Question

Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services (2000 Merrill Lynch Client Satisfaction Survey). Higher ratings on the client satisfaction survey indicate better service with 7 the maximum service rating. Independent samples of service ratings for two financial consultants are summarized here. Consultant A has 10 years of experience, whereas consultant B has 1 year of experience. Use = .05 and test to see whether the consultant with more experience has the higher population mean service rating. Consultant A Consultant B   = 16   = 10   = 6.82   = 6.25   = 0.7   = 0.8 State the null and alternative hypotheses. H0: 1 - 2 Ha: 1 - 2 Compute the value of the test statistic (to 2 decimals). What is the p-value? Use z-table. The p-value is What is your conclusion?

Answer 1

Hypothesis Test for 2 Means

Let be the population mean service rating for Consultant A.

Let be the population mean service rating for Consultant B.

Given:

For Consultant A: = 6.82, s₁ = 0.7, n₁ = 16

For Consultant B: = 6.25, s₂ = 0.8, n₂ = 10

Since s₁/s₂ =0.7/0.8 = 0.875 (it lies between 0.5 and 2) we used the pooled variance.

Since we use the pooled variance, the degrees of freedom = n1 + n2 - 2 = 16 + 10 - 2 = 24

The Hypothesis:

H0: - = 0: The mean service rating for Consultant A is equal to the mean service rating for Consultant B.

Ha: - > 0: The mean service rating for Consultant A is greater than the mean service rating for Consultant B.

This is a Right tailed test.

The Test Statistic:

The p Value: The p value (Right Tail) for Observed statistic = 1.91 is; p value = 0.0281 (Using z table)

The Decision Rule: If the P value is < , Then Reject H0

The Decision: Since P value (0.0281) is < (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the mean service rating for Consultant A is greater than the mean service rating for Consultant B.

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