8. A confidence interval for estimating the mean Distributions1 2The psychologist can be 95% confident that the interval fromv toincludes the unknown population mean u.Normally the psychologist will not know the value of the population mean. But consider the (unrealistic) scenario that a census of legal professionals isconducted. The census reveals that the population mean is u = 41.2.How would the psychologist most likely react to the news?O The psychologist would be surprised that p = 41.2, because that value is outside the confidence interval.O The psychologist would not be surprised that µ = 41.2, because that value is inside the confidence interval.O The psychologist would be surprised that p = 41.2, because that value is inside the confidence interval.O The psychologist would not be surprised that u = 41.2, because that value is outside the confidence interval. An industrial/organizational psychologist wants to improve worker productivity for a client firm, but first she needs to gain a better understanding ofthe life of the typical white-collar professional. Fortunately, she has access to the 2008 Workplace Productivity Survey, commissioned by LexisNexisand prepared by WorldOne Research, which surveyed a sample of 650 white-collar professionals (250 legal professionals and 400 other professionals).One of the survey questions was, "How many work-related emails do you receive during a typical workday?" For the subsample of legal professionals(n = 250), the mean response was M = 36.7 emails, with a sample standard deviation of s = 24.3 emails.The estimated standard error is SM =Use the following Distributions tool to develop a 95% confidence interval estimate of the mean number of work-related emails legal professionalsreceive during a typical workday.Select a DistributionDistributions1

Given Data : Sample Size, n = 250 Sample Mean, x̄ = 36.7 standard deviation, s = 24.3 Significance level, α =1- 0.95 = 0.05Solution : A) The estimated standard error is SM = s/√n = 24.3/√250 = 1.5369B) 95% confidence interval : Degree of freedom df = n-1 = 250 - 1 =249 Point estimate: x̄ = 36.7 critical value at α=0.05 with df =249 is tα/2,df=1.97 .....from student's t distribution table Margin of error: ME=tα/2,df * S/√n =1.97 × 24.3/√250 = 3.0276 Margin of error is 3.0276 Now 95% confidence interval is CI=point estimate ± ME =36.7 ± 3.0276 (36.7-3.0276 , 36.7+3.0276)s = (33.6724, 39.7276) The psychologist can be 95% confident that the interval from 33.6724 to 39.7276 includes the unknown population mean u.C) The psychologist would be surprised that u = 41.2, because that value is outside the confidence interval.