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Unformatted text preview: PROBLEMS 165 has an F distribution with 111 and 112 degrees of freedom. Finally, the square of a
tvariable with 11 degrees of freedom is F1)”. Summarizing: 2 v 2 2 X12/1
Xv: Z, tv =F1yy= . g ‘ XE/V
A special case connects all four pivotal variables: ﬂ=¢=ﬁ=nw Thus, given the Ftable, all the other tables can be generated from it. For completeness, we now summarize the mean and variance of the four ﬁxed
distributions: Distribution Variance Normal Student t (v > 2)
Chisquare 2 _
Fisher’s F 2‘50“ + V2 2) (u2 > 4) 1110/2 — 2)2(V2 — 4) V Note 5.4: Onesided Tests and Onesided Conﬁdence Intervals Corresponding to onesided (onetailed) tests are onesided conﬁdence intervals. A
onesided conﬁdence interval is derived from a pivotal quantity in the same was as
a twosided conﬁdence interval. For example, in the case of a onesample ttest, a
pivotal equation is P[—oo s x” stni,1a]=1—a. s/x/ﬁ Solving for [1. produces a l00(1  a)°/o upper onesided conﬁdence interval for u:
(3c'—t,,_1,1_as/\/r7, 00). Similar intervals can be constructed for all the pivotal variables. . PROBLEMS 1. Rickman et al. [1974] made a study of changes in serum cholesterol and trigly—
ceride levels of subjects following the “Stillman diet.” The diet consists primarily
of protein and animal fats, restricting carbohydrate intake. The subjects followed
the diet with length of time varying from 3 to 17 days. The mean cholesterol
level increased signiﬁcantly from 215 mg/ 100 m1 at baseline to 248 mg/ 100 ml at
the end of the diet. In this problem, we deal with the triglyceride level. 166 . ONE AND TWOSAMPLE INFERENCE Weight (in kg) Triglyceride (in mg1100 ml)
Initial Final Baseline Final 194
122
158
154 93 90
101
99
183
82
100
104
72 . 108
110 a. Make a stemand—leaf diagram of the changes in triglyceride levels. b. Calculate the average change in triglyceride level. Calculate the standard error
of the difference. ' . Test the signiﬁca‘ cc of the average change.
d, Construct a 90% c nﬁdence interval on the difference. e. The authors indicat that subjects (5,6), (7,8), (9,10) and (15,16) were “re
peaters”; that is, the same subjects who followed the diet for two sequences.
Do you think it is reasonable to include their data on the “second time
around” with that of the other subjects? Supposing not, how would you now
analyze the data? Carry out the analysis. Does it change your conclusions? 6 2. The following data are from Dobson et al. [1976]. Thirtysix patients with a con
ﬁrmed diagnosis of phenylketonuria (PKU) were identiﬁed and placed on di
etary therapy before reaching 121 days of age. The children were tested for IQ
(Stanford—Binet) between the ages of four and six; subsequently, their normal
siblings of closest age were also tested with the Stanford—Binet. The fellowing
are the ﬁrst 15 pairs listed in the paper: IQ of PKU case
IQ of sibling IQ of PKU case 110
IQ of sibling 88 a. State a suitable null and an alternative hypotheses with regard to these data. PROBLEMS , 167 b. Test the null hypothesis.
c. State your conclusions.
d. What are your assumptions? e. Discuss the concept of power with respect to this set of data using the fact
that PKU invariably led to mental retardation until the cause was found and
treatment consisting of restricted diet was instituted. f. The mean difference (PKU case — sibling) in IQ for the full 36 pairs was
—5.25; the standard deviation of the difference was 13.18. Test the hypothesis
of no difference in IQ for this whole set of data. 3. Data by Mazze et a1. [1971] deals with the preoperative and postoperative cre
atinine clearance (ml/min) of six patients anesthetized by halothane: Patient Preoperative
Postoperative :1. Why is the paired ttest preferable to the twosample ttest in this case?
b. Carry out the paired ttest and test the signiﬁcance of the difference. e. What is the model for your analysis? ’ (I. Set up a 99% conﬁdence interval on the difference. e. Graph the data by plotting the pairs of values for each patient. 4. Some of the physiological effects of alcohol are well known. A paper by Squires
et al. [1978] assessed the acute effects of alcohol on auditory brainstem potentials
in humans. Six volunteers (including the three authors) participated in the study.
