For questions 20-23, the Pew Foundation\'s Internet & American Life project aske
ID: 3247012 • Letter: F
Question
For questions 20-23, the Pew Foundation's Internet & American Life project asked American cell phone owners whether they have smartphones or not (some didnt know). One demographic category they analyzed was education level:
20. How do we calculate the table of expected values, assuming region and satisfaction are independent? Based on the table below, calculate the expected number of respondents who have a high school eduaction or less, and a smartphone, assuming independence.
a) 558.38 b) 514 c) 432.81 d) 691.24
21. Before combining the "Don't Know" and "Refused" rows, this data could not be tested using the chi-squared test statistic. Why were the rows combined?
a) We can only have 3 or fewer rows for these sorts of tests. b) If the categories were considered separatly, we might expect too many answers in each. c) If the categories were considered separatly, we might expect too few answers in each. d) There is no way to distinguish "don't know" from "refused to answer".
22. What will be true of the chi-squared statistic created above?
a) It must be positive b) It will be larger if the observed table is closer to the expected table c) It would not change if two of the observed cells were switched d) It cannot exceed its critical value
23. When do we reject the null hypothesis for a test like this?
a) If the chi squared value we compute is too small, becasue that shows the tables are different b) If the chi squared value we compute is too large, becasue that shows the tables are different c) If the chi squared value we compute is too small, becasue that shows the tables are similar d) If the chi squared value we compute is too large, becasue that shows the tables are similar
College Grad Some College HS or Less Total Smartphone 632 614 514 1760 No Smartphone 231 315 556 1102 Don't Know/Refused 9 25 111 145 Total 872 954 1181 3007Explanation / Answer
Q20) Expected frequency of High Scholl or less and Smartphone = (1760*1181)/3007 = 691.24
Option D is correct
Q21) Option D is correct
Q22) a) It must be positive because Chi square does not have negative values
Q23) Option B is correct, if Chi square computed is larger than critical value, then we reject the null hypothesis
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