Description
1. (Exercise 1.1 from Agresti (2007)) In the following examples, identify the response variable and the
explanatory variables.
(a) Attitude toward gun control (favor, oppose), Gender (female, male), Mother’s education (high school,
college).
(b) Heart disease (yes, no), Blood pressure, Cholesterol level.
(c) Race (white, nonwhite), Religion (Catholic, Jewish, Protestant), Vote for president (Democrat, Republican, Other), Annual income.
(d) Marital status (married, single, divorced, widowed), Quality of life (excellent, good, fair, poor).
2. (Adapted from Exercise 1.9 from Agresti (2013)) In an experiment on chlorophyll inheritance in maize,
for 1103 seedlings of self-fertilized heterozygous green plants, 854 seedlings were green and 249 were yellow.
Theory predicts the ratio of green to yellow is 2 : 1. Test the hypothesis that 2 : 1 is the true ratio using the
Wald, Score and Likelihood Ratio tests. Report the p-values, and interpret.
3. (Exercise 1.26 from Agresti (2013)) A binomial sample of size n has y = 0 successes.
(a) Show that the confidence interval for π based on the likelihood function is [0, 1−e
−z
2
α/2
/2n
]. For α = 0.05,
use the expansion of an exponential function to show that this is approximately [0, 1.92/n].
(b) For the score method, show that the confidence interval is [0, z2
α/2
/(n + z
2
α/2
)].
4. For a given sample proportion p and standard normal percentile zα/2 show that the end points of the 100(1 −
α)% two-tailed Score confidence interval for binomial parameter π are given by the solutions of the equation
(1 + z
2
α/2
/n)π
2 + (−2p − z
2
α/2
/n)π + p
2 = 0.
Find these solutions and thus derive the so-called Wilson confidence interval for π.
5. The data in the Table 1 is obtained from a multinomial distribution.
Cell 1 2 3 4 5
Probability π1 π2 π3 π4 π5
Frequency 10 13 21 23 29
Table 1: Multinomial Data
(a) Test with α = 0.05 the null hypothesis H0 : π1 = 0.1, π2 = 0.1, π3 = 0.2, π4 = 0.35 by using the Pearson
chi-square test and the likelihood ratio test.
(b) Derive the maximum likelihood estimates of πi
, i = 1, . . . , 5 under the null hypothesis H0 : π1 = π2, π3 =
π4.
1
(c) Test with α = 0.05 the null hypothesis H0 : π1 = π2, π3 = π4 by using the Pearson chi-square test and
the likelihood ratio test.
6. Table 2 gives a random sample of size 150 of the random variable X. Do you think X follows the Poisson
distribution? Use Pearson’s chi-squared test and the likelihood ratio test with (α = 0.05).
Values of X 0 1 2 3 4 5 6 7 8 9
Frequency 5 11 18 29 26 25 15 10 7 4
Table 2: Poisson Data
THE END
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