The latency (delay) in response to an auditory stimulus was measured before
and after an intoxicating dose of alcohol. Seven different peak responses were
identiﬁed. In this exercise, we only discuss latency peak no. 3. The measurements
of the latency of peak (in milliseconds after the stimulus onset) in the six subjects
were as follows: Latency of Peak Before alcohol
After alcohol a. Test the signiﬁcance of the difference at the 0.05 level. b. Calculate the pvalue associated with the observed result. c. Is your pvalue based on a onetail test'or a twotail test? Why? d. As in the previous problem, graph these data and state your conclusion. 168 ONE AND TWOSAMPLE INFERENCE e. Carry out an (incorrect) twosample test and state your conclusions. f. Using the observed sample variances s12 and s22 associated with the set of
readings before and after, calculate the variance of the difference assuming
independence (call this Variance 1). How does this value compare with the
variance of the difference calculated in the ﬁrst part of the problem? (Call
this Variance 2.) Why do you suppose Variance 1 is so much bigger than
Variance 2? The average of the differences is the same as the difference in
the averages. Show this. Hence, the twosample ttest differed from the paired
ttest only in the divisor. Which of the two tests in more powerful in this case,
that is, declares a difference signiﬁcant when in fact there is one? 5. The following data from Schecter et al. [1973] deals with sodium chloride pref
erence as related to hypertension. TWO groups, 12 normal and 10 hypertensive
subjects, were isolated for a week and compared with respect to Na+ intake. The
following are the average daily Na+ intakes: ' Normal 10.2 2.2 0.0 2.6 0.0 43.1
Hypertensive 92.8 54.8 51.6 61.7 250.8 84.5
Normal 45.8 63.6 1.8 0.0 3.7 0.0
Hypertensive 34.7 62.2 11.0 ' 39.1 a. Compare the average daily Na+ intake of the hypertensive subjects with that
of the normal volunteers by means of an appropriate ttest. b. State your assumptions. c. Assuming that the population variances are not homogenous, carry out an
appropriate ttest (see Note 5.2). 6. Kapitulnick et al. [1976] compared the metabolism of a drug, zoxazolamine, in
placentas from 13 women who smoked during pregnancy and 11 who did not. The Nonsmoker 0.18 ' 0.36 ' 0.24 I 0.50 l 0.42 ‘ 0.36 l 0.50 Smoker 0.66 g 0.60 0.96 1.37 1.51 3.56 3.36
Nonsmoker 0.60 0.56 0.36 0.68
Smoker 4.86 ’ 7.50 9.00 10.08 14.76 16.50 a. Calculate the sample mean and standard deviation for each of the two groups. b. Test the assumption that the two sample variances came from a population
with the same variance. c. Carry out the ttest using the approximation to the t procedure discussed in
Note 5.2. What are your conclusions? d. Suppose we agree that the variability (as measured by the standard deviations)
is proportional to the level of the response. Statistical theory then suggests PROBLEMS ~ ‘ 169 Number of children Mean phenylalanine level (mg/d1)
Standard deviation two? Why?
d. Test the hypothesis that the sample variances came from two populations with
‘ the same variance. e. Construct a 95% conﬁdence interval on the difference in the population
phenylalanine levels. f. Interpret the interval constructed in Part (e). g. “This set of data has little power, ” someone says. What does it mean? Interpret
the implications of a Type II error in this example. 4h. What is the interpretation of a Type I error in this example? Which, in
your opinion, is more serious in this example: a Type I error, or a Type II 4yearold PKU child? j. Can you think of some additional information that would make the analysis
more precise? birthweight is a major determinant of SIDS. The following set of data collected by
Dr. D. R. Peterson of the Department of EpidemiologY, University of Washington, consists of the birthweights (in grams) of each of 22 dizygous twins, and each of
19 monozygous twins: 170 10. ONE AND TWO~SAMPLE INFERENCE Dizygous Twins Monozygous Wins a. With respect to the dizygous twins, test the above hypothesis. State the null
hypothesis. I). Make a similar test on the monozygous twins.
c. Discuss your conclusions. a. What is an appropriate test?
b. Set up the appropriate critical region.
c. State your conclusion. d. Suppose the sample size is doubled. State
region for the null hypothesis is changed. precisely how the non—rejection Consider Problem 3.9, dealing with the treatment of essential hypertension. Com
pare treatments A and B by means of an appropriate ttest. Set up a 99% conﬁ
dence interval on the reduction of blood pressure under treatment B as compared
to treatment A. ‘ 0‘s» whence 63:161  Chapter 5 Problems . The sample mean change (ﬁnalbaseline) is 15.56 with sample s.d. 51.79 and s.e. 12.95.
