Confidence Intervals: Calculations and Applications in Statistical Inference

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Confidence Intervals: Calculations and Applicationsin Statistical InferenceWhy is a 99% confidence interval wider than a 95%confidence interval?Solution)The definition of a confidence interval is that it contains the true populationmean. If Ihave a 95% confidence interval, that means I am 95% certainthat the true population mean is in the interval. If I want to be even morecertain, I have to widen the interval. If I can be less certain, I can narrowthe interval.So the widestinterval will be 99%, and the narrowest would be 90%.Example:you're trying to figure out where in the city Comet Donuts is in, but youreally don't know for sure. A desperately hungry person hands you a mapand asks you to show him where it is. If someone forces you to be 99%accurate, are you going to draw a wide or narrow circle on the map? Youcan't afford to be wrong-at 99% you're saying that you'll be wrong onetime out of 100! So you draw a big circle.If the person asking doesn't even like donuts, they're just asking for theheck of it, you can be 90% accurate, so you can take a chance and draw asmall circle. You'll be wrong 10% of the time.12.A person claims to be able to predict the outcome offlipping a coin. This person is correct16/25 times. Computethe 95% confidence interval on the proportion of times thisperson can predict coin flips correctly. What conclusion canyou draw about this test of his ability to predict the future?Solution)WE HAVE GIVEN THAT n = 25 and p = 16/25And we need to construct the 95% C.I. for the proportion of times thisperson can predict coins flips correctly as,ṕ± 1.96 *√(ṕq^/n)=.64± 1.96 *√(.64*.36/25)= .64 ± .1882

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So the 95% C.I. is,(0.4518, 0.8282)So We Are 95 Out Of 100 Attemptsare confident that the values of the samplesare lies b/w (.4518,.8282)15.You take a sample of 22 from a population of test scores,and the mean of your sample is 60.(a) You know the standard deviation of thepopulation is 10.What is the 99% confidence interval on the population mean?Solution)We have given that 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) Now assume that you do not know the populationstandard deviation, but the standard deviation in yoursample is 10. What is the 99% confidence interval on themean now?Solution)Here we have given that n = 22, sample mean = 60 and S = 10Here we assume that the population stranded deviation is unknown, andalso n<30 so we here use 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 psychologymajor at your college studies per night, so you asked 10

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psychology majors to tell you the amount they study. Theytold you the following times: 2, 1.5, 3, 2, 3.5, 1, 0.5, 3, 2, 4.(a)Find the 95% confidence interval on the populationmean.Sol)Here we have given that sample mean = 2.25 n= 10 andσ= 1.0548We use a standard normal distribution and assume that populationstandard deviation is known 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% confidenceinterval on the populationmean.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” whenconstructing a confidence interval for a mean?a. If we took repeated samples, approximately 90% of thesamples would produce the same confidence interval.b. If we took repeated samples, approximately 90% of theconfidence intervals calculated from those samples wouldcontainthe sample mean.c. If we took repeated samples, approximately 90% of the confidence intervalscalculated from those samples would contain the true value of the populationmean.d. If we took repeated samples, the sample mean would equal thepopulation mean in approximately 90% of the samples.

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106.Suppose that a committee is studying whether or not there iswaste of time in our judicial system. It is interested in the meanamount of time individuals waste at the courthouse waiting to becalled for jury duty. The committee randomly surveyed 81 peoplewho recently served as jurors. The sample mean wait time waseight hours with a sample standard deviation of four hours.Solution)i)x ¯ = __________8 hoursii. sx = __________4 hoursiii. n = __________81iv. n–1 = __________80b. Define the random variables X and X ¯ in words.Xisthenumberofhoursapersonwasteswhilewaitingtobecalledforservice.Xbaristheaveragenumberofhourswastedforpeopleinthesample. c.c. Which distribution should you use for this problem? Explain yourchoice.Weshouldusethestandard normal distribution because n is largeenough.d. Construct a 95% confidence interval for the population mean timewasted.(7.12,8.88)i.State the confidence interval.Sol)

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The 95% C.I. for the population mean worth of coupons is,=x (bar) ± Zα/2*S/√n= 8 ± 1.96(4/√81)= 8 ± 0.8711So,(7.129, 8.8711)ii) Sketchiii. Calculate the error bound.EBM=Zα/2*S/√n= 1.96(4/√81)= 0.88e. Explain in a complete sentence what the confidence interval means.Ifwesampled81people100times,wewouldexpect95ofthesamplestohaveaveragesintheinterval.112. In a recent sample of 84 used car sales costs, the samplemean was $6,425 with a standard deviation of $3,156. Assumethe underlyingdistribution is approximately normal.a.Which distribution should you use for this problem? Explainyour choice.We should use a standard normal distribution because n>=30

