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 = $1,391 x¯2 = $1,249 Standard Deviation s1 = 201 s2 = 265 (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 ? equal to .10. H0 with ? equal to .05 H0 with ? equal to .01 H0 with ? 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

x¯1 = $1,391
x¯2 = $1,249

s1 = 201
s2 = 265

a)
H0: muA = muB
Ha: muA > muB

b)
yes two samples are indpeendent

Pooled estimate of variance  
Sp = sqrt(((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2))  
Sp = 235.1872

SE = 235.1872*sqrt(1/11 + 1/11) =100.2841

test statistic,
t = (1391 - 1249)/100.2841
t = 1.416

p-value = 0.0861

Hence 0.1 is the evidence

c)
t-value = 2.0860

CI = (1391 - 1249 - 2.0860*100.2841, 1391 - 1249 + 2.0860*100.2841)

= (-67.1891 , 351.1891 )