According to a Pew Research Center study, in May 2011, 32% 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 351 community college students at random and finds that 130 of them have a smart phone. In testing the hypotheses: H 0 : p = 0.32 versus H a : p > 0.32, she calculates the test statistic as z = 2.0230. Then the p ‑value =
Given that according to a Pew Research Center study, in May 2011, 3 p = 2% 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 n = 351 community college students at random and finds that X= 130 of them have a smart phone.
Thus the hypotheses are:
H 0 : p = 0.32 versus
H a : p > 0.32
Thus based on the hypothesis it will be a right tailed test.
Now the test statistic is given as z = 2.0230. Then the p ‑value is computed using the excel formula for normal distribution which is =1-NORM.S.DIST(2.0230, TRUE), thus the P-value is computed as 0.0215.
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