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26. For the scenario of Example 14.8 assume the same 29. Exercise 27 gives an alternative way of finding
normal prior distribution but assume that the data beta probabilities when software for the beta dis-
set is just one observation ¥ = 118.28 with stan- tribution is unavailable.
dard deviation / Yn = 15/V18 = 3.5355. Use a. Use Exercise 27 together with the F table to
Bayes’ theorem to derive the posterior distribu- obtain a 90% credibility interval for Exercise
tion, and compare your answer with the result of 28(c). [Hint: To find c such that .05 is the
Example 14.8. probability that F is to the left of c, reverse
27. Let X have the beta distribution on (0, 1] with the degrees of freedom and take the reciprocal
of the value for « = .05.]
parameters ot = v,/2 and B = v2/2, where v,/2 5 z “al
. . b. Repeat (a) using software for the beta distribu-
and v2/2 are positive integers. Define Y = ti dc th th It of
(X/a)/[(1 — X)/B]. Show that ¥ has the F distri- ga and compare wath the-result ob (a),
bution with degrees of freedom 1, v2. 30. If x and f are large, then the beta distribution can
28. Ina study by Erich Brandt of 70 restaurant bills, Pesapproximated by: the normal disenibution.usiag
5 . the beta mean and variance given in Section 4.5.
40 of the 70 were paid using cash. We assume a ae : mene :
1. ; aie’ This is useful in case beta distribution software is
random sample and estimate the posterior distri- ‘ Pash
: } ' unavailable. Use the approximation to compute
bution of the binomial parameter p, the population ea 3
: the credibility interval in Example 14.7.
proportion paying cash.
a. Use a beta prior distribution with a =2 and 31. Assume a random sample X;,X2, ... , X, from the
b=2. Poisson distribution with mean 4. If the prior dis-
b. Use a beta prior distribution with a = 1 and tribution for / has a gamma distribution with para-
b=1. meters a and B, show that the posterior distribution
¢. Use a beta prior distribution with a and b very is also gamma distributed. What are its parameters?
small and positive. e
32. Consider a random sample X1, X>,...,X, from the
d. Calculate a 95% credibility interval for p using SSIES A EDSON SAME esc me
" normal distribution with mean 0 and precision t
(c). Is your interval compatible with p = .5? aia 2
7 . ; (use t as a parameter instead of o? = 1/t).
e. Calculate a 95% confidence interval for p using :
_ . Assume a gamma-distributed prior for t and
Equation (8.10) of Section 8.2, and compare : . :
3 show that the posterior distribution of t is also
with the result of (d). ee .
gamma. What are its parameters?
33. The article “Effects of a Rice-Rich Versus Potato- article used a distribution-free test. Use such a test
Rich Diet on Glucose, Lipoprotein, and Cholesterol with significance level .05 to determine whether
Metabolism in Noninsulin-Dependent Diabetics” the true mean cholesterol-synthesis rate differs sig-
(Amer. J. Clin. Nutrit., 1984: 598-606) gives the nificantly for the two sources of carbohydrates.
accompanying data on cholesterol-synthesis rate
for eight diabetic subjects. Subjects were fed a 9°.
standardized diet with potato or rice as the major Subject 1 2 3 4 #5 6 7 8
carbohydrate source. Participants received. both | — AAA AA AA AANA
diets for specified periods of time, with cholesterol- Potato 1.88 2.60 1.38 4.41 1.87 2.89 3.96 2.31
synthesis rate (mmol/day) measured at the end of — Rice 1.70 3.84 1.13 4.97 .86 1.93 3.36 2.15
each dietary period. The analysis presented in this_§ —- -—aAANHSYH
--- Trang 797 ---
784 —cuarrer 14 Alternative Approaches to Inference
mine the CI for the given data, and state the
34, The study reported in “Gait Patterns During Free confidence level.
Choice Ladder Ascents” (Hum. Movement Sci. 37, The single-factor ANOVA model considered in
1983; 187-195) was motivated by publicity'con- Chapter 11 assumed the observations in the ith
cerning the increased accident rate for individuals sample were selected from a normal distribution
climbing ladders. A number of different gait pat- with mean 4 and variance o%, that is,
terns were used by subjects climbing a portable Xj 1 +e) where the c's are normal with
straight ladder according to specified instructions. mean Oan d variance G2, The normality assump-
Allie, aiCGAts Tinie forseven aibpeis Who Uedia tion implies that the F test is not distribution-free.
lateral gait and six subjects who used a four-beat We jidwi assuine iat thé @8 all cone from thé
dingonal gaitate given. same continuous, but not necessarily normal, dis-
Oo tribution, and develop a distribution-free test of
Lateral = .86 1.31 1.64 151 1.53 139 1.09 the null hypothesis that all / j1;’s are identical. Let
Diagonal 1.27 1.82 1.66 85 1.45 1.24 N = CJ, the total number of observations in the
data set (there are J; observations in the ith sam-
a, Carry out a test using % = .05 to see whether ple). Rank these N observations from 1 (the smal-
the data suggests any difference in the true lest) to N, and let R; be the average of the ranks for
average ascent times for the two gaits. the observations in the ith sample. When Ho is
b. Compute a 95% CI for the difference between true, we expect the rank of any particular observa-
the true average gait times. tion and therefore also R, to be (N + 1)/2. The
35. The sign test is a very simple procedure for testing data argues against Ho when some of the Ri’s
hypotheses about a population median assuming differ considerably from (NV + 1)/2. The Krus-
only that the underlying distribution is continuous. kal-Wallis test statistic is
To illustrate, consider the following sample of 20 b waa?
observations on component lifetime (hr): _ RN hY
: ” «away i(®—“F)
17 33 5.1 69 126 144 164
24.6 26.0 26.5 32.1 37.4 40.1 40.5 When Hg is true and either (1) / = 3, all J; > 6 or
415 72.4 80.1 86.4 87.5 100.2 (2)1 > 3, all J; > 5, the test statistic has approxi-
We wish te taacthe hypotticees yy i= 25.0;verms mately a chi-squared distribution with 1 — 1 df.
Hy: fi > 25.0 The test statistic is Y = the number of ‘The accompanying observations on-axial stiff:
observations thaexcesd25. ness index resulted from a study of metal-plate
a. Consider rejecting Ho if Y > 15. What is the connected trusses in which five different plate
value of «(the probability of a type I error) for lengths—4 in., 6 in., 8 in., 10 in., and 12 in. —
this test? (Hint: Think of a “success” as a were used (“Modeling Joints Made with Light-
litetinne: that Exceeds 95.0. THEN Ve the Alls Gauge Metal Connector Plates,” Forest Products
ber of successes in the sample, What kind of a F.y1979: 39-44).
distribution does Y have when ji = 25.02]
b. What rejection region of the form Y > spe #=!@in): 309.2 309.7 311.0 316.8
cifies a test with a significance level as close to 326.5 349.8 409.5
.05 as possible? Use this region to carry out the j= 2(6in.): 331.0 347.2. 348.9 361.0
test for the given data. [Note: The test statistic 381.7 402.1 404.5