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b. Define the random variable X ¯ in words.X ¯ is the average no ofcar sales costsc. Construct a 95% confidence interval for the population meancost of a used car.i. State the confidence interval.We have given that sample maen = 6425 n = 84 and standerddeviation = 3156The 95% C.I. for the population mean is,=sample mean ± 1.96 *σ/√n= 6425 ± 1.96 * 3156 /√84= 6425 ± 674.922S0, (5777.078, 7099.922)ii. Sketch the graph.ii.Calculate the error bound.=EBM=Zα/2*σ/√nEBM = 1.96 * 3156 /√84= 674.922d. Explain what a “95% confidence interval” means for thisstudy.The 95% confidence interval means that we are 95% confidentthat thesample mean lie in the interval (5777.078, 7099.922)Use the following information to answer the next exercise: Aquality control specialist for a restaurant chain takes a

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random sample of size 12 to check the amount of sodaserved in the 16 oz. serving size. The sample mean is 13.30with a sample standard deviation of 1.55. Assume theunderlying population is normally distributed.116. What is the error bound?a. 0.87b. 1.98c. 0.99d. 1.74120.An article regarding interracial datingand marriage recentlyappeared in the Washington Post. Of the 1,709 randomly selectedadults, 315 identified themselves as Latinos, 323 identifiedthemselves as blacks, 254 identified themselves as Asians, and779 identified themselves as whites. In this survey, 86% of blackssaid that they would welcome a white person into their families.Among Asians, 77% would welcome a white person into theirfamilies, 71% would welcome a Latino, and 66% would welcomea black person.a.We are interested in finding the 95% confidence interval forthe percent of all black adults who would welcome a whiteperson into their families. Define the random variables X andP′, in words.X῀is the noof black adults who are well come a whitepersons into their families.p῀is the probability of the no of black adults who are wellcome a white persons into their families.b.Which distribution should you use for this problem? Explainyourchoice.

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We use the standard normal distribution for the populationproportion because the sample size is large enough as>=30.c.Construct a 95% confidence interval.i. State the confidence interval.We have given that n = 1709 , p = 323/1709 = .189The 95% C.I. is.=ṕ± Zα/2√(pq/n= .189 ± 1.96 *√.189*.811 /1709= 0.189± 0.0186So, (0.170, .2076)ii. Sketch the graph.iii.Calculate the error bound.EBM = zα/2√(pq/n)= 1.96 *√.189*.811/1709= 0.0186130.On May 23, 2013, Gallup reported that of the 1,005 peoplesurveyed, 76% of U.S. workers believe that they will continueworking past retirement age. The confidence level forthis studywas reported at 95% with a ±3% margin of error.Solution)a.Determine the estimated proportion from the sample.The estimated proportion from the sample,P^ = 76% = .76b.Determine the sample size.Sample size = n= 1,005c.Identify CL andα.Confidence level = 95%

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α= 5% = 0.05d.Calculate the error bound based on the information provided.EBM = zα/2√(pq/n)= 1.96 *√(.76*0.24/1005)= 0.0264e. Compare the error bound in part d to the margin of error reportedby Gallup. Explain any differences between theValues.The stated ±3 represent the MEB.Whichmeans that those working pastretirement age. And is equal to 0.0264. And seems that there is nodifference b/w EBM and margins of error that is stated in problems.e.Create a confidence interval for the results of this study.The 95% C.I. for the results of study are,=ṕ± Zα/2√(pq/n)= .76 ± 1.96 *√(.76*.24/1005)= .76 ± 0.0264(0.7336, .7864)g. A reporter is covering the release of this study for a local newsstation. How should she explain theconfidence-Interval to heraudience?A reporter is converting the release of this study for a local station, thismeans that if we sampled we would expect that 95 of the samples tohave audience in this intervals (0.7336, .7864)

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STAT200 WEEK 4 HOMEWORKProbability Distributions Problem SetSolutions are provided at the end of the documentP-1The salespeople at Owl Realty sell up to 9 houses per month. The probability distribution ofa salesperson selling n houses in a month is as follows:SalesProbability0.051.102.153.204.1556.107.058.059.05a.What is the probability of selling 5 houses in a month?b.What are the expected value and the standard deviation for the number houses sold by thesalespeople per month?P-2The following data has been recorded regarding the rental history of a high pressurecleaner at a home improvement store.DemandFrequency020

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123227314416Find the expected demand for the high pressure cleaner.

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