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

Answer:

a) Null hypothesis:$\mu_{A} \leq \mu_{B}$

Alternative hypothesis:$\mu_{A} > \mu_{B}$

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

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 $p_v$ 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.

Step-by-step 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:

$t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (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:

$t=\frac{6.82-6.25}{\sqrt{\frac{0.64^2}{16}+\frac{0.75^2}{10}}}}=1.992$  

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 $p_v$ 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.