eriodically, Merrill Lynch customers are asked to evaluateMerrill Lynch financia
ID: 3220904 • Letter: E
Question
eriodically, Merrill Lynch customers are asked to evaluateMerrill Lynch financial consultants and services (2000 MerrillLynch Client Satisfaction Survey). Higher ratings on the clientsatisfaction survey indicate better services, with 7 the maximumservice rating. Independent samples of service ratings for twofinancial consultants are summarized here. Consultant A has 10years of experience, whereas consultant B has 1 year experience.Use = 0.05 and test to see whether the consultant with moreexperience has the higher population mean service rating.
ConsultantA Consultant B
n1=16 n2 =10
x1 =6.82 (with bar overx) x2 = 6.25(with bar over x)
s1 =0.64 s2 = 0.75
a.) State the null and alternative hypothesis. step by step ( i dont know how to differntiate between null and alternative
c.)What is the p-value? ( step by step please)
d.) What is your conclusion?
Explanation / Answer
Given that,
mean(x)=6.82
standard deviation , s.d1=0.64
number(n1)=16
y(mean)=6.25
standard deviation, s.d2 =0.75
number(n2)=10
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.833
since our test is right-tailed
reject Ho, if to > 1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6.82-6.25/sqrt((0.4096/16)+(0.5625/10))
to =1.992
| to | =1.992
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 1.99235 & | t | = 1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 1.9923 ) = 0.03875
hence value of p0.05 > 0.03875,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.992
critical value: 1.833
decision: reject Ho
p-value: 0.03875
the consultant with more experience has the higher population mean service rating
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