CH12-13 exercises solutions

CH12-13 exercises solutions - X X X X X X X X X X X X X X X...

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1 X X X X X X X X X X X X X X X X X AP Statistics Packet 12/13 X Inference for Proportions Inference for a Population Proportion Comparing Two Proportions Inference for Tables: Chi-Square Procedures Test for Goodness of Fit Inference for Two-Way Tables X X X X X
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2 HW #25 6 – 8, 10, 11, 13, 21 12.6 EQUALITY FOR WOMEN? Have efforts to promote equality for women gone far enough in the United States? A poll on this issue by the cable network MSNBC contacted 1019 adults. A newspaper article about the poll said, “Results have a margin of sampling error of plus or minus 3 percentage points.” (a) Overall, 54% of the sample (550 of 1019 people) answered “Yes.” Find a 95% confidence interval for the proportion in the adult population who would say “Yes” if asked. Is the report’s claim about the margin of error roughly right? (Assume that the sample is an SRS.) p = true proportion of adults who would say “Yes” if asked whether efforts to promote equality for women have gone far enough in the U.S. SRS 1019 < 10% of adult population ˆ np = 550 > 10 and ˆ ( 1 ) np - = 469 > 10 One-proportion z interval (0.54)(0.46) 95 % 0.5 4 1.96 1019 (0.51,0.57) CI = We are 95% confident that the true proportion of adults who would say “Yes” if asked whether efforts to promote equality for women have gone far enough in the U.S. is between 51 and 57%. The margin of error is about 3% as stated. (b) The news article said that 65% of men, but only 43% of women, think that efforts to promote equality have gone far enough. Explain why we do not have enough information to give confidence intervals for men and women separately. We weren’t given sample sizes for each gender. (c) Would a 95% confidence interval for women alone have a margin of error less than 0.03, about equal to 0.03, or greater than 0.03? Why? You see that the news article’s statement about the margin of error for poll results is a bit misleading. The margin of error for women alone would be greater than 0.03 since the sample size is smaller.
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3 12.7 TEENS AND THEIR TV SETS The New York Times and CBS News conducted a nationwide poll of 1048 randomly selected 13- to 17-year-olds. Of these teenagers, 692 had a television in their room and 189 named Fox as their favorite television network. We will act as if the sample were an SRS. (a) Use your calculator to give 95% confidence intervals for the proportion of all people in this age group who have a TV in their room and the proportion who would choose Fox as their favorite network. Check that we can use our methods. (Check assumptions.) Check assumptions for TV in room: SRS 1048 < 10% of all 13- to 17-year olds ˆ np = 692 > 10, ˆ ( 1 ) np - = 356 >10 Check assumptions for prefer FOX: SRS 1048 < 10% of all 13- to 17-year olds ˆ np = 189 > 10, ˆ ( 1 - = 859 >10 95 % 0.66 (0.631 , 0.689) C I for TV in room (0.6629(0.3429 = ± 1.96 1048 = 95 % 0.18 (0.157, 0.203) C I for preferring FOX (0.1829(0.8229 = ± 1.96 1048 = (b) The news article says, “In theory, in 19 cases out of 20, the poll results will differ by no more than three percentage points in either direction from what would have been obtained by seeking out all American teenagers.” Explain how your results agree with this statement.
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This note was uploaded on 12/09/2011 for the course STAT 101 taught by Professor O during the Fall '08 term at Lake Land.

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CH12-13 exercises solutions - X X X X X X X X X X X X X X X...

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