You can try a proportional one-sample z-test for this one since this problem is using proportions.
Null hypothesis:
p ≤ .50
Alternate hypothesis:
p > .50
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .52 - .50 -->test value minus population value
divided by
√[(.50)(.50)/1132]
Finish the calculation. Determine the p-value using the z-score you calculate. (The p-value is the actual level of the test statistic.) Compare to the level of significance for a one-tailed test, which is 0.05. Determine whether or not to reject the null and conclude a difference (p > .50). For part c, redo the same process using .53 instead of .52.
(Hint: One null hypothesis will be rejected; one will not.)
I hope this will help get you started.
Webcredible, a UK-based consulting firm specializ- ing in websites, intranets, mobile devices, and applications, conducted a survey of 1,132 mobile phone users between February and April 2009. The survey found that 52% of mo- bile phone users are now using the mobile Internet. (Data extracted from “Email and Social Networking Most Popular Mobile Internet Activities. The authors of the article imply that the survey proves that more than half of all mobile phone users are now using the mobile Internet.
a. Use the five-step p-value approach to hypothesis testing and a 0.05 level of significance to try to prove that more than half of all mobile phone users are now using the mo- bile Internet.
b. Based on your result in (a), is the claim implied by the authors valid?
c. Suppose the survey found that 53% of mobile phone users are now using the mobile Internet. Repeat parts (a) and (b).
d. Compare the results of (b) and (c).
1 answer