Answer to Question #287764 in Statistics and Probability for Tey

Question

The local government often faces the problem of not enough money to pay for the services that it provided. One way to deal is to raise the taxes and another way is to reduce services. When the local poll asked a random sample of 1200 residents, the results show that 52% favouring to raise taxes and 48% favouring to reduce services. Test whether the proportion of residents who choose to raise taxes is different from 50%. Test at 1% significance level. 

Answer 1

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

$H_0:p=0.5$

$H_1:p\not=0.5$

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

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

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

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}=\dfrac{0.52-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{1200}}}$$

$$\approx1.3856$$

Since it is observed that $|z| = 1.3856 \le 2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(Z>1.3856)=0.165869,$ and since $p=0.165869>0.01=\alpha,$ 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 $0.5,$ at the $\alpha = 0.01$ significance level.

Therefore, there is not enough evidence to claim that the proportion of residents who choose to raise taxes is different from 50%, at the $\alpha = 0.01$ significance level.