MS Module 8: Two Population Proportions & Variances – practice problems
(The attached PDF file has better formatting.)
Exercise 8.1: Difference in population proportions
A study on a treatment group vs a control group shows
treatment control
observations 120 100
successes 72 50
The null hypothesis is H0: p1 = p2, where p1 and p2 are the true proportions of success for the two groups.
A. What are the sample proportions of success in the treatment group and the control group?
B. What is the sample proportion of success in the combination of the two groups?
C. What is the sample difference in the proportions of success between the two groups?
D. What is the variance of the sample difference in the proportions of success between the two groups?
E. What is the standard deviation of the sample difference in the proportions of success?
F. What is the z statistic to test the null hypothesis H0: p1 = p2?
G. What is the p value for this two-tailed z test of the null hypothesis H0: p1 = p2?
Part A: The sample proportions of success in the treatment group and the control group are
● Treatment group: 72 / 120 = 60%
● Control group: 50 / 100 = 50%
Part B: The sample proportion of success in the combination of the two groups = 122 / 220 = 55.45455%.
treatment control combined
observations 120 100 220
successes 72 50 122
sample proportion 0.6 0.5 0.554545
Part C: The sample difference in the proportions of success between the two groups is 60% – 50% = 10%.
Part D: The variance of the sample differences in the proportion of success between the two groups is
× (1 – ) × (1/m + 1/n) = 0.554545 × (1 – 0.554545) × (1/120 + 1/100) = 0.004529.
Part E: The standard deviation is the square root of the variance = 0.0045290.5 = 0.067298.
Part F: The z statistic to test the null hypothesis H0: p1 = p2 is the difference in the sample proportions divided by its standard deviation = 10% / 0.067298 = 1.485928.
Part G: The p value for this two-tailed z test of the null hypothesis H0: p1 = p2 is 2 × CDF(–1.4859) = 0.1373.
Exercise 8.2: Type II errors and required sample sizes
The incidence of a disease in an untreated population is 2%. We test whether a vaccine reduces the incidence of the disease, using a 5% significance level. We want enough subjects so that if the incidence level of the disease with the vaccine is 1% or less, the probability of a Type II error is 10% or less.
A. What is zα for this exercise?
B. What is zβ for this exercise?
C. How many subjects are needed for the given values of α and β?
Part A: This scenario is a one-tailed test for α = 5%, so we use zα (not zα/2) = 1.644854.
Part B: The given β is 10%, so zβ = 1.281552.
Part C: For p1 = 2% and p2 = 1% and the given α and β, the number of subjects needed is
n = [ zα √( (p1 + p2)(q1 + q2)/2 ) + zβ √(p1q1 + p2q2) ]2 / (p1 – p2)2 =
(1.644854 × ((2%+1%) × (98% + 99%)/2)0.5 + 1.281552 × (2% × 98% + 1% × 99%)0.5 )2 / (2% – 1%)2 = 2,528.7
Exercise 8.3: Confidence interval for difference in proportions
The observations and success for a treatment group and a control group are
treatment control
observations 120 100
successes 72 50
A. What is the sample difference in the probability of success between the treatment and control groups?
B. What is the sample variance of the difference in the probability of success between the two groups?
C. What is the sample standard deviation of this difference in the probability of success?
D. What is the 95% confidence interval for the null hypothesis that the probability of success is the same for the two groups?
Part A: The sample difference in the probability of success between the treatment and control groups is
72 / 120 – 50 / 100 = 0.60 – 0.50 = 0.10
Part B: The sample variance of the difference in the probability of success between the two groups is
0.60 × (1 – 0.60) / 120 + 0.50 × (1 – 0.50) / 100 = 0.0045
Question: To test the null hypothesis H0: p1 = p2, we used the variance of the combined group:
× × (1/m + 1/n)
where = (mp1 + np2) / (m + n) and = (mq1 + nq2) / (m + n).
Answer: We use the variance of the combined group when we test the null hypothesis H0: p1 = p2.
The confidence interval does not assume that p1 = p2 and we form an interval around the observed difference.
