Chapter 10, Problem T10.12SPT

(a)

To determine

To find out was the company’s adjustment successful or not and carry out appropriate test to support the answer.

Answer to Problem T10.12SPT

No.

Explanation of Solution

It is given in the question that:

  $\begin{array}{l}{x}_1=20\\ x_2=50\\ n_1=100\\ n_2=350\end{array}$

Now the hypotheses will be:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_{\alpha }:p_1>p_2\end{array}$

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

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{20}{100}\\ =0.2\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{50}{350}\\ =0.1429\\ {\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{20+50}{100+350}\\ =\frac{70}{450}\\ =0.1556\end{array}$

Now we will find the value of test statistics:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.2-0.1429}{\sqrt{0.1556(1-0.1556)}\sqrt{\frac{1}{100}+\frac{1}{300}}}\\ =1.39\end{array}$

The P -value is the probability of obtaining the value of the test statistic or a value more extreme. Then, determine the P -value using table A:

  $\begin{array}{l}P=P(Z>1.39)\\ =P(Z<-1.39)\\ =0.0823\end{array}$

If the P -value is smaller than the significance level, reject the null hypothesis, thus we have,

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

Thus we can say that there is no sufficient evidence to support the claim of a difference. Therefore, company’s adjustments will not be successful.

(b)

To determine

To find out which mistake Type I error or Type II error have been made and describe a possible consequence of this error.

Answer to Problem T10.12SPT

Type II error could have been made.

Explanation of Solution

As we have define the hypotheses in the part (a), that is,

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_{\alpha }:p_1>p_2\end{array}$

Now, we know that the Type I error is:

Reject the null hypothesis $H_0$ , when $H_0$ is true.

So we can interpret that the adjustments appear to be effective while it is not effective.

And the consequence of this will be that more cars have a break defect than expected and this might thus cost the company more money.

Now, for Type II error, this error occurs when:

Fail to reject the null hypothesis $H_0$ , when $H_0$ is false.

The interpretation of this will be that the adjustments appear to be not effective while it is effective.

And the consequence for this will be that less cars have a brake defect than expected which might save the company some money. Thus, from part (a), it seems that Type II error could have been made.