Practice Problems for Exam 2
1. The Frizzle fowl is a striking variety of chicken with curled feathers. In a 1930 experiment,
Launder and Dunn crossed Frizzle fowls with a leghorn variety exhibiting straight feathers. The
first generation (F1) produced all slightly frizzled chicks. When the F1 interbred, the following
characteristics were observed in F2:
Feather Type
Count
Frizzled
23
Slightly Frizzled
50
Straight
20
Total
93
The most likely genetic model for these results is that of a single gene locus with two
codominant alleles. Under such a model, we would expect a 1:2:1 ratio in F2.
Research Question – Do these data provide evidence the feather types of the F2
generation follow the 1:2:1 ratio?
a. Determine the null and alternative hypotheses that would be used to test the research
question of interest.
H0: pfrizzled = 0.25
pslightly frizzled = 0.50
pstraight = 0.25
Ha: Two or more differ
b. Calculate the expected counts that would be used in the computation of the test
statistic.
Frizzled  93(0.25) = 23.25
Slightly Frizzled  93(0.50) = 46.5
Straight  93(0.25) = 23.25
c. Using JMP, carry out the analysis. Make sure to clearly state your test statistic, p-value,
and conclusion in terms of the research question.
Test Statistic = 0.7204
p-value = 0.6975
No evidence that the feather types of the F2 generation DO NOT follow the
1:2:1 ratio.
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2. A study was conducted to determine the effects of antiserum on survival rate. One hundred and
eleven mice were divided into two groups; bacteria and antiserum, bacteria only. After
sufficient time had elapsed for an incubation period and for the disease to its course, the
number of dead and alive mice was counted for each group. The data are given below.
Bacteria & Antiserum
Bacteria Only
Total
Dead
13
25
38
Alive
44
29
73
Total
57
54
111
Research Question – Is the proportion of survivors in the bacteria and antiserum group
greater than those in the bacteria only group?
a. Determine the null and alternative hypotheses that would be used to test the research
question of interest.
H0: pbacteria & antiserum ≤ pbacteria only
Ha: pbacteria & antiserum > pbacteria only
b. Using JMP, carry out the analysis. Make sure to clearly state your test statistic, p-value,
and conclusion in terms of the research question. If a test statistic is required, you must
also show the expected counts.
No Test Statistic
p-value = 0.0078
Evidence that the proportion of survivors in the bacteria & antiserum group is
greater than those in the bacteria only group.
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3. Toschke et al. (American Journal of Epidemiology, 2003) collected data on the smoking status of
mothers during pregnancy and whether their child was born premature (37 weeks or fewer of
gestation) or full term. 3970 mothers participated in the study and of the 406 mothers who
were smokers, 36 had premature babies, while 168 of the mothers who were non-smokers had
premature babies.
a. Fill in the contingency table for the scenario.
Smoker
Non-Smoker
Total
Premature Baby
36
168
204
Full Term Baby
370
3396
3766
Total
406
3564
3970
b. Using JMP, find the relative risk of having a premature baby for the smokers compared
to the non-smokers.
c. Interpret the relative risk found in part b.
The risk/probability of a premature baby for smokers is 1.88 times more likely
than the risk/probability of a premature baby for non-smokers.
d. Using JMP, find and interpret the odds ratio for this scenario.
The odds of a full term baby for non-smokers is 1.97 times greater than the
odds of a full term baby for smokers.
e. According to the relative risk found in part b and the odds ratio found in part d, can a
relationship/association between smoking status and having a premature baby be
concluded? Explain.
Yes, because the RR ≠ 1 and the OR ≠ 1.
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4. Sickle-cell anemia is a heredity chronic blood disease that is extremely severe when an
individual carries two copies of the defective gene. It is particularly common in countries
plagued by malaria, a parasitic infection transmitted by mosquitos. A study in Africa tested 543
children for the sickle-cell gene and malaria. Of the children tested, 136 had the sickle-cell gene
and 407 did not possess the gene. Of those with the sickle-cell gene, 36 had heavy malaria
infections whereas 139 of the children without the sickle-cell gene had heavy malaria infections.
Research Question – Is the proportion of children with heavy malaria infections different
for those with the sickle-cell gene compared to those without?
a. Fill in the contingency table for the scenario.
Malaria
No malaria
Sickle-cell
36
100
No Sickle-cell
139
268
Total
175
368
Total
136
407
543
b. Determine the null and alternative hypotheses that would be used to test the research
question of interest.
H0: psickle cell = pno sickle cell
Ha: psickle cell ≠ pno sickle cell
c. Using JMP, carry out the analysis. Make sure to clearly state your test statistic, p-value,
and conclusion in terms of the research question. If a test statistic is required, you must
also show the expected counts.
Expected Counts:
Malaria
No malaria
Sickle-cell
136(0.32) = 43.52
136(0.68) = 92.48
No Sickle-cell 407(0.32) = 130.24 407(0.68) = 276.76
Total
175
368
% overall with malaria = 175/543 = 0.32
% overall without malaria = 368/543 = 0.68
Total
136
407
543
Test Statistic = 2.754
p-value = 0.0970
No evidence that the proportion of children with heavy malaria infections is
different for those with the sickle-cell gene compared to those without.
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