Question

According to a Pew Research Center study, in May 2011, 35% of all American Points: 10 out of 10 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 300 community college students at random and finds that 120 of them have a smart phone. In testing the hypotheses:
H0: p = 0.4 versusHa: p > 0.4,what is the test statistic?z =________________. (Please round your answer to two decimal places.)

Answer 1

Answer:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$  

If we assume the hypothesis of:

H0: p = 0.4 versus Ha: p > 0.4

The statistic for this case would be:

$z=\frac{0.4 -0.4}{\sqrt{\frac{0.35(1-0.35)}{300}}}=0$  

Step-by-step explanation:

Information given

n=300 represent the random sample selected

X=120 represent the number of people that have smart phone

$\hat p=\frac{120}{300}=0.4$ estimated proportion of people with a smart phone

$p_o=0.35$ is the value that we want to test

z would represent the statistic

Hypothesis to test

For this case we want to test if the true proportion is hgiher than 0.35 since thats the claim given:

Null hypothesis:$p\leq 0.35$  

Alternative hypothesis:$p > 0.35$  

The statistic is given by:

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

Replacing the info we got:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$  

If we assume the hypothesis of:

H0: p = 0.4 versus Ha: p > 0.4

The statistic for this case would be:

$z=\frac{0.4 -0.4}{\sqrt{\frac{0.35(1-0.35)}{300}}}=0$