Suppose that a health expert has claimed that 28% of college students smoke cigarettes. To investigate this claim, researchers survey a random sample of college students. Using this sample, a 95% confidence interval for the percentage of college students who smoke is found to be 19% to 24%. On the basis of this interval, explain whether the claim that 28% of all college students smoke is reasonable or not.

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A: The objective of this question is to construct a 90% confidence interval for the population…

Q: A survey administered in a certain socioeconomic group gathered information to assess self-esteem of…

A:

Q: According to a poll, 81% of California adults (409 out of 506 surveyed) feel that education is one…

A: Point estimate=0.81

Q: Suppose a 95% confidence interval was computed based on a sample of 320 trick-or-treaters. If the…

A:

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

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Q: n = 260 and X = 208, construct a 90% confidence interval for the population proportion, p. Give…

A: Given data,x=208n=260p^=xn=208260=45=0.80

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A: We know that a margin of error tells us how many percentage points our results will differ from the…

Q: As part of the Pew Internet and American Life Project,, researchers conducted two surveys in late…

A: Given: n1= 800 ,x1= 584 n2= 2253 ,x2= 1059 At 95.0 % confidence level , z score= 1.96 Need to find…

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A: Given: A survey reveals that 12% of Americans believe Elvis is still alive. A margin of error is 4…

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A: Confidence interval : Confidence interval is a range of values it contains the true population…

Q: If n = 400 and X = 25, construct a 99% confidence interval estimate for the population proportion

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A: Part 1: a) 95% of all students pay between $1372 and $1562 for food. Comment: The interpretation is…

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A: Given: n = 506 x = 406 90% confidence level Level of significance = 0.10

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A: From the above data  We are given that the  95% confidence interval is (10, 10.5). The population…

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A: The sample proportion of the sample of size n can be defined as:  Here, X = Total number of success…

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A: The confidence level is 0.90, the standard deviation is 310 and margin of error E is 50.

Q: If n = 230 and X = 184, construct a 99% confidence interval for the population proportion, p.Give…

A: Given data n = 230 x = 184 sample proportion

Q: You are told that a survey produced a 90% confidence interval for the mean spent on utilities per…

A:

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

A: Solution: It is given here:

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A: Number of Canadian adults in 2005= 75 Number of Canadian adults in 2009 = 120   significance level =…

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A: Given that  X=133 , n=624 , p=20%=0.20

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A: It is given that  Favourable cases, X = 121 Sample size, n = 275 Sample proportion, p^ = X/n…

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A:

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A: Given , n = 1008 x = 598

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A: Hey, since there are multiple subparts posted, we will answer the first three subparts. If you want…

Q: The 95​% confidence interval for the proportion that express a great deal of confidence or quite a…

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Q: According to a recent study only 34% of adults in the United States could name all three branches of…

A: Given data, p=34%=0.34 n=1750 z-value at 95% confidence is Zc=1.96

Q: A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll,…

A:

Q: According to a poll, 83% of California adults (420 out of 506 surveyed) feel that education is one…

A:

Q: You are going to construct a 99% confidence interval for the proportion of hip replacement…

A: