Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financ

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.

ConsultantA: n1=16, s1 =0.6, x1 =6.82 (with bar overx) ; Consultant B: x2 = 6.25(with bar over x), n2 =10, s2 = 0.81

a.) State the null and alternative hypothesis.

b.) compute the value of the test statistic.

c.)What is the p-value?

d.) What is your conclusion?

Explanation / Answer

Consultant A: n1=16, s1 =0.6, x1 =6.82; s12 =0.36

Consultant B: x2 = 6.25, n2 =10, s2 = 0.81, s22 = 0.6561

For testing the equality of population variance:

Ho : The population variances are equal

H1 : The population variance are not equal

f = s22 / s12 = 0.6561/0.36 = 1.8225

f table value at = 0.05 with (10-1,16-1)d.f = 2.59

Since ftab = 2.59 > fcal = 1.8225, There is no evidence to reject null hypothesis. Hence, we can conclude that the population variances are equal.

For testing the equality in mean:

Ho = The mean service rates are equal i.e., µ1=µ2

H1 = The mean service rate of consultant A is greater than consultant B i.e., µ1 > µ2

t = x1 - x2 / [Sp (1/n1 + 1/n2)]

where Sp is the pooled standard deviation, since population variances are equal.

Sp2 = (n1-1)s12 + (n2 -1)s22 / (n1 + n2 -2) = 0.47104

Sp = 0.686321

t = 6.82 – 6.25 /[ 0.686321(0.1625)] =   2.060255

d.f = 16+10-2 = 24

p-value (obtained using software) = 0.025185

t table value for =0.05 at 24 d.f = 1.711

Since, ttab = 1.711 < tcal = 2.0603 and p-value = 0.0251 < 0.05, we reject the null hypothesis and hence conclude that the mean service rate of consultant A is greater than consultant B

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