Question

The mayor of a town has proposed a plan for the annexation of an adjoining bridge. A political study took a sample of 1700 voters in the town and found that 35% 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 more than 32%. Make the decision to reject or fail to reject the null hypothesis at the 0.05 level.

Answer 1

`z=\frac{0.35 -0.32}{\sqrt{(0.32(1-0.32))/(1700)}}=2.652`

`p_v =P(z>2.652)=0.004`

We see that the p value is lower than the significance level of 0.05 so then we have enough evidence to conclude that the true proportion of residents who favor annexation is more than 0.32 at 5% of significance.

Reject the null hypothesis

Explanation:

Information given

n=1700 represent the sample size

`\hat p=0.35` estimated proportion of the residents favored annexation

`p_o=0.32` is the value that we want to check

`\alpha=0.05` represent the significance level

z would represent the statistic

`p_v` represent the p value

Hypothesis to check

We want to analyze if the percentage of residents who favor annexation is more than 32%, so then the system of hypothesis are:

Null hypothesis:
`p\leq 0.32`

Alternative hypothesis:
`p > 0.32`

The statistic to check the hypothesis is given by:

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

With the info given we have this:

`z=\frac{0.35 -0.32}{\sqrt{(0.32(1-0.32))/(1700)}}=2.652`

The test is a right tailed test to then the p value can be founded like this:

`p_v =P(z>2.652)=0.004`

We see that the p value is lower than the significance level of 0.05 so then we have enough evidence to conclude that the true proportion of residents who favor annexation is more than 0.32 at 5% of significance.

Reject the null hypothesis

Answer 2

Null hypothesis: H0 = 0.32

Alternative hypothesis: Ha > 0.32

z = 2.65

P value = P(Z>2.65) = 0.004

Decision: We REJECT the null hypothesis and accept the alternative hypothesis.

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 95% confidence interval) ---- reject Null hypothesis

Z score < Z(at 95% confidence interval) ------ accept Null hypothesis

Explanation:

Given;

n=1700 represent the random sample taken

Null hypothesis: H0 = 0.32

Alternative hypothesis: Ha > 0.32

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 1700

po = Null hypothesized value = 0.32

p^ = Observed proportion = 0.35

Substituting the values we have

z = (0.35-0.32)/√{0.32(1-0.32)/1700}

z = 2.652

z = 2.65

To determine the p value (test statistic) at 0.05 significance level, using a one tailed hypothesis.

P value = P(Z>2.65) = 0.004

Since z at 0.05 significance level is between -1.96 and +1.96 and the z score for the test (z = 2.65) which doesn't falls with the region bounded by Z at 0.05 significance level. And also the one-tailed hypothesis P-value is 0.004 which is lower than 0.05. Then we can conclude that we have enough evidence to reject the null hypothesis, and we can say that at 5% significance level the null hypothesis is invalid, therefore we accept the alternative hypothesis.