. The paired t statistic is 1.20 with 15 df (twosided p—value is 0.2 <p <0.5).
. The 90% conﬁdence interval for the true mean change is 15.56i1.75(12.95) or (38.22,7 .10). . Ho: the mean IQ difference between the PKU case and sibling is zero. The twosided alternative hypothesis is of interest (since the problem does not state a speciﬁc alternative). . The mean change (PKU IQ  sibling IQ) is —4.13 with a standard deviation of 15.90. The paired t statistic is 1.01 with 14 df. The two—sided pvalue is 0.2<p<0.5.
At the 0.05 level of signiﬁcance, we do not reject the null hypothesis. . This test is valid if the IQ difference is normally—distributed, and if the 15 cases in the study are a random sample from the population of PKUsibling pairs.
The paired t statistic is 2.39 with 35 df; the twotailed pvalue is 0.02<p <0.05. Because the same six patients are studied preoperatively and postoperatively. . The mean change (postpre) is 24.3333 with a standard deviation of 42.2122. The paired t statistic is 1.41 with 5 (If; the two—tailed p—value is 0.2 <p <0.5. . The change in creatinine clearance is normallydistributed; the six patients are a random sample of patients anesthetized by halothane. . The 99% conﬁdence interval is 24.3333i4.03(17.23) or (45.12, 93.78). . The mean change (afterbefore) is 0.118333 with a s.d. cf 0.083287. The paired t statistic is 3.48 .with 5 df. Since 3.48 exceeds 2.57 (30.975), we reject the null hypothesis. . The two—tailed p—value is 0.01 <p <0.02. . The (mean, variance) in normal and hypertensive groups are (14.4167, 512.608) and (74.32, 4401.4818). Using the twosample t test, the pooled estimate of variance is 2262.602 and the t
statistic is —2.94. Since 0.005 <p <0.01 (two—sided), we reject H0 at a=0.05. . We assume NA+ is normally distributed with the same variance in both populations. If we test for variance equality, F=8.59>3.59 (Fg’mm) and we reject H0 at a=0.05. Since 8.59 exceeds ’
F9,11,o.999, P<0.002. The (mean, s.d.) in nOnsmokers and smokers are (0.432727, 0.152124) and (5.747692, 5.423356). . The F statistic is 1270.97 with 12,10 df. Since this exceeds F,2,,0,0‘975=3.52, we reject the hypothesis of variance homogeneity at signiﬁcance level a=0.05. Since 1270.97 exceeds Flam”, p <0.002. Let it, 01.2) denote the population mean phenylalanine level in children whose diet is terminated
(continued). We wish to test How, =u2. Since it appears that the focus is on determining if
discontinuing the diet increases the mean phenylalanine level, the alternative of interest is HAzu, > pg. . The pooled estimate of variance is 32.444.  The twosample t statistic is 2.67. With reference to the t7 distribution, 0.01<p <0.025 (onesided).
F=(7.3/4.1)2=3.17 has the F,4 distribution under Hozalz=022. Since F3.4,0.m=9.98, we do not reject.
A 95% conﬁdence interval for the mean difference is (26.916.7):t2.37(3.82) or (1.14, 19.26). If we repeatedly select random samples of sizes 5 and 4 from the "terminated" and "continued"
populations, then 95 % of the conﬁdence intervals we construct will straddle the true mean difference.
Type I error occurs if we conclude diet termination has an adverse effect, when it does not; Type 11
error occurs if we conclude termination has no effect, when it does. A Type 11 error is more serious.
Do not terminate the diet of your child. ' .I In dizygous twins, the mean difference (SID  NonSid) is 43.9091 with a standard deviation of 365.1833. The paired t statistic is —0.564 with 21 df (p>0.5). ' In monozygous twins, the mean difference (SID  NonSid) is 59.7368 with a standard deviation of 370.1531. The paired t statistic is ~0.7035 with 18 df (0.2<p<0.5).
In both groups of twins, there is no evidence that birthweight is a major determinant of SIDS. ...
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 Summer '08
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 Statistics, Biostatistics

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