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

Explanation / Answer

Given that,
mean(x)=6.82
standard deviation , s.d1=0.58
number(n1)=16
y(mean)=6.25
standard deviation, s.d2 =0.7
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.83
since our test is right-tailed
reject Ho, if to > 1.83
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6.82-6.25/sqrt((0.3364/16)+(0.49/10))
to =2.15
| to | =2.15
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.83
we got |to| = 2.15401 & | t | = 1.83
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.154 ) = 0.02982
hence value of p0.05 > 0.02982,here we reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.15
critical value: 1.83
decision: reject Ho
p-value: 0.02982

we have enouth evidence to support the claim,consultant with more experience has the higher population service rating

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