Question

The mayor of a town has proposed a plan for the construction of an adjoining bridge. A political study took a sample of 1700 voters in the town and found that 66% 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 63%.

A) Testing at the 0.02 level, is there enough evidence to support the strategist's claim?

Answer 1

`z=\frac{0.66 -0.63}{\sqrt{(0.63(1-0.63))/(1700)}}=2.562`

`p_v =P(z>2.562)=0.0052`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.02` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that at 2% of significance the proportion of voters that favored construction is higher than 0.63 or 63%.

Explanation:

Data given and notation

n=1700 represent the random sample taken

`\hat p=0.66` estimated proportion of voters that favored construction

`p_o=0.63` 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that percentage of residents who favor construction is more than 63%.:

Null hypothesis:
`p \leq 0.63`

Alternative hypothesis:
`p > 0.63`

When we conduct a proportion test we need to use the z statisitc, 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`.

Calculate the statistic

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

`z=\frac{0.66 -0.63}{\sqrt{(0.63(1-0.63))/(1700)}}=2.562`

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>2.562)=0.0052`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.02` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that at 2% of significance the proportion of voters that favored construction is higher than 0.63 or 63%.