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

Answer:

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

$p_v =P(z>2.562)=0.0052$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.02$ we have $p_v$ 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%.

Step-by-step 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:  

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

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

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 $p_v$ 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%.