Statistics: Exam Study Guide
A statistics practitioner took a random sample of 50 observations from a population with a standard deviation of 25 and computed the sample mean to be 100. a.Use an interval estimate to estimate the population mean with 90% confidence.[mu-1.64*SD/√n,mu+1.64*SD/√n] [100-1.64*25/√50,100+1.64*25/√50][94.2017,105.7983]-90% Confidence b.95% confidence. [mu-1.96*SD/√n,mu+1.96*SD/√n] [100-1.96*25/√50,100+1.96*25/√50][93.0704,106.9296]-95% Confidence c.99% confidence. [mu-2.57*SD/√n,mu+2.57*SD/√n] [100-2.57*25/√50,100+2.57*25/√50][90.9137,109.0863]-99% Confidence
The mean of a sample of size n=35 was calculated as xbar=503.4. The sample was randomly drawn from a population with a standard deviation of 15. A researcher wishes to perform the following hypothesis test: H0 :mu500 a.Determine the t-statistic of the above test. Xbar-mu/(SD/√n)=t-statistic 503.4 – 500/(15/√35)=3.4/2.535=1.340978 b.Determine the pvalue of the above test. 1-(Chart of Tstat)=1-.9099=.0901=pvalue c.Suppose a larger sample size n=75, and sample mean remains xbar=503.4 Determine the pvalue of the hypothesis test described above. 503.4 – 500/(15/√75)=3.4/1.732=1.96299 1-.9750=.025=pvalue d.Determine with sample size n=125 503.4 – 500/(15/√125)=3.4/1.3416=2.53421 1-.9943=.0057=pvalue e.Relationship between sample size and pvalue? As sample size increases,pvalue decreses. Inverse relationship.
A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a preliminary study, he guesses that the standard deviation of the population of weight losses is about 10 pounds. How large a sample should he take to estimate the mean weight loss to within 2 pounds with 90% confidence? 1.645(90%)*10/√n=2 2√n=(1.645*10/2)2 n=67.6506 Sample size should be 67.65 or 68 to estimate within 2 pounds having a90%confidencelevel
Fightmaster and Associates Real Estate Inc. advertises that the mean selling time of a residential home is 40 days or less. A sample of 50 recently
The mean of a sample of size n=35 was calculated as xbar=503.4. The sample was randomly drawn from a population with a standard deviation of 15. A researcher wishes to perform the following hypothesis test: H0 :mu500 a.Determine the t-statistic of the above test. Xbar-mu/(SD/√n)=t-statistic 503.4 – 500/(15/√35)=3.4/2.535=1.340978 b.Determine the pvalue of the above test. 1-(Chart of Tstat)=1-.9099=.0901=pvalue c.Suppose a larger sample size n=75, and sample mean remains xbar=503.4 Determine the pvalue of the hypothesis test described above. 503.4 – 500/(15/√75)=3.4/1.732=1.96299 1-.9750=.025=pvalue d.Determine with sample size n=125 503.4 – 500/(15/√125)=3.4/1.3416=2.53421 1-.9943=.0057=pvalue e.Relationship between sample size and pvalue? As sample size increases,pvalue decreses. Inverse relationship.
A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a preliminary study, he guesses that the standard deviation of the population of weight losses is about 10 pounds. How large a sample should he take to estimate the mean weight loss to within 2 pounds with 90% confidence? 1.645(90%)*10/√n=2 2√n=(1.645*10/2)2 n=67.6506 Sample size should be 67.65 or 68 to estimate within 2 pounds having a90%confidencelevel
Fightmaster and Associates Real Estate Inc. advertises that the mean selling time of a residential home is 40 days or less. A sample of 50 recently