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Question
1
The personnel manager of a large insurance company
would like to evaluate the effectiveness of four different
sales training programs. Eight new recruits are assigned
to each of the four training programs A to D. The
scores of the members in each group on an exam are recorded
below. In answering the questions below you must support
your answers with specific numerical output from the Excel
analysis.
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A
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B
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C
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D
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66
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72
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61
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63
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74
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51
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60
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61
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82
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59
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57
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76
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75
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62
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60
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84
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73
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74
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81
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58
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97
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64
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55
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65
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87
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78
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70
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69
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78
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63
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71
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80
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a)
Is there a significant difference between the four groups?
First we
get the mean and standard variance of each group as follows.
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mean
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std
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std2
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Std2(a/b)
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Std2(a/c)
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Std2(a/d)
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Std2(b/c)
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Std2(b/d)
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Std2(c/d)
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A
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79.000
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9.592
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92.000
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1.184
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1.184
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1.022
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1.000
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0.863
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0.863
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B
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65.375
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8.815
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77.696
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C
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64.375
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8.815
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77.696
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D
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69.500
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9.487
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90.000
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And we find
all the ratios of sample variance from different two groups
are in the interval (F(0.025,8,8)=0.22557,F(0.975,8,8)=4.43326).
So we can prove that the variance of each group is equal.
The groups
variance:
 =9.211;  = 9.211;  =9.539;  =8.815;  =9.157; =9.157;
-t(0.025,14)=t(0.975,14)=2.145;
 =  =2.958> t(0.975,14)=2.145;
 =  =3.176> t(0.975,14)=2.145;
 =  =1.992
 =  =0.227
 = =-0.902
 =  =-1.119
So we can
easily find that the mean of A and B and the mean of A and C
are significant different.
b)
Which individual groups differ from each other?
Please refer
to part a)
c)
What is the within groups variance and why is it part of the
calculations?
The within
groups variance is  .
When sample size is equal.
Question 2
An industrial psychologist
wishes to study the effects of motivation on sales in a particular
firm. Of 24 new salespersons, 12 are paid an hourly rate
and 12 are paid a commission. The 24 individuals are randomly
assigned to two groups. The following data represent the
sales volume (in thousands of dollars) achieved during the first
month on the job. Hair removal
a)
Using a 0.05 level of significance is there evidence of a significant
difference in the two groups.
First calculate
the Mean and the standard variance for each group as follows:
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Mean
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STD
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STD2
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STD2(hourly/commission)
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Hourly
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227.58
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13.84
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191.546
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0.411
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Commission
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247.92
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21.59
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466.128
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Then test
for the equal variance. We find that STD2(hourly/commission)=0.411
is in the interval (http://www.chinafashuo.com/spacer.gif)=0.30513,F(http://www.chinafashuo.com/spacer.gif)=3.27728).So
the variance of the two groups are equal.
So  =328.837;
and  =-2.7475<t(0.025,22)=-2.074
so there
is evidence of a significant difference in the two groups.
b)
Explain what would happen to the findings if the variance of
both groups were much larger. Hint: Consult the formula
for computing the t or z value in a difference of means test.
You could rerun the analysis and compare outputs to answer this
question.
If the sample
size is very large the  would be very small, but
the t(0.025,n-2) would be larger when n is larger. So we will
not change our conclusion in part b) when n is very large.
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Hourly
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Commission
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256
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224
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212
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261
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239
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254
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216
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228
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222
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273
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|
236
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234
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|
207
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285
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219
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225
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228
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237
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225
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232
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241
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277
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230
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245
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Question 3
As you know from
your marketing studies brand awareness is a key consideration
in promotion. An advertising manager wants to know whether the
campaign the company is currently running is achieving an average
awareness score of 80 or above. He distributes 50 questionnaires
to potential buyers. The questionnaire measures brand awareness
on a 0-100 scale. The mean score is 82.3 with a standard deviation
of 14.1.
a)
Set up the null and alternative hypotheses.
H0:  ;H1: 
b)
Determine at the 0.05 level of significance whether the advertising
manager has achieved his brand awareness objective.
As we know
 .Under H0,  =  =1.1534<t(0.95,49)=2.010.
So we should
not reject H0.
c)
Rerun the analysis in part b but with a sample size of 150.
If the conclusion drawn from the analysis is different you must
explain why. Hint look at the formula for z or t in a test of
a single mean and/or compare your two outputs very carefully.
If n=150,
 =  =2.00>t(0.95,149)=1.65514.
So we should
reject H0.
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