Q2 Periodirally, Merill Lyach customes are asked to evaluate Merill Lynch finaic
ID: 3320257 • Letter: Q
Question
Q2
Periodirally, Merill Lyach customes are asked to evaluate Merill Lynch finaicalsultarits and services on point scale 7 is the maximun). You are g with two csultats you are tying to copare. Consultant A has 10 yes of experience, whereas consultant B has 1 year of experience. You want to test the claim hat the consultant with more experience has the higher mean service rating. So, you colleet independent samples of service ratings fot these two financial eonsultants. Cosan A has 22 surveys with a mean of 6.68 and standard deviation of0.44. Consultant B has 19 suveys with a mcan of 6.55 and standard deviation of 0.47. You wish to test the claim at a significance level ofa 0.005 a. What is the test statisti fo this saople? test statistic = Round to 4 decimal places b What is the p-value for this sample? p-value = Round to 4 decimal places. c. The p-value is... less than (or eqnal to) a O greater than at i This test statistic leads to a decision to eject the nuil accept the null fail to reject the null e. As such, the final conclusion is that. There is sufficient evidence to warrant rejection of the clairn that the first population mean is greater than the second population mca T ere is not rfficient evidence to warrant rejection atthe claim that the first popilaton mean greater than the second population mean The sample dala support the clam dal Lhe lust populalon mear1 1s glealet dan die second population nean. There is not sufficient sample evidence to support the claim that the first population mean is greater than the second population mean.Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1< 2
Alternative hypothesis: 1 > 2
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.005. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees offreedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.143
DF = 39
t = [ (x1 - x2) - d ] / SE
t = 2.45
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means produced a t statistic of 2.45.
Therefore, the P-value in this analysis is 0.0094
Interpret results. Since the P-value (0.0094) is less than the significance level (0.05), we have to reject the null hypothesis.
The sample data support the claim that the first population mean is greater than the second population mean.
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