Chapter 22, Problem 25E

(a)

To determine

To conduct: a suitable hypothesis and give conclusion.

Answer to Problem 25E

  $\begin{array}{l}{H}_0:p_W-p_B=0\\ H_1:p_W-p_B\ne 0\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}{n}_W=3121\\ y_W=94\\ n_B=606\\ y_B=20\end{array}$

Formula used:

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_W+y_B}{n_w+n_B}\\ S{E}_{Pooled}\left({\widehat{p}}_W-{\widehat{p}}_B\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_w}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_B}}\end{array}$

Calculation:

Here, there is required to check whether the probability of multiple births is racially distinct. Let $p_W$ represent the proportion of white women's multiple births, and $p_B$ represent the proportion of black women's multiple births. $p_W$ should not be equivalent to $p_B$ if there is some racial difference in the probability of multiple births. So, our hypotheses are going to be

  $\begin{array}{l}{H}_0:p_W-p_B=0\\ H_1:p_W-p_B\ne 0\end{array}$

Using Normal model and conduct a two-proportion z-test.

  $\begin{array}{l}{\widehat{p}}_w=\frac{94}{3132}=0.0300\\ {\widehat{p}}_B=\frac{20}{606}=0.0330\end{array}$

Pool the sample data

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{y_W+y_B}{n_w+n_B}\\ =\frac{94+20}{3132+606}\\ =\frac{114}{3738}\\ =0.0305\end{array}$

Using the pooled SE to estimating the $SD\left(p_W-p_B\right)$ .therefore,

  $\begin{array}{c}S{E}_{Pooled}\left({\widehat{p}}_W-{\widehat{p}}_B\right)=\sqrt{\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_w}+\frac{{\widehat{p}}_{Pooled}{\widehat{q}}_{Pooled}}{n_B}}\\ =\sqrt{\frac{0.0305\left(1-0.0305\right)}{3132}+\frac{0.0305\left(1-0.0305\right)}{606}}\\ =\sqrt{\frac{0.0305\left(0.9695\right)}{3132}+\frac{0.0305\left(0.9695\right)}{606}}\\ =0.0076\end{array}$

The observed difference is

  $\begin{array}{c}{\widehat{p}}_W-{\widehat{p}}_B=0.0300-0.0330\\ =-0.0030\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{\left({\widehat{p}}_W-{\widehat{p}}_B\right)-0}{S{E}_{Pooled}\left({\widehat{p}}_W-{\widehat{p}}_B\right)}\\ =\frac{-0.0030-0}{0.0076}\\ =-0.39\end{array}$

The p-value is

  $\begin{array}{c}\text{P-value =2}P\left(z<-0.39\right)\\ =2\left(0.3476\right)\\ =0.6951\end{array}$

To consider the null hypothesis, the P-value = 0.6951 is high enough. It might assume that the possibility of multiple births does not make any racial difference.

(b)

To determine

To find: the type of error that comment, if the conclusion is false.

Answer to Problem 25E

Type II

Explanation of Solution

It is accepted the null hypothesis in part (a). If the inference is incorrect, this implies believing the null hypothesis when it is false. So, it produced an error in type II.