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$x_1=20$

$x_2=50$

$n_1=100$

$n_2=350$

Determine the hypothesis:

$H_0:p_1=p_2$

$H_a:p_1>p_2$

The sample proportion is calculated by dividing the number of successes by the sample size:

localid="1650383335009" $$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1}=\frac{20}{100}=0.2 \\ {\hat{p}}_2=\frac{x_2}{n_2}=\frac{50}{350}鈮�0.1429 \\ {\hat{p}}_p=\frac{x_1+x_2}{n_1+n_2}=\frac{20+50}{100+350}=\frac{70}{450}鈮�0.1556 \end{gathered}$$

Determine the value of the test statistic:

localid="1650383319593" $$\begin{gathered} z=\frac{{\hat{p}}_1-{\hat{p}}_2}{\sqrt{{\hat{p}}_p\left(1-{\hat{p}}_p\right)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \\ =\frac{0.2-0.1429}{\sqrt{0.1556(1-0.1556)}\sqrt{\frac{1}{100}+\frac{1}{350}}} \\ 鈮�1.39 \end{gathered}$$

The $P-$value is the chance of getting the test statistic's result, or a number that is more severe. Using table A, calculate the$P-$value:

localid="1650383420208" $$\begin{gathered} P=P(Z>1.39) \\ =P(Z<-1.39) \\ =0.0823 \end{gathered}$$

Reject the null hypothesis if the P-value is less than the significance level:

$P>0.05鈬�\text{Fail to reject}{H}_0$

Therefore, there is not sufficient evidence to support the claim of a difference.

Determine the hypothesis:

$H_a:p_1>p_2$

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

The change appears to be effective, but it is in fact ineffective.

As a result, more cars than projected have a braking issue, potentially costing the corporation extra money.

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

Interpretation: While the change looks to be ineffective, it actually is.

As a result, fewer cars have brake defects than expected, potentially saving the company money.