A sample of 40 investment customers serviced by an accountmanager are found to h

Question

A sample of 40 investment customers serviced by an accountmanager are found to have had an average of $23,000 in transactionsduring the past year, with a standard deviation of $8,500. A sampleof 30 customers serviced by another account manager averaged$28,000 in transactions, with a standard deviation of $11,000.Assuming the population standard deviations are equal, use the 0.05level of significance in testing whether the population means couldbe equal for customers serviced by the two account managers. Usingthe appropriate statistical table, what is the most accuratestatement we can make about p-value for this test? Construct andinterpret the 95% confidence interval for the difference betweenthe population means.

Explanation / Answer

Given n1=40, xbar1=23,000, s1=8,500           n2=30,xbar2=28,000, s2=11,000 =0.05, Z(0.025)=1.96 (check normal table) The hypothesis is Ho:1=2 Ha:2 not equal 2 The test statistic is Z=(xbar1 - xbar2)/(s1^2/n1+s2^2/n2) =(23000-28000)/sqrt(8500^2/40+11000^2/30) =-2.07 The p-value is 2*P(Z< - 2.07) = 0.0385 (check normal table) Since Z= -2.07< -1.96, we reject Ho. The 95% CI is (xbar1 - xbar2)±Z*(s1^2/n1+s2^2/n2) -->(23000-28000) ±1.96*sqrt(8500^2/40+11000^2/30) -->(-9736.385, -263.6150)

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