Question

Periodically, customers of a financial services company are asked to evaluate the company's financial consultants and services. 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 α = 0.05 and test to see whether the consultant with more experience has the higher population mean service rating. Consultant A Consultant B n1 = 16 n2 = 10 x1 = 6.85 x2 = 6.28 s1 = 0.65 s2 = 0.75

(a) State the null and alternative hypotheses.

H0:

Ha:

(b) Compute the value of the test statistic. (Round your answer to three decimal places.)

(c) What is the p-value? (Round your answer to four decimal places.)

(d) What is your conclusion? Reject H0. There is sufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Do not reject H0. There is insufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Reject H0. There is insufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Do not Reject H0. There is sufficient evidence to conclude that the consultant with more experience has a higher population mean rating.

Answer 1

To test null hypothesis Ho M1 - U2 = 0 against alternative hypothesis H1 H1-2 > 0 Let či denotes the sample mean of first sample, si denotes the sample standard deviation of first sample and ni denotes the sample size of first sample and 22 denotes the sample mean of second sample, S2 denotes the sample standard deviation of second sample and nz denotes the sample size of second sample. Here, 21 = 6.8, si = 0.6, n1 = 16,2 = 6.3, s2 = 0.8, n2 = 10 The test statistic can be written as: (C1 – C2) - 0 t = + which under Ho follows a t distribution with 17 df. where We reject Ho at 0.05 level of significance if P-value < 0.05 or if tobs > t0.05,df Now, The value of the test statistic = 1.982609 associated degrees of freedom = 17 The critical value =1.739607 P-value = P(t17 > tobs) = P(t17 > 1.982609) = P(t17 > 1.982609) = 1 -0.968095 = 0.031905 Since p-value < 0.05 and tobs > tcritical = 1.739607, so we reject Ho at 0.05 level of significance Hence, we can conclude that mean for first population is signiifcantly greater than the mean for second population.

d) Conclusion - Reject H0. There is sufficient evidence to conclude that the consultant with more experience has a higher population mean rating.