Part C: The sample standard deviation of the difference in the probability of success between the two groups is 0.00450.5 = 0.067082
Part D: The z value for a two-sided 95% confidence interval is 1.959964. The 95% confidence interval is
● Lower bound: 0.1 – 1.959964 × 0.067082 = -0.031478
● Upper bound: 0.1 + 1.959964 × 0.067082 = 0.231478
Exercise 8.4: Difference of variances
Groups #1 and #2 are normally distributed.
● σ21 = the variance of Group #1
● σ22 = the variance of Group #2
The null hypothesis is H0: σ21 = σ22.
● A sample from Group #1 is {1, 3, 5, 7, 9, 11}.
● A sample from Group #2 is {12, 13, 14, 15, 16, 17, 18, 19}.
(To illustrate the procedure, these samples have uniform distributions with different variances, not normal distributions with the same variance.)
A. What are the variances of the samples from Group 1 and Group 2?
B. What is the ratio of the variances?
C. What is the distribution of this ratio of variances?
D. What is the p value for testing the null hypothesis?
E. What is the 95% confidence interval for the ratio of the variances of the two groups?
F. What is the 95% confidence interval for the ratio of the standard deviations of the two groups?
Part A: The mean of Group 1 is (1 + 3 + 5 + 7 + 9 + 11) / 6 = 6.00.
The sample variance of Group 1 is
( (1 – 6)2 + (3 – 6)2 + (5 – 6)2 + (7 – 6)2 + (9 – 6)2 + (11 – 6)2) / (6 – 1) = 14.00
The mean of Group 2 is (12 + 13 + 14 + 15 + 16 + 17 + 18 + 19) / 8 = 15.50
The sample variance of Group 2 is
((12 – 15.5)2 + (13 – 15.5)2 + (14 – 15.5)2 + (15 – 15.5)2 + (16 – 15.5)2 + (17 – 15.5)2 + (18 – 15.5)2 + (19 – 15.5)2) / (8 – 1) = 6.00
Part B: The ratio of these variances is 14 / 6 = 2.333333.
Part C: If both groups are normally distributed and their variances are equal, the ratio of the sample variances has an F distribution with (m-1) and (n-1) degrees of freedom.
The textbook explains that if X1, … , Xm is a random sample from a normal distribution with variance σ21 and Y1, …, Yn is another random sample (independent of the Xi’s) from a normal distribution with variance σ22 and s21 and s22 are the two sample variances, then the random variable F = (s21 / σ21) / (s22 / σ22) has an F distribution with degrees of freedom υ1 = m-1 and υ2 = n-1 (equation 10.8).
Part D: The p value for testing the null hypothesis H0: σ21 = σ22 is F2.3333, 5, 7 = 0.14989.
Question: Which of the two groups is the numerator of the ratio of the variances? The exercise here uses the variance of Group 1 as the numerator and the variance of Group 2 as the denominator. If we use the variance of Group 2 as the numerator and the variance of Group 1 as the denominator, the ratio of the variances is
6 / 14 = 0.428571, which is much different from 14 / 6 = 2.333333.
Answer: The critical values for the F distribution have the relation Fα, s, t = 1 / (Fα, t, s )
Interchanging Group 1 with Group 2 replaces F by 1/F and interchanges the degrees of freedom s and t. The statistical tests (p values and confidence intervals) have the same results.
Part F: We look up (or compute) two critical F values for the 95% confidence interval (α = 0.05):
● Fα/2, s, t = F0.025, 5, 7 = 5.285237
● Fα/2, t, s = F0.025, 7, 5 = 6.853076
The bounds for the 95% confidence interval divide or multiply the ratio of the sample variances by the critical F values:
● Lower bound: [ σ21 / σ22 ] × 1 / Fα/2, s, t = 2.333333 / 5.285237 = 0.441481
● Upper bound: [ σ21 / σ22 ] × Fα/2, t, s = 2.333333 × 6.853076 = 15.990508
Part G: The bounds for the 95% confidence interval for the ratio of the standard deviations are the square roots of the bounds for the 95% confidence interval for the ratio of the variances:
● Lower bound: 0.4414810.5 = 0.664440
● Upper bound: 15.9905080.5 = 3.998813