MS Module 5 Hypothesis testing of proportions practice exam questions
(The attached PDF file has better formatting.)
The average proportion of death from a disease is 79% . A study tests whether a drug reduces the proportion of death from the disease. Of 100 subjects who are given the drug, 73 die from the disease. Let μ be the expected proportion of death from the disease among subjects given the drug.
● The null hypothesis is H0: the expected proportion of subjects dying μ = μ0 = 79%
● The one-tailed alternative hypothesis is Ha: the expected proportion dying μ < μ0
The null hypothesis is tested at a 1% significance level and the true incidence of death with the drug is 68%.
Question 5.1: Sample mean
What is the incidence of death from the disease in the sample?
Answer 5.1: 73 / 100 = 73%
Question 5.2: Standard deviation
What is the standard deviation of the incidence of death in the sample if the null hypothesis is true?
Answer 5.2: (79% × (1 – 79%) / 100)0.5 = 0.040731
(variance of proportion = p(1-p)/n; use proportion assumed in null hypothesis)
Question 5.3: z value
What is the z value used to test the null hypothesis?
Answer 5.3: (73% – 79%) / 0.040731 = -1.4731
( z value = sample mean – μ0 (mean assumed in the null hypothesis) / standard deviation of the sample mean)
Question 5.4: p value for one-tailed alternative hypothesis
What is the p value for the one-tailed alternative hypothesis?
Answer 5.4: Φ(-1.4731) = 0.0704
Interpolating in the statistical tables:
Φ(1.47) = 0.9292
Φ(1.48) = 0.9306
Φ(–1.4731) = 1 – ( (1.4731 – 1.47) × 0.9306 + (1.48 – 1.4731) × 0.9292) / (1.48 – 1.47) = 0.0704
Question 5.5: p value for two-tailed alternative hypothesis
What is the p value for the two-tailed alternative hypothesis?
Answer 5.5: 2 × 0.0704 = 0.1408 (0.1407 if computations carried to more decimal places)
Question 5.6: Expected value and variance of the z value
What are the expected value and variance of the z value if the null hypothesis is true?
Answer 5.6: μ = 0; σ2 = 1
(If the null hypothesis is true, the z value has a standard normal distribution: mean = zero and variance = 1)
Question 5.7: Standard deviation of sample mean
What is the standard deviation of the sample mean if the true incidence of death with the drug is 68%?
Answer 5.7: (68% × (1 – 68%) / 100)0.5 = 0.046648
Question 5.8: Expected value of the z value
What are the expected value of the z value for testing the null hypothesis if the true incidence of death with the drug is 68%?
Answer 5.8: (68% – 79%) / 0.040731 = -2.7006
Question 5.9: Standard deviation of the z value
What is the standard deviation of the z value for testing the null hypothesis if the true incidence of death with the drug is 68%?
Answer 5.9: 0.046648 / 0.040731 = 1.1453
Question 5.10: Probability of Type II error for one-tailed test
What is the probability of a Type II error for the one-tailed test?
Answer 5.10: The probability of a Type II error when the true proportion is pʹ for the one-tailed test is
1 – Φ[ (p0 – pʹ – zα × (p0 (1 – p0) / n)0.5 ) / ( (pʹ(1 – pʹ) / n)0.5 ]
(p0 – pʹ – zα × (p0 (1 – p0) / n)0.5 ) / ( (pʹ(1 – pʹ) / n)0.5 = (79% – 68% – 2.326 × 0.040731) / 0.046648 = 0.3271
zα = 2.326 (lower limit of right 1% tail)
1 – Φ(0.3271) = 0.3718
Interpolating in the statistical tables:
Φ(0.32) = 0.6255
Φ(0.33) = 0.6293
1 – Φ(0.3271) = 1 – ( (0.3271 – 0.32) × 0.6293 + (0.33 – 0.3271) × 0.6255) / (0.33 – 0.32) = 0.3718
Question 5.11: Probability of Type II error for two-tailed test
What is the probability of a Type II error for the two-tailed test?
Answer 5.11: The probability of a Type II error when the true proportion is pʹ for the two-tailed test is
Φ[ (p0 – pʹ + zα/2 × (p0 (1 – p0) / n)0.5 ) / ( (pʹ(1 – pʹ) / n)0.5 )
– Φ[ (p0 – pʹ – zα/2 × (p0 (1 – p0) / n)0.5 ) / ( (pʹ(1 – pʹ) / n)0.5 )
(p0 – pʹ + zα/2 × (p0 (1 – p0) / n)0.5 ) / ( (pʹ(1 – pʹ) / n)0.5 = (79% – 68% + 2.576 × 0.040731) / 0.046648 = 4.6073
(p0 – pʹ – zα/2 × (p0 (1 – p0) / n)0.5 ) / ( (pʹ(1 – pʹ) / n)0.5 = (79% – 68% – 2.576 × 0.040731) / 0.046648 = 0.1088
zα/2 = 2.576 (lower limit of right 0.5% tail)
Interpolating in the statistical tables:
Φ(4.6073) ≈ 1.000
1 – Φ(0.1088) = 0.4567
Φ(0.10) = 0.5398
Φ(0.11) = 0.5438
1 – Φ(0.1088) = 1 – ( (0.1088 – 0.10) × 0.5438 + (0.11 – 0.1088) × 0.5398) / (0.11 – 0.10) = 0.4567
Question 5.12: Observations needed for one-tailed test
How many observations are needed for a one-tailed test if α = 1% and β = 5%?
Answer 5.12: The number of observations needed for a one-tailed test =
( (zα × (p0 (1 – p0))0.5 + zβ × (pʹ(1 – pʹ))0.5 ) / (p0 – pʹ) )2
For α = 1% and β = 5%, this equals
((2.326 × (0.79 × (1 – 0.79))0.5 + 1.645 × (0.68 × (1 – 0.68) )0.5) / (0.79 – 0.68) )2 = 243.006
The next highest integer is 244.
Question 5.13: Observations needed for two-tailed test
How many observations are needed for a two-tailed test if α = 1% and β = 5%?
Answer 5.13: The number of observations needed for a two-tailed test =
( (zα/2 × (p0 (1 – p0))0.5 + zβ × (pʹ(1 – pʹ))0.5 ) / (p0 – pʹ) )2
For α = 1% and β = 5%, this equals
((2.576 × (0.79 × (1 – 0.79))0.5 + 1.645 × (0.68 × (1 – 0.68) )0.5) / (0.79 – 0.68) )2 = 272.724
The next highest integer is 273.
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