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Stat 571 Second Midterm Exam November 25, 2008
Name:
- The exam is open book and open notes.
- Do all your work in the spaces provided. If you need additional space for your work, indicate clearly where the additional work can be found.
- The parts within a problem are not necessarily sequential.
- To receive full credit, you must show your work.
- Do not dwell too long on any one question. Answer as many questions as you can.
For instructor’s use:
1 16
2 28
3 22
4 16
5 18
Total 100
- A study in rats investigated whether regularly consuming tuna increases life span. 20 randomly selected rats were fed a tuna-heavy diet for three years, and another 30 were fed regular rat food. At the end of 3 years, it was observed whether the rat was alive or dead. Overall, 35 of the 50 rats lived less than 3 years, and 15 lived beyond 3 years. This results in the following partial table:
Life span < 3 years ≥ 3 years Tuna Diet 20 Regular Diet 30 35 15 50
(a) What assumptions are necessary for chi-square to be a valid test? Verify them.
(b) What are the expected counts for this test?
(c) The chi-squared statistic for this data is 8.0357. What is the p-value for the test? What do you conclude?
- Multiple choice questions: check all that apply.
(a) To investigate the role of genetic diversity, randomly chosen plots were planted with 24 eelgrass shoots. With the ‘no diversity’ treatment (A), all 24 shoots in a plot had the same genotype. With the ‘high diversity’ treatment (B), 8 genotypes of shoots were represented in the plot. At the end of the season, the total number of shoots was counted in each plot. The left and middle graphs show the normal probability plot for plots in samples A and B. On the right graph, only one boxplot corresponds to sample B.
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Normal Q−Q Plot, Treatment A
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The correct boxplot for sample B is: B1 B2 B3. Explain briefly.
Based on the normal probability plots above, the ecologist decides to not use a t-test to compare the mean number of shoots in the two treatments, because:
the sample size is too small in both samples the plots show a lack of independence the plots show a strong skew in the data the plots show a mild skew in the data
Instead of using a t-test to compare the mean number of shoots in the ‘no diversity’ (A) vs. the ‘high diversity’ (B) treatments, the researcher can use: the Mann-Whitney test Levene’s test
The researcher performed the Mann-Whitney test and obtained t = 109 (t∗^ was 109 and t∗∗^ was 131). She concluded that there is evidence no evidence that genetic diversity affects the abundance of shoots. Explain your answer very briefly.
(b) Levene’s test applied to the data below gives a p-value: sample 1 60 61 62 64 68 sample 2 80 81 82 84 88 Explain briefly (almost no calculations please).
p >. 20 . 05 < p <. 20 p < .05.
- Data on wing spans of two sub-species, A and B, of butterflies was collected. Samples had nA = 12 and nB = 9 butterflies. In those samples, means were ¯yA = 13.5, ¯yB = 9.19, and standard deviations sA = 21.6, sB = 8.7. The investigator wants to use a t-test to test H 0 : μA = μB at the 10% level, and thus first checks that the data is normal —it is. The investigator then performs Levene’s test to check equality of variances. The p-value on Levene’s test is 0.0034. Given this information, perform an appropriate test of the hypothesis by computing a p-value and making a conclusion. Justify your choice of test. The adjusted degree of freedom is 15.3 for the t-test that does not assume equal variances, if you choose to use that test.
- The World Health Organization was reported 42 human cases of avian flu (H5N1) in Indonesia in 2007. Among these reported cases, there were 37 deaths. Based on this data, build a 90% confidence interval for the probability of survival in humans, after avian flu infection in Indonesia.
State and assess any assumptions you made.
