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

Answer:

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

$p_v =P(Z>2.25)=0.012$  

Step-by-step explanation:

1) Data given and notation

n=366 represent the random sample taken

X=160 represent the people with smartphones

$\hat p=\frac{160}{366}=0.437$ 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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.