Chapter 10.2, Problem 17E

Comparing Financial Consultant Ratings. 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.

1. a. State the null and alternative hypotheses.
2. b. Compute the value of the test statistic.
3. c. What is the p-value?
4. d. What is your conclusion?

To determine

State the null and alternative hypotheses.

Answer to Problem 17E

Null hypothesis:$H_0:{\mu }_1-{\mu }_2\le 0$

Alternative hypothesis:$H_a:{\mu }_1-{\mu }_2>0$

Explanation of Solution

Calculation:

The results from two independent samples of service ratings for financial consultant A and financial consultant B are as follows:

Consultant A	Consultant B
$n_1=16$	$n_2=10$
${\overline{x}}_1=6.82$	${\overline{x}}_2=6.25$
$s_1=0.64$	$s_2=0.75$

Furthermore, it is given that consultant A has 10 years of experience and consultant B has 1 year experience and $\alpha =0.05$.

State the hypotheses:

The test hypotheses are as follows:

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2\le 0$

That is, the consultant with more experience has the lower population mean service rating.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2>0$

That is, the consultant with more experience has the higher population mean service rating.

To determine

Find the value of test statistic.

Answer to Problem 17E

The value of the test statistic is 1.99.

Explanation of Solution

Calculation:

Test statistic:

The test statistic for hypothesis test of ${\mu }_1-{\mu }_2$ when ${\sigma }_1$ and ${\sigma }_2$ are unknown is as follows:

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

Substitute ${\overline{x}}_1$ as 6.82, ${\overline{x}}_2$ as 6.25, $s_1$ as 0.64, $s_2$ as 0.75, $n_1$ as 16, and $n_2$ as 10 in the above formula.

$\begin{array}{c}t=\frac{\left(6.82-6.25\right)-0}{\sqrt{\frac{{0.64}^2}{16}+\frac{{0.75}^2}{10}}}\\ =\frac{0.57}{\sqrt{\frac{\text{0}\text{.4096}}{16}+\frac{\text{0}\text{.5625}}{10}}}\\ =\frac{0.57}{\sqrt{\text{0}\text{.0256}+\text{0}\text{.05625}}}\\ =\frac{0.57}{\text{0}\text{.2861}}\end{array}$

  $=1.99$

Thus, the test statistic is 1.99.

To determine

Find the p-value.

Answer to Problem 17E

The p-value is 0.032.

Explanation of Solution

Degrees of freedom:

For t distribution with two independent random sample, the degrees of freedom is obtained as follows:

$df=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{1}{n_1-1}{\left(\frac{s_1^2}{n_1}\right)}^2+\frac{1}{n_2-1}{\left(\frac{s_2^2}{n_2}\right)}^2}$

Substitute $s_1$ as 0.64, $s_2$ as 0.75, $n_1$ as 16, and $n_2$ as 10 in the above formula.

$\begin{array}{c}df=\frac{{\left(\frac{{0.64}^2}{16}+\frac{{0.75}^2}{10}\right)}^2}{\frac{1}{16-1}{\left(\frac{{0.64}^2}{16}\right)}^2+\frac{1}{10-1}{\left(\frac{{0.75}^2}{10}\right)}^2}\\ =\frac{{\left(0.0256+0.05625\right)}^2}{\frac{1}{15}{\left(0.0256\right)}^2+\frac{1}{9}{\left(0.05625\right)}^2}\\ =\frac{\text{0}\text{.0067}}{\text{0}\text{.00004+0}\text{.0004}}\\ =\frac{\text{0}\text{.0067}}{\text{0}\text{.00044}}\end{array}$

     $\begin{array}{c}=15.2\\ \approx 15\end{array}$

Software procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =T.DIST.2T(1.99,15) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

Thus, the p-value is 0.032.

To determine

Obtain the conclusion whether the consultant with more experience has a higher population mean service rating or not.

Answer to Problem 17E

There is sufficient evidence to conclude that the consultant with more experience has the higher population mean service rating.

Explanation of Solution

Rejection rule:

If $p\text{-value}\le \alpha$, reject the null hypothesis.

If $p\text{-value}>\alpha$, do not reject the null hypothesis.

Conclusion:

From part (c), the p-value is 0.032.

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.032\right)<\alpha \left(=0.05\right)$.

From the rejection rule, the null hypothesis is rejected.

Therefore, the null hypothesis $H_0$ is rejected.

Therefore, there is sufficient evidence to conclude that the consultant with more experience has the higher population mean service rating.