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 = $1,587 x¯2 = $1,083 Standard Deviation s1 = 205 s2 = 283

(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.

(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.)

(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.)

Answer 1

Given that,
mean(x)=1587
standard deviation , s.d1=205
number(n1)=22
y(mean)=1083
standard deviation, s.d2 =283
number(n2)=22
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.1
from standard normal table,right tailed t α/2 =1.3
since our test is right-tailed
reject Ho, if to > 1.3
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (21*42025 + 21*80089) / (44- 2 )
s^2 = 61057
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=1587-1083/sqrt((61057( 1 /22+ 1/22 ))
to=504/74.5026
to=6.7649
| to | =6.7649
critical value
the value of |t α| with (n1+n2-2) i.e 42 d.f is 1.3
we got |to| = 6.7649 & | t α | = 1.3
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: right tail -ha : ( p > 6.7649 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
i. level of significance =0.10
test statistic: 6.7649
critical value: 1.3
decision: reject Ho
p-value: 0
ii.
level of significance =0.05
Given that,
mean(x)=1587
standard deviation , s.d1=205
number(n1)=22
y(mean)=1083
standard deviation, s.d2 =283
number(n2)=22
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.68
since our test is right-tailed
reject Ho, if to > 1.68
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (21*42025 + 21*80089) / (44- 2 )
s^2 = 61057
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=1587-1083/sqrt((61057( 1 /22+ 1/22 ))
to=504/74.5026
to=6.7649
| to | =6.7649
critical value
the value of |t α| with (n1+n2-2) i.e 42 d.f is 1.68
we got |to| = 6.7649 & | t α | = 1.68
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: right tail -ha : ( p > 6.7649 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 6.7649
critical value: 1.68
decision: reject Ho
p-value: 0
iii.
level of significance =0.01
Given that,
mean(x)=1587
standard deviation , s.d1=205
number(n1)=22
y(mean)=1083
standard deviation, s.d2 =283
number(n2)=22
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.42
since our test is right-tailed
reject Ho, if to > 2.42
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (21*42025 + 21*80089) / (44- 2 )
s^2 = 61057
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=1587-1083/sqrt((61057( 1 /22+ 1/22 ))
to=504/74.5026
to=6.7649
| to | =6.7649
critical value
the value of |t α| with (n1+n2-2) i.e 42 d.f is 2.42
we got |to| = 6.7649 & | t α | = 2.42
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: right tail -ha : ( p > 6.7649 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 6.7649
critical value: 2.42
decision: reject Ho
p-value: 0
iv.
level of significance =0.001
Given that,
mean(x)=1587
standard deviation , s.d1=205
number(n1)=22
y(mean)=1083
standard deviation, s.d2 =283
number(n2)=22
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.001
from standard normal table,right tailed t α/2 =3.3
since our test is right-tailed
reject Ho, if to > 3.3
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (21*42025 + 21*80089) / (44- 2 )
s^2 = 61057
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=1587-1083/sqrt((61057( 1 /22+ 1/22 ))
to=504/74.5026
to=6.7649
| to | =6.7649
critical value
the value of |t α| with (n1+n2-2) i.e 42 d.f is 3.3
we got |to| = 6.7649 & | t α | = 3.3
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: right tail -ha : ( p > 6.7649 ) = 0
hence value of p0.001 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 6.7649
critical value: 3.3
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that type A training results in higher mean weekly sales than does type B training.
c.
TRADITIONAL METHOD
given that,
mean(x)=1587
standard deviation , s.d1=205
number(n1)=22
y(mean)=1083
standard deviation, s.d2 =283
number(n2)=22
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (21*42025 + 21*80089) / (44- 2 )
s^2 = 61057
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 61057 * (1/22+1/22) )
=74.5
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 42 d.f is 2.018
margin of error = 2.018 * 74.5
= 150.35
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (1587-1083) ± 150.35 ]
= [353.65 , 654.35]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=1587
standard deviation , s.d1=205
sample size, n1=22
y(mean)=1083
standard deviation, s.d2 =283
sample size,n2 =22
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 1587-1083) ± t a/2 * sqrt( 61057 * (1/22+1/22) ]
= [ (504) ± 150.35 ]
= [353.65 , 654.35]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [353.65 , 654.35]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion