STA
6166 UNIT 3 Section 3 Answers

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1. In a study of children aged 0 to15, concern is focused on the presence
or absence of the carrier for Streptococcus Pyogenesis
and the relationship between the presence of the carrier and tonsil size.

Tonsils present. Not enlarged

Tonsils enlarged

Presence of the carrier 
15

19

Absence of the carrier 
380

360

Total number 
395

379

H_{0}: p_{Not}  p_{Enlarged }= 0
H_{A}: p_{Not}  p_{Enlarged }not equal to 0
T.S.:
RR: Reject if z > z_{a/2}=z_{0.025 } = 1.96
Here, the test statistic value is 0.823 which is less than 1.96 causing us to NOT reject the null hypothesis and conclude that there is no difference in carrier rates for children with enlarged tonsils versus children without enlarged tonsils. This suggests that children with enlarged tonsils have no higher incidence of the carrier of Streptococcus Pyogenesis than do children without enlarged tonsils.
b.
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.
Drug 
Effective 
Total 
A 
26 
45 
B 
14 
28 
C 
10 
17 
D 
22 
90 
Use these data to run a chisquare test to determine whether the effectiveness
of all drugs are the same.This is equivalent to testing that all effectiveness
proportions are the same.
These data represent a a 2 by 4 contingency table only not in traditional format. We could write the table as follows instead.
Effective


Yes

No


Drug 
A

26

19

B

14

14


C

10

7


D

22

68

We will perform the Chi Square analysis using Excel to make the calculations. A copy of the spreadsheet follows. Note that we first compute the row and column proportions which are not particularly important to computing the Chi Square statistic but will be useful in interpreting it. Next we compute the expected values, then the Chi Square values and in that table we sum all the cells in the table to get the overall Chi Square statistic.
Frequency  Effective 

Drug  Yes  No  Total  
A  26  19  45  
B  14  14  28  
C  10  7  17  
D  22  68  90  
Total  72  108  180  
Column Proportions 
Effective


Drug  Yes  No  
A  0.3611  0.1759  
B  0.1944  0.1296  
C  0.1389  0.0648  
D  0.3056  0.6296  
Total  1  1  
Row Proportions 
Effective


Drug  Yes  No  Total  
A  0.5778  0.4222  1  
B  0.5000  0.5000  1  
C  0.5882  0.4118  1  
D  0.2444  0.7556  1  
Total  
Expectations 
Effective


Drug  Yes  No  Total  
A  18.0  27.0  45  
B  11.2  16.8  28  
C  6.8  10.2  17  
D  36.0  54.0  90  
Total  72  108  180  
Differences 
Effective


Drug  Yes  No  Total  
A  8  8  0  
B  2.8  2.8  0  
C  3.2  3.2  0  
D  14  14  0  
Total  0  0  0  
Chi Square Values 
Effective


Drug  Yes  No  Total  
A  3.5556  2.3704  
B  0.7000  0.4667  
C  1.5059  1.0039  
D  5.4444  3.6296  
Total  18.67647  
Overall Chi Square Value  
DF = (r1)(c1) =  3  
Type I error rate=  0.05  
Chi Square Critical value=  7.815  
We find from above that the Chi Square test statistic value is 18.676 and the critical value for the rejection region is 7.815. Hence we reject the null hypothesis and conclude that there effectiveness is not independent of drug. This suggests that different drugs have different effectiveness. The Chi Square values in each cell tell us which drug by effectiveness combinations we should be looking at. The larger the value, the more the observed count differs from the expected count. Clearly we should be looking at drug D and drug A. Drug D is clearly less effective than expected, whereas Drug A is slightly more effective than expected. From the row percentages we see that Drugs A, B and C are 50% or more effective, whereas Drug D is only half that effective at 24%. The most effective drug was C, but it is also the one for which we have the least information.