Question

In a recent poll of 3011 adults, 73% said that they use the Internet. A newspaper claims that more than 75% of adults use the Internet. Use a 0.05 significance to test the claim. Find the P-value and state an initial conclusion.

0.0057; fail to reject the null hypothesis

0.9943; reject the null hypothesis

0.0057; reject the null hypothesis

0.9943; fail to reject the null hypothesis

Answer 1

`z=\frac{0.73 -0.75}{\sqrt{(0.75(1-0.75))/(3011)}}=-2.534`

`p_v =P(z<-2.534)=0.0057`

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.

So then the best option woudl be:

0.0057; reject the null hypothesis

Explanation:

Data given and notation

n=2011 represent the random sample taken

`\hat p==0.73` estimated proportion of adults who use the Internet

`p_o=0.75` 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 0.75.:

Null hypothesis:
`p\geq 0.75`

Alternative hypothesis:
`p < 0.75`

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{0.73 -0.75}{\sqrt{(0.75(1-0.75))/(3011)}}=-2.534`

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 left tailed test the p value would be:

`p_v =P(z<-2.534)=0.0057`

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.

So then the best option woudl be:

0.0057; reject the null hypothesis