FREC/STAT 608
Guided Exercise 2
1. Geneticists have identified E2F1 transcription factor as an important component of cell proliferation
control. The researchers induced DNA synthesis in two batches of serum-starved cells. In one group of
92 cells (treatment), cells were micro-injected with the E2F1 gene. A control group of 158 cells was not
exposed to E2F1. After 30 hours, researchers determined the number of altered growth cells in each
batch. The data are given below.
Altered
Not Altered
Row Total
E2F1
41
51
92
Control
15
143
158
Column Total
56
194
250
a. What is the probability of altered growth cells given the cells were injected with the E2F1 gene?
P(A|E2F1) = 41/92 = .4457 = 44.57%
b. What is the probability of altered growth cells given the cells were not injected (i.e., the Control)
P(A|Control) = 15/158 = .0949 = 9.49%
c. What is the odds of altered to not altered for cells injected with E2F1?
OddsE2F1 = 41/51 = .8039
d. What is the odds of altered to not altered for the control cells?
OddsControl = 15/143 = .1049
e. What is the odds ratio of altered to not altered for E2F1 injectd cells compared to the Control cells?
Express this odds ratio in words.
Odds Ratio= .8039/.1049 = 7.66
Cells injected with E2F1 were more than 7.66 times as likely to have altered cells
f.
Generate the Expected Frequencies for this table.
Altered
Not Altered
E2F1
20.6
71.4
92
Control
35.4
122.6
158
56
194
250
Column Total
Cell
Cell
Cell
Cell
1,1,
1,2,
2,1,
2,2,
= (56*92)/250 = 20.6
= (194*92)/250 = 71.4
= (56*158)/250 = 35.4
= (194*158)/250 = 122.6
Row Total
Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag
is normally distributed with a mean of 5 pounds per square inch and a standard deviation of 1.5
pounds per square inch. What proportion of the bags produced have a breaking strength of:
a. Less than 3.17 pounds per square inch?
Z = (3.17 – 5)/1.5 = -1.22
P(<= Z) = .5 - .3888 = .1112
b. At least 3.6 pounds per square inch?
Z = (3.6 – 5)/1.5 = -.9333
P(>=Z) = .3238 + .5 = .8238
c. Between 5 and 5.5 pounds per square inch?
Z = (5.5 – 5)/1.5 = .3333
P(5<Z<5.5) = .1293
d. Between 3.2 and 4.2 pounds per square inch?
Z = (3.2 – 5)/1.5 = -1.20
P(5<= |Z|) =
.3849
Z =(4.2-5)/1.5 = -.5333
P(5<= |Z|) =
.2019
Answer = .3849 - .2019 =
.1830
e. Between what two values symmetrically distributed around the mean will 95% of the breaking
strengths fall? Be careful here! With the normal distribution we need to be more precise than 2
standard deviations.
Use 1.96 since this is the z-value associated with .95/2 = .4775
5 ± 1.96(1.5) = 2.06 to 7.94
f.
Answer Part E if the standard deviation is really 1.0 pounds per square inch.
5 ± 1.96(1.0) = 3.04 to 6.96
If the company wanted to reduce the 95% range of breaking strength they should seek to reduce
the variability in their product.
3. You have been hired as a consultant to provide analysis for the Personnel Department at ZTel
company, a large communications company. Every applicant of ZTel must take a standardized exam, and
the hire or no-hire decision depends in part on this exam. The exam was purchased from a company
which says the exam is distributed approximately normal with:
µ = 525
σ = 55
The current interview policy has two phases. The first phase separates all applicants into one of three
categories:
Automatic Interview
Maybe Interview
Automatic Rejects
score of 600 or above
score of 500 to 600
score less than 500
The Maybe group are passed on to a second phase where their previous experiences, education, special
skills, and other factors are taken into consideration in whether to grant an interview.
No one at the company can remember why the values of 600 and 500 were used as the standards for
automatic interview or rejection, and most likely there were decided arbitrarily by a former Personnel
Manager. The current Personnel Manager of Ztel needs to know the following:
My answers will have more precision than your answers – don’t worry if your z-values have less
precision. Your answer should be close.
a. The probability associated with the current standard of being automatically rejected - what proportion of
the applicants are automatically rejected?
Z = (500-525)/55 = -.4545
P(X <= -.4545)= .5 -.1753
Automatic Reject < 500 = .5 - .1753 = .3247
b. The probability associated with the current standard of being automatically interviewed - what
proportion of the applicants are automatically interviewed?
Z = (600-525)/55 = 1.364
P(X >= 1.364) = .5 - .4137
Automatic Interview > 600 = .5 - .4131 = .0863
c. The manger notices that applicants that score between 535 and 580 tend to be good hires, having both
good skills and a higher probability of accepting an offer to the company. She would like to give this
group a higher priority in the second phase of evaluation. What percentage of the applicants should
she expect to fall within this range?
Z = (580-525)/55 = 1.000
Z = (535-525)/55 = .182
P(Z) = .3413
P(Z) = .0721
P (535 <= X <=580) = .3413 - .0721 = .2692
26.9% or about 27 percent are in the “sweet Spot”
d. The manager would prefer that the exam score for automatically interview would be set at 15% and the
automatic rejection would be set at 20%. What are the exam values in this distribution associated with
these probabilities (in this case, round to whole numbers)?
For the top 15% automatically interviewed, it would be at the 85 percentile, Z = 1.04
1.04 = (X-525)/55 =
(1.04*55)+525 = 582.2
582
For the bottom 20% it would be at the 20 percentile, Z = -.8416
-.675 = (X-525)/55 =
-.8416*55+525 = 487.87
478.71
Summarize your results as a recommendation to your client.
There isn’t a correct answer here.
One could argue that it is far better to tie the standards of automatic acceptance and rejection for
interviews to probabilities rather than some vague standards that cannot be defended. At least
when we use probabilities we can adjust the standards easily if conditions suggest they need to be
changed.
EXTRA:
In the study, published in the American Journal of Obstetrics and Gynecology, researchers interviewed
1,063 women in the San Francisco area who became pregnant between 1996 and 1998 about their
caffeine intake. While 164 of the women drank 200 mg of caffeine or more daily, 635 consumed some
caffeine but less than 200 mg. The remaining 264 women said they didn’t consume any caffeine.
Generate the expected frequencies for the following table:
Table: The effects of Caffeine on Miscarriages
Miscarriage
Caffeine
Yes
No
None
32
232
264
< 200 mg
99
536
635
>200 mg
41
123
164
172
891
1063
EXPECTED FREQUENCIES
Table: The effects of Caffeine on Miscarriages
Miscarriage
Caffeine
Yes
No
42.7
221.3
264
< 200 mg
102.7
532.3
635
>200 mg
26.5
137.5
164
172
891
None
1063