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

Answer:

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

$p_v =P(Z>1.945)=0.0259$  

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.  

Step-by-step 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:  

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

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

3) Calculate the statistic  

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

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

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.