Question

Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services. 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 α = .05 and test to see whether the consultant with more experience has the higher population mean service rating.Consultant A: n = 16, x = 6.82, s = 0.64Consultant B: n = 10, x = 6.25, s = 0.75a. State the null and alternative hypotheses.b. Compute the value of the test statistic.c. What is the p-value?d. What is your conclusion?

Answer 1

a) Null hypothesis:
`\mu_(A) \leq \mu_(B)`

Alternative hypothesis:
`\mu_(A) > \mu_(B)`

b)
`t=\frac{6.82-6.25}{\sqrt{(0.64^2)/(16)+(0.75^2)/(10)}}}=1.992`

c)
`p_v =P(t_((24))>1.992)=0.0289`

d) If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the more experience consultant A have a significant higher rate compared to the consultant B with less experience at 5% of significance.

Explanation:

1) Data given and notation

`\bar X_(A)=6.82` represent the mean for the sample of Consultant A

`\bar X_(B)=6.25` represent the mean for the sample of Consultant B

`s_(A)=0.64` represent the sample standard deviation for the sample of Consultant A

`s_(B)=0.75` represent the sample standard deviation for the sample of bonsultant B

`n_(A)=16` sample size selected for the Consultant A

`n_(B)=10` sample size selected for the Consultant B

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the Consultant A (more experience) is higher than the mean for the Consultant B, the system of hypothesis would be:

Null hypothesis:
`\mu_(A) \leq \mu_(B)`

Alternative hypothesis:
`\mu_(A) > \mu_(B)`

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{\bar X_(A)-\bar X_(B)}{\sqrt{(s^2_(A))/(n_(A))+(s^2_(B))/(n_(B))}}` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

`t=\frac{6.82-6.25}{\sqrt{(0.64^2)/(16)+(0.75^2)/(10)}}}=1.992`

Part c: P-value

The first step is calculate the degrees of freedom, on this case:

`df=n_(A)+n_(B)-2=16+10-2=24`

Since is a one side test the p value would be:

`p_v =P(t_((24))>1.992)=0.0289`

Part d: Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and the more experience consultant A have a significant higher rate compared to the consultant B with less experience at 5% of significance.