Question

In the book Business Research Methods, Donald R. Cooper and C. William Emory (1995) discuss a manager who wishes to compare the effectiveness of two methods for training new salespeople. The authors describe the situation as follows: The company selects 22 sales trainees who are randomly divided into two equal experimental groups—one receives type A and the other type B training. The salespeople are then assigned and managed without regard to the training they have received. At the year’s end, the manager reviews the performances of salespeople in these groups and finds the following results: A Group B Group Average Weekly Sales x¯1 = $1,593 x¯2 = $1,017 Standard Deviation s1 = 209 s2 = 255 (a) Set up the null and alternative hypotheses needed to attempt to establish that type A training results in higher mean weekly sales than does type B training. H0: µA - µB = versus Ha: µA - µB > (b) Because different sales trainees are assigned to the two experimental groups, it is reasonable to believe that the two samples are independent. Assuming that the normality assumption holds, and using the equal variances procedure, test the hypotheses you set up in part a at level of significance .10, .05, .01 and .001. How much evidence is there that type A training produces results that are superior to those of type B? (Round your answer to 3 decimal places.) t = H0 with alpha equal to .10. H0 with alpha equal to .05 H0 with alpha equal to .01 H0 with alpha equal to .001 evidence that µA - µ B > 0 (c) Use the equal variances procedure to calculate a 95 percent confidence interval for the difference between the mean weekly sales obtained when type A training is used and the mean weekly sales obtained when type B training is used. Interpret this interval. (Round your answer to 2 decimal places.) Confidence interval [ , ]

Answer 1

GIVEN:

Sample size of group A

Sample size of group B

Sample average weekly sales of group A

Sample average weekly sales of group B

Sample standard deviation of group A

Sample standard deviation of group B

(a) HYPOTHESIS:

The hypothesis is given by,

(That is, the mean weekly sales of type A training is not significantly different from the mean weekly sales of type B training)

(That is, the mean weekly sales of type A training results is higher than the mean weekly sales of type B training)

(b) LEVEL OF SIGNIFICANCE:

Let the significance level

TEST STATISTIC:

which follows t distribution with degrees of freedom

where pooled standard deviation is,

CALCULATION:

The pooled standard deviation is,

  

Now

CRITICAL VALUE:

The right tailed t critical value with degrees of freedom at significance level is .

The right tailed t critical value with degrees of freedom at significance level is .

The right tailed t critical value with degrees of freedom at significance level is .

The right tailed t critical value with degrees of freedom at significance level is .

DECISION RULE:

CONCLUSION:

Since the calculated t statistic value (5.79) is greater than the t critical values (1.3253, 1.7247, 2.5280, 3.5518) at significance levels 0.10, 0.05, 0.01 and 0.001, we reject the null hypothesis and conclude that the mean weekly sales of type A training results is higher than the mean weekly sales of type B training. That is, there is sufficient evidence to prove that the type A training results in higher mean weekly sales than does type B training.

(c) 95% CONFIDENCE INTERVAL FOR DIFFERENCE IN TWO POPULATION MEANS:

The formula for 95% confidence interval for difference in two population means is,

where is the t critical value with degrees of freedom at 95% confidence level

where pooled standard deviation is,

CRITICAL VALUE:

The right tailed t critical value with degrees of freedom at significance level is .

CALCULATION:

The pooled standard deviation is,

  

Now the 95% confidence interval for difference in two population means is,

\

CONCLUSION:

The 95% confidence interval for difference in two population means is . Since the value specified by the null hypothesis (0) is not in the interval , the null hypothesis can be rejected at the 0.05 level. Thus we can conclude that the mean weekly sales of type A training results is higher than the mean weekly sales of type B training. That is, there is sufficient evidence to prove that the type A training results in higher mean weekly sales than does type B training.