A random sample of 100 of a certain popular car model last year found that 20 had a certain minor defect in the brakes. The car company made an adjustment in the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year's model found that 50 had the minor brake defect.

(a) Was the company's adjustment successful? Carry out an appropriate test to support your answer.

(b) Describe a Type I error and a Type Il error in this setting, and give al possible consequence of each.

$x_1=20$

$x_2=50$

$n_1=100$

$n_2=350$

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

Determine the value of the test statistic:

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

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:

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

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.