Question

In a recent poll of 750 randomly selected​ adults, 589 said that it is morally wrong to not report all income on tax returns. Use a 0.05 significance level to test the claim that 70​% of adults say that it is morally wrong to not report all income on tax returns. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

Answer 1

z= 5.08

The p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .

Explanation:

1) Data given and notation

n=750 represent the random sample taken

X=589 represent the adults that said that it is morally wrong to not report all income on tax returns

`\hat p=(589)/(750)=0.785` estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

`p_o=0.7` 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)

pv 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 70% of adults say that it is morally wrong to not report all income on tax returns.:

Null hypothesis:
`p=0.7`

Alternative hypothesis:
`p \\eq 0.7`

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`.

3) Calculate the statistic

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

`z=\frac{0.785 -0.7}{\sqrt{(0.7(1-0.7))/(750)}}=5.08`

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 bilateral test the p value would be:

`p_v =2*P(z>5.08)=0.000000377`

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who say's that it is morally wrong to not report all income on tax returns differs from 0.7 or 70% .