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

`z=\frac{0.75 -0.72}{\sqrt{(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%

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{(p_o (1-p_o))/(n)}}` (1)

Replacing the data given we got:

`z=\frac{0.75 -0.72}{\sqrt{(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%