Biostatistics for the Health Sciences: Inferences Regarding Proportions - Exercises questions answers

**10.1** Give definitions of the following terms in your
own words:

a. Sample proportion

b. Population proportion

c. Binomial variable

d. Bernoulli trial

e. Continuity correction

f. Confidence interval for a proportion

**10.2** Peripheral neuropathy is a complication of
uncontrolled diabetes. The num-ber of cases of peripheral neuropathy among a
control group of 35 diabetic patients was 12. Among a group of 11 patients who
were taking an oral agent to prevent hyperglycemia, there were three cases of
peripheral neu-ropathy. Is the proportion of patients with peripheral
neuropathy compara-ble in both groups? Perform the test at *a* = 0.05.

**10.3** Construct exact 95% confidence intervals for the
proportion of patients with peripheral neuropathy in the medication group and
the proportion of pa-tients in the control group in the previous exercise.
Construct two confi-dence intervals for each proportion, one with correction
for continuity and the other without correction for continuity.

**10.4** Referring to Exercise 10.2, construct an
approximate 95% confidence inter-val for the difference between the proportions
of patients affected by pe-ripheral neuropathy in the control group and in the
medication group.

**10.5** A dental researcher investigated the occurrence of
edentulism (defined in the research study as loss of two or more permanent
teeth, not including loss of prophylactically extracted wisdom teeth) in a rural
Latin American village. A total of 34 out of 100 sampled adults had lost at
least two teeth. A study of a U.S. city found that the rate of loss of at least
two teeth was 14%. Was the proportion of persons who had edentulism higher in
the Latin American village than in the U.S. city? Conduct the test at the *a* = 0.05 level.

**10.6** Calculate an exact 95% confidence interval for the
proportion of edentulous persons in the Latin American village (refer to
Exercise 10.5.)

**10.7** For the data in Exercise 10.5, compute a 99%
confidence interval using the normal approximation with continuity correction.
Is the result close to the exact interval found in Exercise 10.6? Explain why
or why not.

**10.8** In a British study of social class and health, a
total of 171 out of 402 lower social class persons were classified as
overweight. The percent of over-weight persons in the general population was
39%. Based on these findings, would you assert that low social class is related
to being overweight? Test this hypothesis at the *α* = 0.01 level.

**10.9** A longitudinal study of occupational status and
smoking behavior among women reported at baseline that 170 per 1000
professional/managerial women were nicotine dependent. The corresponding rate
among blue collar women was 310 per 1000. At the *a* = 0.05
level, determine whether there is a significant difference in nicotine
dependence between the proportion of women who are classified as
professional/managerial workers in compari-son to those who are classified as
blue collar workers. Then compute the ap-proximate 99% confidence interval for
the difference between these two proportions.

**10.10** An epidemiologic study examined risk factors
associated with pediatric AIDS. In a small study of 30 cases and 30 controls, a
positive history of substance abuse occurred among 11 of the cases and 6 of the
controls. Based on these data, can the investigator assert that substance abuse
is sig-nificantly associated with pediatric AIDS at the *a* = 0.05 level? Compute the approximate 95% confidence interval for the
difference between the proportions of substance abuse found in the case and
control groups.

**10.2** *Z*’* *= (*W _{1} *–

**10.6 **The number of Latin American patients with
edentulism is** ***x*** **= 34. The sample** **size is *n* = 100. The confidence level 1 – *a* = 0.90.
The formula for the confidence in-terval is by Clopper–Pearson [{1 + (100 – 34
+ 1)*F*(0.95:200 – 68 + 2, 68)/34}^{–1},
{1 + (100 – 34)/{35*F*(0.95:68 + 2,
2(100 – 34)}^{–1}] = [1/{1 + 67*F*(0.95:134,
68)/34}, 1/{1 + 66/(35*F*(0.95:70,
132))}]. *F*(0.95:134, 68) 1.45 and *F*(0.95:70, 132) 1.42. So the interval is
[0.259, 0.430]

**10.8 **The sample proportion** ***p*** **= 171/402 = 0.425. We are testing the hypothesis** **that *p* = 0.39 against the alternative *p* > 0.39 that the proportion overweight in the lower social
class in Britain is higher than for the general British population. Take *Z = *(0.425 – 0.39)/{ √ (0.39) √ (0.61)/ √402} = 0.035/0.02433 = 1.439. This is non-significant. For a one-sided
test at the 0.01 significance level, the critical *Z* = 2.33.

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