Chapter 10, Problem 12PT

To determine

To test whether the proportion of brake problem in this year is less compared to the previous year.

Explanation of Solution

Given the following information:

  $\begin{array}{l}{n}_1=100{\text{ n}}_2=350\\ x_1=20x_2=50\end{array}$

  $x_1$ , $x_2$ are the proportion of car brakes problem during the previous year and this year from a random samples.

The test procedure:

The null and alternative hypothesis is given by

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_1:p_1>p_2\end{array}$

The test statistic is given by,

  $Z=\frac{p_1-p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}~N(0,1)$

Reject the null hypothesis if $Z>Z_{\alpha }$

Calculation:

On substituting the given values the test statistic is given by

  $\begin{array}{l}Z=\frac{p_1-p_2}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}\\ Where{p}_1=\frac{x_1}{n_1}=\frac{20}{100}=0.2\\ p_2=\frac{x_2}{n_2}=\frac{50}{350}=0.1428\\ \therefore Z=\frac{0.2-0.1428}{\sqrt{\frac{0.2(1-0.2)}{100}+\frac{0.1428(1-0.1428)}{350}}}\\ \text{ =}\frac{0.0572}{\sqrt{0.0016+0.0003}}\\ \text{ =1}\text{.3123}\end{array}$

The critical value at 0.05 level of significance is $Z_{0.05}=1.65$

Since $Z>1.65$ , we do not reject the null hypothesis and conclude that the proportion of brake problem in this year is not less compared to the previous year.

To determine

To describe the Type 1 error and Type 2 error in this problem.

Explanation of Solution

Type 1 error is the mistakes in which we reject the proportion of brake failure in the last year and this year is are same even though it is true.

Type 2 error is the mistakes in which we reject the proportion of brake failure in the last year and this year is are same even though it is not true.