Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financ
ID: 3228777 • Letter: P
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.
State the null and alternative hypotheses.
H0: 1 - 2 greater than, greater than or equal to, less than, less than or equal to, equal to, not equal to, Item 1
Ha: 1 - 2 greater than, greater than or equal to, less than, less than or equal to, equal to, not equal to, Item 2
Compute the value of the test statistic (to 2 decimals).
What is the p-value?
The p-value is than .005, between .005 and .01, between .01 and .025, between .025 and .05, between .05 and .10, between .10 and .20, greater than .20,
What is your conclusion?
The consultant with more experience has a higher population mean rating
Cannot conclude the consultant with more experience has a higher rating
Consultant A Consultant B = 16 = 10 = 6.82 = 6.25 = 0.7 = 0.8Explanation / Answer
Back-up Theory
Model:
Let X = Service Rating of Consultant A
Y = Service Rating of Consultant B
We assume X ~ N(µ1, 12) and Y ~ N(µ2, 22) where 12 = 22 = 2 ,say and it is unknown.
We have a sample of n1 observations on X and n2 observations on Y
Hypothesis:
Since the question specifically says, ‘test to see whether the consultant with more experience has the higher population mean service rating’,
Null Hypothesis: H0: µ1 = µ2 Vs Alternative: HA: µ1 > µ2
Test Statistic:
t = (Xbar - Ybar)/[s{(1/n1) + (1/n2)}]
where Xbar and Ybar are sample means of X and Y respectively and
s2 = {(n1 - 1)s12 + (n2 - 1)s22}/(n1 + n2 - 2), s1 and s2 being the respective sample standard deviations.
Calculations:
Given
n1 =
16
n2 =
10
Xbar =
6.82
Ybar =
6.25
s1 =
0.7
s2 =
0.8
s^2 =
0.54625
s =
0.739087
t =
1.91
p-value
Distribution:
Under H0, t ~ tn1 + n2 – 2 i.e., t24 and hence p-value =P(t24 > 1.91) = 0.034 i.e.,
p-value is between .025 and .05.
Conclusion
H0 is rejected at 5% level of significance since p-value of tcal < 0.05
i.e.,
The consultant with more experience has a higher population mean rating
DONE
Given
n1 =
16
n2 =
10
Xbar =
6.82
Ybar =
6.25
s1 =
0.7
s2 =
0.8
s^2 =
0.54625
s =
0.739087
t =
1.91
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