Answer to Question #286332 in Statistics and Probability for swe

Question

5. A manufacturer believes exactly 8% of its product contain at least one minor flaw. Suppose a company researcher wants to test this belief. The researcher randomly selects a sample f 200 products inspects each item and determine that 33 items have at least one minor flaw. Test at 10% level of significance.  

Answer 1

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

$H_0:p=0.08$

$H_1:p\not=0.08$

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

Based on the information provided, the significance level is $\alpha = 0.1,$ and the critical value for a two-tailed test is $z_c = 1.6449.$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.165-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{200}}}\approx4.4309$$

Since it is observed that $|z| = 4.4309 > 1.6449=z_c ,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p = 2P(Z>4.4309)=0.0000094,$ and since $p=0.0000094<0.1=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$

is different than $0.08,$ at the $\alpha = 0.1$ significance level.