Answer to Question #347492 in Statistics and Probability for jouuuuuu

Question

Determine the ff. problem as: a.)One-tailed (Right tailed or left tailed) or Two-tailed b.)Alpha value c.)Critical value It is believed that in the coming election, 65% of the voters in the province of Pampanga will vote for the administration candidate for governor. Suppose 713 out of the 1 150 randomly selected voters indicate that they would vote for the administration candidate. At 0.10 level of significance, find out whether the percentage of voters for the administration candidate is different from 65% 

Answer 1

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

$H_0:p=0.65$

$H_a:p\not=0.65$

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

b) Based on the information provided, the significance level is $\alpha = 0.10 ,$ and the critical value for a two-tailed test is $z_c =1.6449.$

c) The rejection region for this two-tailed test is $R = \{z: |z|>1.6449\}.$

d)The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{713/1150-0.65}{\sqrt{\dfrac{0.65(1-0.65)}{1150}}}=-2.1329$$

Since it is observed that $|z| =2.1329>1.6449= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(Z<-2.1329)=0.032933,$ and since $p=0.032933<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is different than 0.65, at the $\alpha = 0.10$ significance level.