The problem asks us to perform a pooled t-test to determine if the average monthly sales of Product Line A are significantly greater than those of Product Line B, given the sample sizes, sample means, and sample standard deviations for each product line. We are to assume equal population variances and use a significance level of 0.01. Finally, we need to state the decision regarding the management's aim.

1. Problem Description

The problem asks us to perform a pooled t-test to determine if the average monthly sales of Product Line A are significantly greater than those of Product Line B, given the sample sizes, sample means, and sample standard deviations for each product line. We are to assume equal population variances and use a significance level of 0.

0

1. Finally, we need to state the decision regarding the management's aim.

2. Solution Steps

a. State the null and alternative hypotheses:

Null hypothesis ($H_0$): The average monthly sales of Product Line A is less than or equal to the average monthly sales of Product Line B.

$H_0: \mu_A \le \mu_B$

Alternative hypothesis ($H_1$): The average monthly sales of Product Line A is greater than the average monthly sales of Product Line B.

$H_1: \mu_A > \mu_B$

b. Given Data:

Product Line A:

$n_A = 15$

$\bar{x}_A = 1400$

$s_A = 140$

Product Line B:

$n_B = 20$

$\bar{x}_B = 1250$

$s_B = 180$

Significance level: $\alpha = 0.01$

c. Calculate the pooled variance ($s_p^2$):

$s_p^2 = \frac{(n_A - 1)s_A^2 + (n_B - 1)s_B^2}{n_A + n_B - 2}$

$s_p^2 = \frac{(15 - 1)(140)^2 + (20 - 1)(180)^2}{15 + 20 - 2}$

$s_p^2 = \frac{14(19600) + 19(32400)}{33}$

$s_p^2 = \frac{274400 + 615600}{33}$

$s_p^2 = \frac{890000}{33} \approx 26969.69697$

d. Calculate the t-statistic:

$t = \frac{(\bar{x}_A - \bar{x}_B) - (\mu_A - \mu_B)}{s_p \sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$

Since we are testing if $\mu_A > \mu_B$, we assume $\mu_A - \mu_B = 0$ under the null hypothesis.

$t = \frac{1400 - 1250}{\sqrt{26969.69697} \sqrt{\frac{1}{15} + \frac{1}{20}}}$

$t = \frac{150}{\sqrt{26969.69697} \sqrt{\frac{4 + 3}{60}}}$

$t = \frac{150}{\sqrt{26969.69697} \sqrt{\frac{7}{60}}}$

$t = \frac{150}{\sqrt{26969.69697} \times \sqrt{0.1166667}}$

$t = \frac{150}{\sqrt{31464.9798}}$

$t = \frac{150}{177.384} \approx 0.8456$

e. Determine the critical value:

Degrees of freedom = $n_A + n_B - 2 = 15 + 20 - 2 = 33$

Significance level $\alpha = 0.01$, one-tailed test.

Using a t-table or calculator, the critical value $t_{0.01, 33} \approx 2.448$

f. Decision Rule:

If $t > t_{critical}$, reject the null hypothesis.

If $t \le t_{critical}$, fail to reject the null hypothesis.

g. Conclusion:

Since $0.8456 \le 2.448$, we fail to reject the null hypothesis.

b. Decision regarding the management's aim:

The pooled t-test does not provide enough evidence to support the claim that the average monthly sales of Product Line A are significantly greater than those of Product Line B at the 0.01 significance level. Therefore, we cannot confirm the management's aim based on the given data and the chosen significance level.

3. Final Answer

a. The t-statistic is approximately 0.

8

4

5

6. b. The decision is: Fail to reject the null hypothesis. There is not enough evidence to support the management's aim that the average monthly sales of Product Line A are significantly greater than those of Product Line B at the 0.01 significance level.