Chapter 10.1, Problem 21E

(a)

To determine

To find: the significant is the observed difference and carries out an appropriate test to support the answer.

Answer to Problem 21E

Not significant

Explanation of Solution

Given:

  $\begin{array}{c}{x}_1=3396\\ n_1=19541\\ x_2=4929\\ n_2=29294\end{array}$

Formula used:

  $\widehat{p}=\frac{x}{n}$

  ${\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}$

  $z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{{\widehat{p}}_p\left(1-{\widehat{p}}_p\right)\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}}$

Calculation:

The hypotheses:

  $\begin{array}{c}{H}_0:p_1-p_2=0\\ H_1:p_1-p_2\ne 0\end{array}$

The sample proportion is

  $\begin{array}{c}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{3396}{19541}\\ =0.174\end{array}$

  $\begin{array}{c}{\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{4929}{29294}\\ =0.168\end{array}$

  $\begin{array}{c}{\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{3396+4929}{19541+29294}\\ =0.170\end{array}$

The test statistic:

  $\begin{array}{c}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{{\widehat{p}}_p\left(1-{\widehat{p}}_p\right)\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}}\\ =\frac{0.174-0.168}{\sqrt{0.168\left(1-0.168\right)\sqrt{\frac{1}{19541}+\frac{1}{29294}}}}\\ =1.74\end{array}$

The P-value is

  $\begin{array}{c}P=P\left(\text{Z < }-\text{ 1}\text{.74 or Z > 1}\text{.74}\right)\\ =2\times P\left(Z<-1.74\right)\\ =2\times 0.0409=0.0818\end{array}$

If the P-value is lesser than the significance level, therefore reject the null hypothesis:

  $P>0.05\Rightarrow \text{ Fail to reject }{H}_0$

There is not sufficient proof to help the claim of a difference.

(b)

To determine

To Explain: the Type I and Type II error in this setting and find the more serious one.

Answer to Problem 21E

Type II error

Explanation of Solution

Type I error: Reject the null hypothesis $H_0$ , when $H_0$ is true.

Consequence: There is no enough difference in both groups, where it displays that there is significance. Therefore the design is random, where it displays no random.

Type II error: Failing to reject the null hypothesis $H_0$ , when $H_0$ is false.

Consequence: There is a enough difference in both groups, where it displays that there is no enough difference. Therefore the design is not random, when it displays random.

Type II error is worse, the reason is that it would be assumed that the people were arbitrary assigned to every group and every group seems to be the same, where it is not the case.