Question

Suppose a poll is taken that shows that 765 out of 1500 randomly​ selected, independent people believe the rich should pay more taxes than they do. Test the hypothesis that a majority​ (more than​ 50%) believe the rich should pay more taxes than they do. Use a significance level of 0.05.

Answer 1

`z=\frac{0.51 -0.5}{\sqrt{(0.5(1-0.5))/(1500)}}=0.775`

`p_v =P(z>0.775)=0.219`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` 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 5% of significance the proportion of interest is not significantly higher than 0.5

Explanation:

Data given and notation

n=1500 represent the random sample taken

X=765 represent the successes

`\hat p=(765)/(1500)=0.51` estimated proportion of successes

`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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:

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{(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{0.51 -0.5}{\sqrt{(0.5(1-0.5))/(1500)}}=0.775`

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>0.775)=0.219`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` 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 5% of significance the proportion of interest is not significantly higher than 0.5