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 a = .05 and test to see whether the consultant with more experience has the higher population mean service a. State the null and alternative hypotheses. b. Compute the value of the test statistic (to 2 decimals). 1.99 O c. What is the p-value? d. What is your conclusion?

Answer 1

Explanation:

Hello!

The objective is to compare the mean rating on the client satisfaction survey of two consultants A and B.

Consultant A

X₁: Rating on the client satisfaction survey given to consultant A

n₁= 16

X[bar]₁= 6.82

S₁= 0.64

Consultant B

X₂: Rating on the client satisfaction survey given to consultant B

n₂= 10

X[bar]₂= 6.25

S₂= 0.75

a. The claim is that consultant A, which is more experienced, has a higher average service rate than consultant B, which has less experience, then the hypotheses are:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

b. If both variables have a normal distribution and the population's variances are unknown but equal, the statistic for this test is a t-test for independent samples with pooled sample variance.

`t_(H_0)= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{(1)/(n_1) +(1)/(n_2) } } ~~t_(n_1+n_2-2)`

`Sa^2= ((n_1-1)S_1^2+(n_2-1)S_2^2)/(n_1+n_2-2) = (15*(0.64^2)+9*(0.75^2))/(16+10-2)= 0.47`

Sa= 0.68

`t_(H_0)= \frac{6.82-6.25-0}{0.68*\sqrt{(1)/(16) +(1)/(10) } }= 2.079`

c. The p-value is the probability of obtaining a value as extreme as the value of the statistic. As the test it is one-tailed and has the same direction (right), symbolically:

P(t₂₄≥2.079)= 1 - P(t₂₄<2.079)= 1 - 0.9758= 0.0242

d. The p-value:0.0242 is less than α:0.05, then the decision is to reject the null hypothesis.

Using a significance level of 5%, there is significant evidence to say that the average service rate of consultant A is higher than the average service rate of consultant B.

I hope this helps!