Answer to Question #350552 in Statistics and Probability for Khali

Question

A toy company wishes to put a new toy in the market for the Christmas season. It does not want to do so if the manufacturing process produces more than 10% of toys that do no work. In a test run producing 80 of the toys, 10 are found to be defective. Based on the sample evidence, should the company continue to put a new toy in the market? Use 0.05 level of significance.

Answer 1

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:p\le0.1$

$H_a:p>0.1$

This corresponds to a right-tailed test, for which a z-test for one population proportion will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a right-tailed test is $z_c = 1.6449.$

The rejection region for this right-tailed test is $R = \{z: z>1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{10/80-0.1}{\sqrt{\dfrac{0.1(1-0.1)}{80}}}=2.3570$$

Since it is observed that $z=2.3570> z_c = 1.6449,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z>2.3570)=0.009212,$ and since $p =0.009212< 0.05,$ it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greaterthan $0.1,$ at the $\alpha = 0.05$ significance level.