13 Periodically, MerillLynch customers are asked to evaluare Merrill Lynch finan
ID: 3319962 • Letter: 1
Question
13
Periodically, MerillLynch customers are asked to evaluare Merrill Lynch financial consultants and services on a 7-point scale (7 is the maximumm). You are a manager with two consulrants you are trying to compare. Conaultant A has 10 years of experience, whereas consultant Bhas 1 year of experience You want to test the claim that the consulrant with more experience has the higher mean service rating So, you collect independent samples of service ratings for these two financil consultants Consltant A has 11 surveys with a mean of 6.55 and standard deviation of 051. Consultant B has 10 surveys with a mean of 6.4 and standard deviation of 0.55. You wish to test the claim at a significance level afa 0.01 a. What is the test staristic for this sample? test statistic- Round to 4 decimal places. b. What is the p-value for this sauple? p value- Round to 4 decimal places. c. The p valu. less than (or equal to)a greater than d. This test statistic lcads to a decision to reject the mll accept the mall fail to reject the null e. As such, the final conclusion is that There is sufficient evidence to waran rejection of the claim that the first population mean is greater tan the second population mean . here not sti cient eric ence to warrant re ection ofthe claim that the first popilation mean reater than the secon population mean The sauple data suppor tclai tht th first populatio me is greala than the secod populatiuea here ts not sufficient sample evidence to support the claim that the first popu ation mean reater 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: A< B
Alternative hypothesis: A > B
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.2322
DF = 19
t = [ (x1 - x2) - d ] / SE
t = 0.65
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 0.65. We use the t Distribution Calculator to find P(t > 0.65)
Therefore, the P-value in this analysis is 0.2617
Interpret results. Since the P-value (0.2617) is greater than the significance level (0.05), we cannot reject the null hypothesis.
Fail to reject the null hypothesis.
There is sufficient evidence to warrant rejection of the claim that the first population mean is greater than the second population mean.
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