Statistical Methods and Confidence Interval Analysis: Assignments on Sampling, Estimation, and Hypothesis Testing

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1Statistical Methods and Confidence Interval Analysis: Assignments onSampling, Estimation, and Hypothesis TestingAssignment-3-solution:(Chs.7and 8):Due byMidnight ofSunday,October14th, 2012:drop box3:Total70pointsTrue/False(1pointeach)Chapter 71.A sample size of1000is large enough to conclude that the sampling distributionof pis a normal distribution, when the estimate of the population proportion is.996.FALSEHere, n(1-p) =1000*0.004= 4 which is less than 10or even 5 (theliberal rulesuggested in the book)2.The standard deviation of all possible sample proportionsdecreasesas thesample sizedecreases.FALSEIt increases when n gets smaller.3.If the population is normally distributed then the sample meanisnormallydistributedfor anysample size.TRUE(Instructionson Ch 7, property 4)4.The reason sample variance has a divisor of n-1 rather than n is that it makes thestandard deviation an unbiased estimator of the population standard deviation.FALSESample variance is unbiased but sample standard deviation is not.5.The mean of the sampling distribution ofis always equal to the mean of thesampled population.TRUEChapter 86.First a confidence interval is constructed without using the finite populationcorrection factor. Then, for the same identical data, a confidence interval isconstructed using the finite population correction factor.The width of the intervalwith the finite population correction factor iswiderthan the confidence intervalwithout the finite population correction factor.

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2False.Look at the formula for finite population correction. It reduces thestandard deviation.7.When the population is normally distributed and the population standarddeviationis unknown, then for any sample size n, the sampling distribution of𝑿ഥis based on the t distribution.TRUEWhen we substitute the estimated standard deviation in the formula,thedistribution becomes t-distribution instead of Z distribution.8.When the level of confidence and sample standard deviation remain the same, aconfidence interval for a population mean based on a sample of n=200 will bewiderthan a confidence interval for a population mean based on a sample of n=150.False9.When the level of confidence and the sample size remain the same, a confidenceinterval for a population meanμwill benarrower, when the sample standarddeviation s is small than when s islarge.True10.When the level of confidence and sample proportionpremain the same, aconfidence interval for a population proportion p based on a sample of n=100 willbenarrowerthan a confidence interval for p based on a sample of n=400.FALSE11.The sample mean, the sample proportion and the sample standard deviation areall unbiased estimators of the corresponding population parameters.FALSETheSample Standard Deviation is not an unbiased estimator (Clearly stated inInstructions on Ch8page6)12.Assuming the same level of significance, as the sample size increases, thevalue oft/2approaches the value ofz/2.TRUEMultipleChoices(2points each)Chapter 71.A manufacturing company measures the weight of boxes before shipping themto the customers. If the box weights have a population mean and standard deviationof 90 lbs. and 24 lbs. respectively, then based on a sample size of 36 boxes, theprobability that the average weight of the boxes willexceed94 lbs. is:A.15.87%

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3B.84.13%C.34.13%D.56.36%E.16.87%The𝛔𝑿ഥ= 24/√36= 4.Therefore the Z score is (94-90)/4 = 1.0P(Z≥1.0) =1-0.8413= 0.1587 or 15.87%.2.The internal auditing staff of a local manufacturing company performs a sampleaudit each quarter to estimate the proportion of accounts that are delinquent morethan 90 days overdue. The historical records of the company show that over thepast 8 years 13 percent of the accounts are delinquent. For this quarter, the auditingstaff randomly selected 250 customer accounts. What is the probability that morethan 40 accounts will be classified as delinquent?A.42.07%B.92.07%C.7.93%D.40.15%E.90.15%Hereπ= 0.13, n = 250, and 1-π= 0.87. We have nπ=32.5 and n(1-π) = 217.5,both greater than 10, and nπ(1-π) = 28.275 > 10. So,normal approximationwithout continuity correction is appropriate. The standard error ofp =σp=ඥ𝟎.𝟏𝟑∗𝟎.𝟖𝟕/𝟐𝟓𝟎= 0.0213Next, 40 accounts out of 250 in proportion is 40/250 = 0.16.The questionrefers to more than 40. Therefore, the question is P(p≥0.16)?Using the standardization processwith μp= 0.13 (the population proportion)andσp= 0.0213, we have P(Z≥𝟎.𝟏𝟔ି𝟎.𝟏𝟑𝟎.𝟎𝟐𝟏𝟑)= P(Z≥1.41) =1-P(Z≤1.41) =1-0.9207= 0.0793from the table or MegaStat.3.If we have a sample size of 100 and the estimate of the population proportion is0.10, the mean of the sampling distribution of thesample proportion is:A.0.009B.0.10

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