ANIMAL STUDIES [15 points] Experimental studies of cancer often use strains of a
ID: 3394502 • Letter: A
Question
ANIMAL STUDIES [15 points]
Experimental studies of cancer often use strains of animals that have a naturally high incidence of tumors. In one such experiment, tumor-prone mice were kept in a sterile environment; one group of mice was maintained entirely germ-free, while another group was exposed to the intestinal bacterium, Escherichia coli. The table below shows the incidence of liver tumors.
For the germ-free, sterile environment mice, there were a total of 49 mice and 39% of them developed tumors.
For the E.coli exposed group of mice, there were a total of 13 mice and 62% of them developed tumors.
1. [4 points] Begin by filling in the two-way table.
Treatment
Germ-free
E.coli
Outcome
Develop tumors
Do Not Develop Tumors
2. [3 points] How strong is the evidence that tumor incidence is significantly higher in mice exposed to E.coli? That is, is there a relationship between whether mice form tumors, or not, and whether they were exposed to clean or bacteria-laden environments? Identify what part of your analysis you’re using to reach your conclusion.
3. [3 points] How would the result from question (2) change if the percentages (39% and 62%) of mice with tumors were the same but the sample sizes were doubled (n=98 for germ-free, and n=26 for E.coli)?
4. [3 points] How would the result from question (2) change if the percentages (39% and 62%) of mice with tumors were the same but the sample sizes were tripled (n=147 for germ-free, and n=39 for E.coli)?
Treatment
Germ-free
E.coli
Outcome
Develop tumors
Do Not Develop Tumors
Explanation / Answer
No of Mice given germ free treatment = 49
No of mice developed tumor = 0.39 * 49
No of mice that did not develop tumor = 0.61 * 49
No of Mice given E.coli treatment = 13
No of mice developed tumor = 0.62 * 13
No of mice that did not develop tumor = 0.38 * 13
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2) Proportion P1 = Proportion of mice developing tumor (germ free treatment) = 0.39
Proportion P2 = Proportion of mice developing tumor (E.coli treatment) = 0.62
Pooled proportion p = (p1 * n1 + p2 * n2) / (n1 + n2) = (0.39 * 49 + 0.62 * 13) / (49 + 13) = 0.004
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } = sqrt{ 0.004 * ( 1 - 0.004 ) * [ (1/49) + (1/13) ] } = 0.02
z = (p2 - p1) / SE = (0.62 - 0.39)/0.02 = 11.2
critical z value for right tailed test at alpha = 0.05 is 1.65
Since test stat > critical value we reject the null and conclude that tumor incidence is significantly higher in mice exposed to E.coli
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3)
Pooled proportion p = (p1 * n1 + p2 * n2) / (n1 + n2) = (0.39 * 98 + 0.62 * 26) / (98 + 26) = 0.0044
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } = sqrt{ 0.004 * ( 1 - 0.004 ) * [ (1/98) + (1/26) ] } = 0.015
z = (p2 - p1) / SE = (0.62 - 0.39)/0.015 = 15.8
critical z value for right tailed test at alpha = 0.05 is 1.65
Since test stat > critical value we reject the null and conclude that tumor incidence is significantly higher in mice exposed to E.coli
Result is the same
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4)
Pooled proportion p = (p1 * n1 + p2 * n2) / (n1 + n2) = (0.39 * 147 + 0.62 * 39) / (147 + 39) = 0.002
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } = sqrt{ 0.002 * ( 1 - 0.002 ) * [ (1/147) + (1/39) ] } = 0.007
z = (p2 - p1) / SE = (0.62 - 0.39)/0.007 = 31.8
critical z value for right tailed test at alpha = 0.05 is 1.65
Since test stat > critical value we reject the null and conclude that tumor incidence is significantly higher in mice exposed to E.coli
Result is the same
Germ Free E Coli Develop Tumors 19 8 Do not develop Tumors 30 5Related Questions
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