Chapter 10.2, Problem 46E

(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 distribution of red is skewed to the right because most of the dots lie near the left of the dot plot and the distribution of yellow is skewed to the left because most of the dots lie near the right of the dot plot.

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

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

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 $95\%$ confidence interval for the difference of true mean lengths of these two varieties of flowers.

Answer to Problem 46E

There is $95\%$ confidence that the mean length of H. caribaea red is between $(2.5538,4.4822)$ millimeters higher than the mean length of H. caribaea yellow.

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=39.698\\ {\overline{x}}_2=36.18\\ n_1=23\\ n_2=15\\ s_1=1.786\\ s_2=0.975\\ c=0.95\end{array}$

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:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (23-1,15-1)\\ =14\end{array}$

Then the t -value will be as:

  $t_{\alpha /2}=2.145$

Thus the confidence interval be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(39.698-36.18)-2.145\times \sqrt{\frac{{1.786}^2}{23}+\frac{{0.975}^2}{15}}\\ =3.518-0.9642\\ =2.5538\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(39.698-36.18)+2.145\times \sqrt{\frac{{1.786}^2}{23}+\frac{{0.975}^2}{15}}\\ =3.518+0.9642\\ =4.4822\end{array}$

Thus, we conclude that there is $95\%$ confidence that the mean length of H. caribaea red is between $(2.5538,4.4822)$ millimeters higher than the mean length of H. caribaea yellow.