Answer to Question #224327 in Statistics and Probability for wae

Question

hypothesis testing

The sponsor of a television ‘special’ expected that at least 40 percent of the viewing audience would watch the show in a particular metropolitan area. For a random sample of 100 households with television sets turned on, 30 are viewing the ‘special’. Can the sponsor’s assumption that at least 40 percent of the households would watch the program be rejected at the 5 percent level of significance?

Answer 1

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

$H_0:p\geq 0.4$

$H_1: p<0.4$

This corresponds to a left-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 left-tailed test is $z_c=-1.6449.$

The rejection region for this left-tailed test is $R=\{z:z<-1.6449\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}=\dfrac{0.3-0.4}{\sqrt{\dfrac{0.4(1-0.4)}{100}}}\approx-2.041$$

Since it is observed that $z=-2.041<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=P(Z<-2.041)=0.0206<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is less than 0.4, at the $\alpha=0.05$ significance level.