Question

The mayor of a town has proposed a plan for the annexation of a new bridge. A political study took a sample of 1000 voters in the town and found that 42% 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 39%. Testing at the 0.02 level, is there enough evidence to support the strategist's claim?

Answer 1

`z=\frac{0.42 -0.39}{\sqrt{(0.39(1-0.39))/(1000)}}=1.945`

`p_v =P(Z>1.945)=0.0259`

If we compare the p value obtained with the significance level given
`\alpha=0.02` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 2% of significance the proportion of residents who favored annexation is not significantly higher than 0.39.

Explanation:

1) Data given and notation

n=1000 represent the random sample taken

`\hat p=0.42` estimated proportion of residents who favored annexation

`p_o=0.39` is the value that we want to test

`\alpha=0.02` represent the significance level

Confidence=98% or 0.98

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 0.39:

Null hypothesis:
`p\leq 0.39`

Alternative hypothesis:
`p > 0.39`

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.42 -0.39}{\sqrt{(0.39(1-0.39))/(1000)}}=1.945`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.02`. The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(Z>1.945)=0.0259`

If we compare the p value obtained with the significance level given
`\alpha=0.02` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 2% of significance the proportion of residents who favored annexation is not significantly higher than 0.39.