-1,n_2-1\right) \\ =min(9-1,9-1) \\ =8 \end{gathered}$$

Determine $t^{*}$by using table $B$:

$t^{*}=1.860$

The confidence interval for${\mu }_1-{\mu }_2$ are:
$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(5.5-0.2333)-1.860\times \sqrt{\frac{2.{5169}^2}{9}+\frac{3.{2920}^2}{9}} \\ =2.6975 \end{gathered}$$

Then,

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=(5.5-0.2333)+1.860\times \sqrt{\frac{2.{5169}^2}{9}+\frac{3.{2920}^2}{9}} \\ =7.8539 \end{gathered}$$

Hence, the mean gap is between $2.6975$and $7.8539$with the $90\%$of confidence.

To explain that the interval in part (b) suggest that red wine is more effective than white wine.

The percent changes in level for the subjects in both groups:

Getting the result from part (b): $2.6975,7.8539$
The confidence interval does not have zero and is completely above zero, the mean differential is more likely to be positive, and the interval suggests that red wine is more successful than white wine.

Hence, the answer is yes, the interval in part (b) suggest that red wine is more effective than white wine.