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Homework answers / question archive / A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12
A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means?
Question (2):
Ms. Lisa Monnin is the budget director for Nexus Media, Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information(Please refer the table in attached question file).
At the .10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? What is the p-value?
Question (3):
The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random and their weekly incomes before and after the plan were recorded(Please refer the table in attached question file)
Was there a significant increase in the typical salesperson's weekly income due to the innovative incentive plan? Use the .05 significance level. Estimate the p-value, and interpret it.
Solution (1):
Sample Mean : Population I = 350 , Population II = 342
Sample Standard Deviation : Population I = 12 , Population II = 15
An unbiased estimate of the common population standard deviation = 13.68
Standard Error of the difference between two means = 4.8455
Value of t = 1.6510
Solution (2)
Sales sample variance = 124.92
Standard Error = 7.315
Value of t = 1.6678
At 11 degrees of freedom and at 0.10 significance level the computed value of t ( 1.6678) is greater than the table value of t ( 1.363 ), therefore , the difference between the two means is significant.
Hence we reject the Null hypothesis and accept the Alternate Hypothesis.
Hence it is concluded that the mean daily expenses of sales staff is more than the mean daily expenses of the audit staff.
Solution (3):
Value of t = 2.2035
At 11 degrees of freedom, at 0.05 significance level the calculated value of t (2.2035) is greater than the table value of t (1.796). Therefore, we reject the Null hypothesis and accept the Alternate hypothesis.
Hence, we conclude that there is a significant increase in weekly income after the innovative plan.
PFA
