Chapter 10.2, Problem 44E

(a)

To determine

To Explain: That the centres are similar or different from the two groups and the spreads are identical or distinct from the two groups.

Answer to Problem 44E

Centers: different

Spread: different

Explanation of Solution

Given:

  

The centre lies between the dot most to the left and the dot most to the right, approximately in the centre. Then it is noticed that the centre is more to the right than to the left of the upper dot plot, so the centres seem to be different. The spread is the difference between the leftmost dot and the rightmost dot. Then it is noticed that the spread tends to be more than the spread for the bottom dot plot for the top dot plot, and therefore the spread tends to be different.

(b)

To determine

To construct: and explain a 95 percent confidence interval for the difference in the mean length of these two types flowers.

Answer to Problem 44E

(2.5538, 4.4828)

Explanation of Solution

Given:

  

Formula used:

For the confidence interval

  $\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Calculation:

The computed mean is

  $\begin{array}{c}{\overline{x}}_1=39.6983\\ {\overline{x}}_2=36.18\end{array}$

The computed variance is

  $\begin{array}{c}{s}_1=1.7865\\ s_2=0.9753\end{array}$

Finding the degrees of freedom is

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

Determine $t^{\ast }$ with $df=14$ and $c=95\%$

  $t^{\ast }=2.145$

Confidence interval for ${\mu }_1-{\mu }_2$ are:

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(39.6983-36.18\right)-2.145\times \sqrt{\frac{{1.7865}^2}{23}+\frac{{0.9753}^2}{15}}\\ =2.5538\end{array}$

  $\begin{array}{l}\left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(39.6983-36.18\right)+2.145\times \sqrt{\frac{{1.7865}^2}{23}+\frac{{0.9753}^2}{15}}\\ =4.4828\end{array}$

There is 95% trust that there is a significant difference between 2.5538 and 4.4828.

(c)

To determine

To Explain: The interval confirms the researcher's assumption that the average length of the two flower varieties is different.

Explanation of Solution

From the part (b):

(2.5538. 4.4828)

The confidence interval does not include 0, which means that the mean difference is more likely to be non-zero and that the interval also supports the assumption that the two varieties have different lengths.