ANIMAL STUDIES 15 points Experimental studies of cancer ofte

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)?

Solution

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