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Given

$x_1=3396$

$n_1=19541$

$x_2=4929$

role="math" $n_2=29294$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2鈮�0$

The sample proportion is the number of successes divided by the sample size:

${\hat{p}}_1=\frac{x_1}{n_1}=\frac{3396}{19541}鈮�0.174$

${\hat{p}}_2=\frac{x_2}{n_2}=\frac{4929}{29294}鈮�0.168$

${\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{3396+4929}{19541+29294}鈮�0.170$

Determine the value of the test statistic:

localid="1650450817963" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.174-0.168}{\sqrt{0.168(1-0.168)}\sqrt{\frac{1}{19541}+\frac{1}{29294}}} \\ 鈮�1.74 \end{gathered}$$

The $p$-value is the probability of obtaining the value of the test statistic, or a value more extreme. Determine the $p$-value using table A:

localid="1650450836161" $$\begin{gathered} P=P(Z<-1.74\text{or}Z>1.74) \\ =2脳P(Z<-1.74) \\ =2脳0.0409 \\ =0.0818 \end{gathered}$$

If the $p$-value is smaller than the significance level, reject the null hypothesis:

$P>0.05鈬�\text{Fail to reject}{H}_0$

Given

$x_1=3396$

$n_1=19541$

$x_2=4929$

$n_2=29294$

Determine the hypothesis

$H_0:p_1-p_2=0$

$H_a:p_1-p_2鈮�0$

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

As a result, there is no significant difference between the two groups, even if it appears that there is. Type II error: Failing to reject the null hypothesis$H_0$, when $H_0$is false.

As a result, there is a considerable difference in both groups, even if it looks that there is none. As a result, while the design appears random, it is not.

The Type Il mistake is even more serious.