Answer to Question #341262 in Statistics and Probability for Clouds

Question

A SHS student conducted a survey to test the claim that "less than half of all the adults are annoyed by the violence on television". Suppose that from a poll of 2,400 surveyed adults, 1,152 indicated their annoyance with television violence. Test this claim using 0.10 level of significance

a. State the null and alternative hypothesis

b. Determine the significance level

c. Select the test statistic to be used

d. Compute the test statistic value

e. Determine the critical value

f. Sketch the rejection region

g. Draw the conclusion

Answer 1

a.

$H_0: p\ge 0.5$

$H_a:p<0.5$

b.The significance level is $\alpha = 0.10.$

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

d. The z-statistic is computed as follows:

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

$$=\dfrac{1152/2400-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{2400}}}=-1.9596$$

e. Based on the information provided, the significance level is $\alpha = 0.10,$ and the critical value for a left-tailed test is $z_c = -1.2816.$

f. The rejection region for this right-tailed test is $R = \{z: z <-1.2816\}.$

g. Since it is observed that $z =-1.9596< -1.2816= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-1.9596)=0.025021,$ and since $p=0.025021<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is less than 0.5, at the $\alpha = 0.10$ significance level.