Question

A member of the House of Representatives wants to determine the proportion of voters in her district who favor a flat income tax. A random sample of 200 voters in her district showed 89 in favor. (Brase & Brase, Understandable Statistics, test item file p.8-4).a. Is there evidence at the 5% level that the majority of voters oppose the tax?

Answer 1

Answer:

$z=\frac{0.555 -0.5}{\sqrt{\frac{0.5(1-0.5)}{200}}}=1.56$  

$p_v =P(Z>1.56)=0.05938$  

If we compare the p value obtained with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of voters in her district who oppose a flat income tax is not significantly higher than 0.5 or 50%.  

Step-by-step explanation:

1) Data given and notation

n=200 represent the random sample taken

X=89 represent the voters in her district who favor a flat income tax

$\hat p=1-\frac{89}{200}=0.555$ estimated proportion of voters in her district who oppose a flat income tax

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 majority of voters oppose the tax:  

Null hypothesis:$p\leq 0.5$  

Alternative hypothesis:$p > 0.5$  

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:  

$z=\frac{0.555 -0.5}{\sqrt{\frac{0.5(1-0.5)}{200}}}=1.56$  

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 $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z>1.56)=0.05938$  

If we compare the p value obtained with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 5% of significance the proportion of voters in her district who oppose a flat income tax is not significantly higher than 0.5 or 50%.