Answer to Question #340254 in Statistics and Probability for skum

Question

A manufacturer claims that only 4% of his products supplied by him are defective. A random sample of 600 products contains 36 defectives. Test the claim of the manufacturer.

Answer 1

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

$H_0:p\le 0.04$

$H_a:p>0.04$

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:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{N}}}=\dfrac{\dfrac{36}{600}-0.04}{\sqrt{\dfrac{0.04(1-0.04)}{600}}}=2.5$$

Since it is observed that $z = 2.5 > 1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z>2.5)=0.00621,$ and since $p=0.00621<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than 0.04, at the $\alpha = 0.05$ significance level.