Answer to Question #336755 in Statistics and Probability for Bee

Question

Test the hypotheses, given p=0.42, p is not equal to 0.42, sample size=150, sample proportion=0.45, alpha =0.05

Answer 1

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

$p=0.42$

$p\not=0.42$

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 $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z: |z| > 1.96\} .$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.45-0.42}{\sqrt{\dfrac{0.42(1-0.42)}{150}}}$$

$$\approx0.7444$$

Since it is observed that $|z| = 0.744 \le1.96= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p =2 P(Z>1.7444)=0.456635,$ and since $p = 0.4566 35>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population proportion $p$ is different than 0.42, at the $\alpha = 0.05$ significance level.