Answer to Question #336215 in Statistics and Probability for Myr

Question

It has been reported that 40% of the adult population participate in computer hobbies during their leisure time. A random sample of 180 adults found that 65 engaged in computer hobbies. at a= 0.01, is there sufficient evidence to conclude that the proportion differs from 40%?

Answer 1

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

$H_0:p=0.4$

$H_a:p\not=0.4$

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{\dfrac{65}{180}-0.4}{\sqrt{\dfrac{0.4(1-0.4)}{180}}}\approx-1.0650$$

The p-value is $p =2P(Z<-1.0650)= 2(0.143438)$

$=0.286876.$

Based on the information provided, the significance level is $\alpha = 0.01,$ and the critical value for a two-tailed test is $z_c = 2.5758.$

The rejection region for this two-tailed test is $R = \{z: |z|>2.5758\}.$

Since it is observed that $|z| = 1.0650 <2.5758= z_c ,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is $p =0.286876,$ and since $p = 0.286876 \ge 0.01=\alpha,$ 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 0.40, at the $\alpha = 0.01$ significance level.