In: Statistics and Probability
According to a Pew Research Center study, in May 2011, 39% of all American adults had a smartphone (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 342 community college students at random and finds that 139 of them have a smartphone. Then in testing the hypotheses: H0 : p = 0.39 versus Ha : p > 0.39
What is the test statistic?
z =
(Please round your answer to two decimal places.)
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.39
Alternative hypothesis: P > 0.39
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).
S.D = sqrt[ P * ( 1 - P ) / n ]
S.D = 0.02637
z = (p - P) / S.D
z = 0.62
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.62.
Thus, the P-value = 0.268.
Interpret results. Since the P-value (0.268) is greater than the significance level (0.05), we have to accept the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that percentage is higher among community college students.