STA
6166 UNIT 3 Section 3 Answers

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1. A project manager for an engineering firm submits a bid for engineering
design for two projects. The firm formally evaluates the chances that each
bid will be accepted by surveying its engineers. The following table summarizes
the collective assessment from this survey.
Results of bids 
Project A

Project B

Accepted 
28

25

Refused 
29

24

Total number of bits submitted 
57

49

a. Test whether the chances that each bid was accepted are different. Use a=.05.
H_{0}: p_{A}  p_{B }= 0
H_{A}: p_{A}  p_{B }not equal to 0
T.S.:
RR: Reject if z > z_{a/2}=z_{0.025 } = 1.96
Conclusion: Do NOT reject the null hypothesis and conclude that the proportion saying the bids will be accepted is the same for the two projects. Note that from this we would also conclude that the proportion suggesting the bids will not be accepted for the two projects should also be considered the same.
b. Calculate a 95% confidence interval for the difference of chances that each bid was accepted.
The 95% confidence interval is computed using the equation:
or [0.01897±1.96(0.09739)] or [0.2098, 0.1719]
2. A study plans to assess the impact of several factors
involving the heat treatment of leaf springs. In this process, a conveyor
system transports leaf spring assemblies through a hightemperature furnace.
After heating, a highpressure press induces curvature in the medal. Once
the spring leaves the press, an oil quench cools it to near ambient temperature.
An important quality characteristic of this process is the resulting free
height of the spring. The researchers are interested in examining four factors
known to affect this free height. For each spring tested, the final classification
is whether the spring is below the specified minimal spring height. Results
of numerous tests are given in the following table.
Factors 
Low level 
Total 
1.High heat temp. 
14 
32 
2.heating
time 
23 
48 
3.Transfer
time 
10 
22 
4.Hold
down time 
2 
5 
Use these data to run a Chisquare test to determine whether a finding of low level is independent of the factors used. Use a=.05.
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.
Spring Height


Low  Acceptable  
Factor 
High Heat Temp

14  18 
Heating Time

23  25  
Transfer Time

10  12  
Hold Down Time

2  3 
We will use SAS to analyze these data. The SAS program follows. Note that each factor by spring height combination forms an observation in the data set and the response is the number found to be in that combination. The TABLE statement in PROC FREQ procedure computes the 2 x 4 table, the WEIGHT statement tells us that for each factor x height combination there are NUMBER responses and the "/CHISQ" option on the TABLE statement request the forma Chi Square test.
data spring; input factor $ height $ number ; datalines; hht l 14 hht a 18 ht l 23 ht a 25 tt l 10 tt a 12 hdt l 2 hdt a 3 ; datalines; proc freq data=spring; table factor*height / chisq; weight number; run;
The output from this program is as follows:
The FREQ Procedure Table of factor by height factor height Frequency Expected  Percent  Row Pct  Col Pct a l  Total  hdt  3  2  5 < observed frequencies  2.7103  2.2897  < expected frequencies  2.80  1.87  4.67  60.00  40.00   5.17  4.08   hht  18  14  32  17.346  14.654   16.82  13.08  29.91  56.25  43.75   31.03  28.57   ht  25  23  48  26.019  21.981   23.36  21.50  44.86  52.08  47.92   43.10  46.94   tt  12  10  22  11.925  10.075   11.21  9.35  20.56  54.55  45.45   20.69  20.41   Total 58 49 107 54.21 45.79 100.00 Statistics for Table of factor by height Statistic DF Value Prob  ChiSquare 3 0.2096 0.9760 < This is the Chi Square test statistic Likelihood Ratio ChiSquare 3 0.2102 0.9759 and pvalue MantelHaenszel ChiSquare 1 0.0763 0.7824 Phi Coefficient 0.0443 Contingency Coefficient 0.0442 Cramer's V 0.0443 WARNING: 25% of the cells have expected counts less than 5. ChiSquare may not be a valid test. Sample Size = 107
The Chi Square value of 0.2096 has an associated pvalue of 0.976 which is much greater than the 0.05 level chosen for the Type I error rate. This suggests that we do NOT reject the null hypothesis of independence. This is further supported when we consider the very small differences between the observed and expected frequencies in the table. Note that the results of the Chi Square analysis is thrown into doubt since a number of the cells have expected counts less than 5. We could consider redoing the analysis without this category to see if it makes a difference in the results. When we do this we calculate a Chi Square statistic value of 0.1386 with associated pvalue of 0.933. Our conclusions do not change.
Concluding that the level of factor is independent of the height of spring suggests that none of these factors can be used to control the height of the spring. We must continue to look for other factors.