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tα/2, n−1=t0.01/2,60−1=2.662Mean±t0.01/2,n−1(σ√n)¿23.5−(2.662∗(12.2√60))¿19.30731479¿23.5+(2.662∗(12.2√60))¿27.6926852119.30731479¿¿, 27.69268521) b.What is the margin of error at the 95% confidence level?tα/2, n−1=t0.05/2,60−1=2.662Marginof Error=z0.05/2, n−1(σ√n)¿(2.001∗(12.2√60))¿3.141601468
Unit 6 Exercises56.The average score on an employee satisfaction survey is found to be 3.75 (out of 5). The sample standard deviation is 1.24. These two statistics are calculated from a random sample of 25 employees in a mid-size professional firm. (4 points)a.With 95% confidence, what is the margin of error?tα2, n−1¿t0.052,25−1¿2.064Marginof Error=tα/2, n−1(σ√n)2.064∗(1.24√25)¿¿¿0.511872b.What is the 95% confidence interval estimate of the population mean score?Mean±Margin of Error¿3.75+0.511872¿4.261872¿3.75−0.511872¿3.2381283.238128,4.261872¿¿) 7.Annual starting salary for graduates of the BBA program in a Canadian University is expected to be between $35,000 and $55,000. A survey involving a random sample of graduates from this program is planned and the first order of business is to decide on an appropriate sample size. As usual, the 95% level of confidence will be used. (6 points)a.What is the planning value for the population standard deviation to be used in the sample size formula for estimating the population-mean starting salary?Approximate Mean=Range4¿(55000−35000)4¿1250
Unit 6 Exercises6b.How large should the sample size be if the desired margin of error is $1000?Sample¿¿(zα/2σE)2z1250/2=1.96¿(1.96∗12501000)2¿6.0025c.How large should the sample size be if the desired margin of error is $ 500?