Periodically, Merrill Lynch customers are asked to evaluate Merrill Lynch financial consultants and services (2000 Merrill Lynch Client Satisfaction Survey). 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 = .05 and test to see whether the consultant with more experience has the higher population mean service rating.
ConsultantA Consultant Bn1=16 n2 =10x1 =6.82 (with bar overx) x2 = 6.25(with bar over x)s1 =0.64 s2 = 0.75a.) State the null and alternative hypothesis.b.) compute the value of the test statistic.c.)What is the p-value?d.) What is your conclusion?

d) If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the mean of the consultant with more experience is significantly higher than the mean for the consultant with less experience.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{A}=6.82[/tex] represent the mean for the sample A

[tex]\bar X_{B}=6.25[/tex] represent the mean for the sample B

[tex]s_{A}=0.64[/tex] represent the sample standard deviation for the sample A

[tex]s_{B}=0.75[/tex] represent the sample standard deviation for the sample B

[tex]n_{A}=16[/tex] sample size for the consultant A

[tex]n_{B}=10[/tex] sample size for the consultant B

t would represent the statistic (variable of interest)

a.) State the null and alternative hypothesis.

We need to conduct a hypothesis in order to check if the the consultant (A) with more experience has the higher population mean service rating , the system of hypothesis would be:

Null hypothesis:[tex]\mu_{A} \leq \mu_{B}[/tex]

Alternative hypothesis:[tex]\mu_{A} > \mu_{B}[/tex]

If we analyze the size for the samples both are less than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}[/tex] (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.

b.) compute the value of the test statistic

We can replace in formula (1) the results obtained like this:

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

c.)What is the p-value?

For this case we have a significance level provided [tex]\alpha=0.05[/tex]. We can calculate the degrees of freedom, on this case:

[tex]df=n_{A}+n_{B}-2=16+10-2=24[/tex]

Since is a unilateral right tailed test the p value would be:

[tex]p_v =P(t_{(24)}>1.992)=0.0289[/tex]

d.) What is your conclusion?

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the mean of the consultant with more experience is significantly higher than the mean for the consultant with less experience.