Question

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.(a) A newspaper article states that only a minority of the Americans who decide not to go to college do so because they cannot afford it and uses the point estimate from this survey as evidence. Conduct a hypothesis test to determine if these data provide strong evidence supporting this statement.(b) Would you expect a confidence interval for the proportion of American adults who decide not to go to college because they cannot afford it to include 0.5? Explain.

Answer 1

a)
`z=\frac{0.48 -0.5}{\sqrt{(0.5(1-0.5))/(331)}}=-0.727`

`p_v =P(z<-0.727)=0.234`

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 American adults who do not have a four-year college degree and are not currently enrolled in school is not significantly lower than 0.5

b) Since we fail to reject the null hypothesis then we can't reject the hypothesis that the true proportion can be 0.5 or higher so then we can say that we can expect 0.5 in the confidence interval for this case.

Explanation:

Data given and notation

n=331 represent the random sample taken

`\hat p=0.48` estimated proportion of people who decided not go to college

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

`\alpha=0.05` represent the significance level assumed

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Part a: Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of the Americans who decide not to go to college do so because they cannot afford is less than 0.5.:

Null hypothesis:
`p \geq 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{0.48 -0.5}{\sqrt{(0.5(1-0.5))/(331)}}=-0.727`

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

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

`p_v =P(z<-0.727)=0.234`

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 American adults who do not have a four-year college degree and are not currently enrolled in school is not significantly lower than 0.5

Part b

Since we fail to reject the null hypothesis then we can't reject the hypothesis that the true proportion can be 0.5 or higher so then we can say that we can expect 0.5 in the confidence interval for this case.