DPHS 568 Biostatistics in Dentistry Summer 2007
Answers for homework #3
1.
a. sign test two-sided p-value = 0.210. Do not reject null hypothesis of no
change.
b. Wilcoxon signed rank test
c. Sum of negative ranks = 91, sum of positive ranks = 185. Using the "Case
2" normal approximation formulas on page 124 of coursepack, I get mu =
138, signma squared = 1049, and p-value = 0.151. Using SPSS, I get p=
0.147. It appears that in SPSS they don't use the -1/2 "continuity
correction". Either way, do not reject.
2. mean difference (inoculated – uninoculated) = 0.55.
a. assuming equal variances T= 2.355, p-value = 0.034. Reject null
hypothesis of no difference at significance level 0.05.
b. Use the Wilcoxon rank-sum test.
c. Sum of rank 47 versus 89, p=0.028. Reject null hypothesis of no
difference.
3.
a. reject null hypothesis if there are 15 or more positive differences
b. power = 0.617
c. significance level = P(reject|p=.5) = P(X > 15 | p=0.5) = 0.0207.
4.
a. two-sample test for difference in proportions Z=2.76, p-value=.006.
Reject null hypothesis of no difference between genders.
b. (0.0434, 0.2566)
c. Fisher’s Exact test
5. These are paired dichotomous data. Use McNemar’s test.
a. H0: proportion of failing implants of type A is the same as the proportion
of failing implants of type B.
H1: proportions of failing implants differ by type.
a. Two-sided p-value = 0.1670.
b. Do not reject null hypothesis.
c. H1: Proportion of failing implants of type B is higher than of type A. One
sided p-value = 0.0835. Do not reject at 0.05 significance level.
6.
a.
b. r = 0.794
c. line of best fit y = -4.471 + 0.610x (“y” is urinary Hg and x is number of
occlusal amalgam surfaces). See scatterplot above for drawn line.
d. t statistic = 3.195. Compare to t(6). P-value = 0.019. Reject the null
hypothesis of no association between number of occlusal amalgam
surfaces and urinary Hg.
7.
a.
b.
c.
d.
2.319 + 0.067*20 = 3.659
3.659 + 2.45*0.260 = (3.022, 4.296)
c) 3.659 + 2.45*1.364 = (0.317, 7.001)
d) The confidence intervals for parts b and c estimate different values.
Part c confidence interval is bigger because it is basically predicting the
value of a single observation. This is much more more variable than an
average of many values.