STA
6166 UNIT 3 Section 3 Answers

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1. A company uses a preemployment test to screen applicants for sales jobs.
We are interested in whether the screning test is effective. In an experiment,
a random samples of applicants who pass the test
and a second sample of those who do not pass the screening test are all employed.
The number of employees who successfully completed the training program in
these two samples was recorded as follows.
Experiment Result 
Applicant who pass the preemployment
test

Applicant who do not pass the preemployment
test

Employees completed in the training
program 
67

15

Employees failed in the training
program 
28

22

Total number of Employees 
95

37

a. Test whether the proportions of employee who successfully completed the training program is the same for the group who initially passed the preemployment test to those who did not initially pass the preemplyment test. For this test use P(Type I error)=a=.05.
H_{0}: p_{Pass}  p_{Not }= 0
H_{A}: p_{Pass}  p_{Not }not equal to 0
T.S.:
RR: Reject if z > z_{a/2}=z_{0.025 } = 1.96
Here, the test statistic value is 3.203 which is greater than 1.96 causing us to reject the null hypothesis and conclude that there is a significant difference between the completion rates for applicants who successfully pass the preemployment test and those who do not. Clearly, the preemployment test has some ability to screen applicants and increase completion rates for subsequent training.
b. Calculate a 95% confidence interval for the above difference.
The 95% confidence interval is computed using the equation:
[0.2988 ± 1.96(0.09329)] or [0.2988 ± 0.1828] or [0.1159, 0.48165]
Note that the 95% CI does not contain zero, a good indication that the difference is not statistically equal to zero.
Four
types of advertisement 
Customers
made purchases 
Total

1.Advertisement
A 
27 
39 
2.Advertisement
B 
29 
51 
3.Advertisement
C 
17 
31 
4
Advertisement D 
9 
23 
Can we determine from this whether there is a difference in the purchase fraction across the four groups?
These data represent a 2 by 4 contingency table but is presented in odd format. More commonly you would expect to see a table like the one below.
Purchase


Yes

No


Advertisement 
A

27

12

B

29

22


C

17

14


D

9

14

We will use SPSS to analyse these data. We create the following dataset.
The Chi Square analysis is part of the ANALYZE > DESCRIPTIVE STATISTICS > CROSSTABS options, but first we have to tell SPSS that the frequency counts for the advertisement by purchase combinations is in the count column. This is done with the DATA > WEIGHT CASES option. In the resulting dialog we set Weight Case By [COUNT]. Once this is done, we are ready to specify options in the ANALYZE > DESCRIPTIVE STATISTICS > CROSSTABS dialog box. Set Rows to Advent, Cols to Purchuse, Statistics to Chi Square, and in the Cells dialog check Observed, Expected, Row, Column, Total. The resulting output follows.
PURCHASE  Total  

no  yes  
ADVERT  A  Count  12  27  39 
Expected Count  16.8  22.2  39.0  
% within ADVERT  30.8%  69.2%  100.0%  
% within PURCHASE  19.4%  32.9%  27.1%  
% of Total  8.3%  18.8%  27.1%  
B  Count  22  29  51  
Expected Count  22.0  29.0  51.0  
% within ADVERT  43.1%  56.9%  100.0%  
% within PURCHASE  35.5%  35.4%  35.4%  
% of Total  15.3%  20.1%  35.4%  
C  Count  14  17  31  
Expected Count  13.3  17.7  31.0  
% within ADVERT  45.2%  54.8%  100.0%  
% within PURCHASE  22.6%  20.7%  21.5%  
% of Total  9.7%  11.8%  21.5%  
D  Count  14  9  23  
Expected Count  9.9  13.1  23.0  
% within ADVERT  60.9%  39.1%  100.0%  
% within PURCHASE  22.6%  11.0%  16.0%  
% of Total  9.7%  6.3%  16.0%  
Total  Count  62  82  144  
Expected Count  62.0  82.0  144.0  
% within ADVERT  43.1%  56.9%  100.0%  
% within PURCHASE  100.0%  100.0%  100.0%  
% of Total  43.1%  56.9%  100.0% 
Value  df  Asymp. Sig. (2sided)  

Pearson ChiSquare  5.434(a)  3  .143 
Likelihood Ratio  5.484  3  .140 
N of Valid Cases  144  
a 0 cells (.0%) have expected count less than 5. The minimum expected count is 9.90. 
The Chi Square statistic computed for this table is 5.434 with associated pvalue of 0.143. This is larger than the 0.05 Type I error rate we chose to use initially hence we would NOT reject the null hypothesis of independence between advertisement and purchase decision. This suggests that the decision to purchase does not seem to depend on any particular advertisement, although, the difference between observed and expected counts in the D advertisement is the largest observed in the table.