Answer to Question #348490 in Statistics and Probability for Ellaine

Question

It is claimed that last year, 65% of the people believed that there was an improvement in the country’s economy.

Suppose this year, only 270 out of the 450 people randomly selected believe that there is an improvement in the country’s

economy. Does this indicate that there is a decrease in the number of people who believes that there is an improvement in

the country’s economy? Use 0.05 significance level.

(Use critical value method)

Answer 1

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

$H_0:p\ge 0.65$

$H_a:p<0.65$

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:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{270/450-0.65}{\sqrt{\dfrac{0.65(1-0.65)}{450}}}=-2.2237$$

Since it is observed that $z =-2.2237<-1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(Z<-2.2237)= 0.013084,$ and since $p=0.013084<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is less than 0.65, at the $\alpha = 0.05$ significance level.