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Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Week 5 - Document preview page 1

Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Week 5 - Page 1

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Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Week 5

Practice problems on confidence intervals and statistical inference techniques.

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Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Week 5 - Page 1 preview imageConfidence Intervals and Statistical Inference: Homework Assignments andSolutionsHamiltonWeek 5 Homework4.Why is a 99% confidence interval wider than a 95% confidence interval?Solution)The definition of a confidence interval is that itcontains the true population mean. If I have a95% confidence interval, that means I am 95% certain that the true population mean is in theinterval. If I want to be even more certain, I have to widen the interval. If I can be less certain, Ican narrow the interval.So the widest interval will be 99%,and the narrowest would be 90%.12.A person claims to be able to predict the outcome of flipping a coin. This person is correct16/25 times. Compute the 95% confidence interval on theproportion of times this person canpredict coin flips correctly. What conclusion can you draw about this test of his ability topredict the future?Solution)WE HAVE GIVEN THAT n = 25 and p = 16/25And we need to constructthe 95% C.I. for the proportion of times this person can predict coinsflips correctly as,± 1.96 *(q^/n)=.64± 1.96 *(.64*.36/25)= .64 ± .1882So the 95% C.I. is,(0.4518,0.8282)So We Are 95 OutOf 100Attemptsare confident thatthe values of the samples are lies b/w(.4518,.8282)15.You take a sample of 22 from a population of test scores, and the mean of your sample is60.(a) You know the standard deviation of the population is 10. What is the 99% confidenceinterval on the populationmean?Solution)
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Confidence Intervals and Statistical Inference: Homework Assignments and Solutions Week 5 - Page 3 preview imageWe have giventhat n = 22, sample mean =60 andσ= 10The 99% C.I. for the population mean is,= sample mean ± 2.58*σ/n= 60 ± 2.58* 10 /22= 60 ± 5.501So, (54.499, 65.501)(b) Nowassume that you do not know the population standard deviation, but the standarddeviation in your sample is 10. What is the 99% confidence interval on the mean now?Solution)Here we have given that n =22,sample mean = 60 and S = 10Here we assume that the populationstrandeddeviation isunknown, and also n<30 so we hereuse the student t distribution as,= sample mean± tα/2,vs /n-1Where v= n-1 degree of freedom and v= 22-1 = 21= 60± 2.831* 10/21= 60±6.178So,(53.822, 66.178)18.You were interested in how long the average psychology major at your college studies pernight, so you asked 10 psychology majors to tell you the amount they study. They told you thefollowing times: 2, 1.5, 3, 2, 3.5, 1, 0.5, 3, 2, 4.(a)Find the 95% confidence interval on the population mean.Sol)Here we have given that sample mean = 2.25 n= 10 andσ= 1.0548We use astandardnormal distributionandassume that populationstandarddeviation isknown soThe 95 % C.I. for the population mean is,= sample mean ± 1.96 *σ/n= 2.25 ±1.96* 1.0548 /10= 2.25 ± .6538= (1.596, 2.904)(b)Find the 90% confidence interval on the population mean.THE 90% C.I. IS.= samplemean ± 1.645 *σ/n= 2.25 ± 1.645 * 1.0548/10= 2.25 ± .549So, (1.701, 2.799)100.What is meant by the term “90% confident” when constructing a confidence interval for amean?
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