Question

According to a Pew Research Center study, in May 2011, 38% of all American adults had a smart phone (one which the user can use to read email and surf the Internet). A communications professor at a university believes this percentage is higher among community college students. She selects 366 community college students at random and finds that 160 of them have a smart phone. Then in testing the hypotheses: H 0 : p = 0.38 versus H a : p > 0.38, what is the test statistic? z = . (Please round your answer to two decimal places.)

Answer 1

`z=\frac{0.437 -0.38}{\sqrt{(0.38(1-0.38))/(366)}}=2.25`

`p_v =P(Z>2.25)=0.012`

Explanation:

1) Data given and notation

n=366 represent the random sample taken

X=160 represent the people with smartphones

`\hat p=(160)/(366)=0.437` estimated proportion of people with smartphones

`p_o=0.38` is the value that we want to test

`\alpha` represent the significance level

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 true proportion of people with smatrphones is higher than 0.38:

Null hypothesis:
`p \leq 0.38`

Alternative hypothesis:
`p > 0.38`

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{0.437 -0.38}{\sqrt{(0.38(1-0.38))/(366)}}=2.25`

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 assumed is
`\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>2.25)=0.012`

So the p value obtained was a very low value and using the significance level assumed
`\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 true proportion of people with smatphones is higher than 0.38.