Question

The mayor of a town has proposed a plan for the annexation of an adjoining community. A political study took a sample of 900 voters in the town and found that 75% of the residents favored annexation. Using the data, a political strategist wants to test the claim that the percentage of residents who favor annexation is above 72%. Testing at the 0.05 level, is there enough evidence to support the strategist's claim

Answer 1

Answer:

$z=\frac{0.75 -0.72}{\sqrt{\frac{0.72(1-0.72)}{900}}}=2.00$  

Now we can calculate the p value. Since is a bilateral test the p value would be:  

$p_v= P(Z>2) =0.0228$

Since the p value is lower than the significance level of 0.05 we have enough evidence to conclude that the true proportion of residents favored annexation is higher than 0.72 or 72%

Step-by-step explanation:

Information given

n=900 represent the random sample selected

$\hat p=0.75$ estimated proportion of residents favored annexation

$p_o=0.72$ is the value that we want to test

represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

The political strategist wants to test the claim that the percentage of residents who favor annexation is above 72%.:  

Null hypothesis:$p\leq 0.72$  

Alternative hypothesis:$p > 0.72$  

The statistic for this case is given by:

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

Replacing the data given we got:

$z=\frac{0.75 -0.72}{\sqrt{\frac{0.72(1-0.72)}{900}}}=2.00$  

Now we can calculate the p value. Since is a bilateral test the p value would be:  

$p_v= P(Z>2) =0.0228$

Since the p value is lower than the significance level of 0.05 we have enough evidence to conclude that the true proportion of residents favored annexation is higher than 0.72 or 72%