Question

he mayor of a town has proposed a plan for the construction of a new community. A political study took a sample of 900 voters in the town and found that 81% of the residents favored construction. Using the data, a political strategist wants to test the claim that the percentage of residents who favor construction is more than 78%. Testing at the 0.02 level, is there enough evidence to support the strategist's claim?

Answer 1

Answer:

$z=\frac{0.81 -0.78}{\sqrt{\frac{0.78(1-0.78)}{900}}}=2.173$  

The p value for this case is given by this probability:

$p_v =P(z>2.173)=0.0149$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of people who favored te construction is significantly higher than 0.78 or 78%

Step-by-step explanation:

Information given

n=900 represent the random sample taken

$\hat p=0.81$ estimated proportion of residents who favored the construction

$p_o=0.78$ is the value to test

$\alpha=0.02$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is more than 0.78 or not.:  

Null hypothesis:$p\leq 0.78$  

Alternative hypothesis:$p > 0.78$  

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Replacing the info we got:

$z=\frac{0.81 -0.78}{\sqrt{\frac{0.78(1-0.78)}{900}}}=2.173$  

The p value for this case is given by this probability:

$p_v =P(z>2.173)=0.0149$  

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of people who favored te construction is significantly higher than 0.78 or 78%