"> (Answered) 1) The report of a sample survey of 1014 adults says, "with 95% confidence, between 9% and 15% of all Americans expect to spend more on gifts this - Tutorials Prime

## (Answered) 1) The report of a sample survey of 1014 adults says, "with 95% confidence, between 9% and 15% of all Americans expect to spend more on gifts this

1) The report of a sample survey of 1014 adults says, "with 95% confidence, between 9% and 15% of all Americans expect to spend more on gifts this

Â 1)Â Â Â Â Â The report of a sample survey of 1014 adults says, â€œwith 95% confidence, between 9% and 15% of all Americans expect to spend more on gifts this year than last yearâ€ Â What does the phrase â€œ 95% confidenceâ€ mean? Â A.Â Â Â Â Â 9% to 15 % of all Americans will spend 95% of what they spend last year.B.Â Â Â Â Â Â There is a 95% chance that the percent who expect to spend more is between 9% and 15% C.Â Â Â Â Â Â The method used to get the interval 9% to 15%, when used over and over again, produces intervals which includes the true population proportion 95% of the time. D.Â Â Â Â Â 95% of all Americans will spend between 9% and 15% more than what they spend last year. 2)Â Â Â Â Â A survey of 30 random shoppers found a 95% confidence interval that shoppers will spend between \$24.50 and \$28.30. A second survey surveyed 25 random shoppers. A.Â Â Â Â Â The second interval will be narrower because it is likely that there are less outliers. B.Â Â Â Â Â Â The second interval will be narrower because the sample size is smaller and this creates less uncertainty. C.Â Â Â Â Â Â The second interval will be wider because it is likely that there are more outliers. D.Â Â Â Â Â The second interval will be wider because the sample size is smaller and this creates more uncertainty3)Â Â Â Â Â A sample of 125 randomly selected students, found that the proportion of students planning to travel home for Thanksgiving is 0.67. What is the standard deviation of the sampling distribution? Round to 3 decimal digits. Show work. 4)Â Â Â Â Â A survey of 439 citizens found that 386 of them favor a new bill introduced by the city. We want to find a 95% confidence interval for the true proportion of the population who favor the bill. What is the lower limit of the interval? Round to 3 decimal digits and show work. Â Â Â Â

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This question was answered on: Jan 02, 2020

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