## MS Mod 5 Hypothesis testing of proportions practice exam questions

 Author Message NEAS Supreme Being Group: Administrators Posts: 4.2K, Visits: 1.2K 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 < μ0The null hypothesis is tested at a 1% significance level and the true incidence of death with the drug is 68%.Question 5.1: Sample meanWhat is the incidence of death from the disease in the sample?Answer 5.1: 73 / 100 = 73%Question 5.2: Standard deviationWhat 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 hypothesisWhat is the p value for the one-tailed alternative hypothesis?Answer 5.4: Φ(-1.4731) = 0.0704Interpolating in the statistical tables: Φ(1.47) = 0.9292Φ(1.48) = 0.9306Φ(–1.4731) = 1 – ( (1.4731 – 1.47) × 0.9306 + (1.48 – 1.4731) × 0.9292) / (1.48 – 1.47) = 0.0704Question 5.5: p value for two-tailed alternative hypothesisWhat 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 valueWhat 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 meanWhat 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 = 0.046648Question 5.8: Expected value of the z valueWhat 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 = -2.7006Question 5.9: Standard deviation of the z valueWhat 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 = 1.1453Question 5.10: Probability of Type II error for one-tailed testWhat 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 = 0.3271zα = 2.326 (lower limit of right 1% tail)1 – Φ(0.3271) = 0.3718Interpolating in the statistical tables: Φ(0.32) = 0.6255Φ(0.33) = 0.62931 – Φ(0.3271) = 1 – ( (0.3271 – 0.32) × 0.6293 + (0.33 – 0.3271) × 0.6255) / (0.33 – 0.32) = 0.3718Question 5.11: Probability of Type II error for two-tailed testWhat 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 = 0.1088zα/2 = 2.576 (lower limit of right 0.5% tail)Interpolating in the statistical tables: Φ(4.6073) ≈ 1.0001 – Φ(0.1088) = 0.4567Φ(0.10) = 0.5398Φ(0.11) = 0.54381 – Φ(0.1088) = 1 – ( (0.1088 – 0.10) × 0.5438 + (0.11 – 0.1088) × 0.5398) / (0.11 – 0.10) = 0.4567Question 5.12: Observations needed for one-tailed testHow 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) )2For 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 = 243.006The next highest integer is 244.Question 5.13: Observations needed for two-tailed testHow 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) )2For 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 = 272.724The next highest integer is 273. Attachments MS Module 5 Hypothesis testing of proportions practice exam questions.pdf (254 views, 57.00 KB) tud55886 t Junior Member Group: Forum Members Posts: 11, Visits: 151 Can you help me with Question 5.9: Standard deviation of the z value. I‌ don't understand which formula is used for solving it. NEAS Supreme Being Group: Administrators Posts: 4.2K, Visits: 1.2K +x tud55886 - 10/16/2019 10:13:44 AMCan you help me with Question 5.9: Standard deviation of the z value. I‌ don't understand which formula is used for solving it. See the variance of the z value in Example 9.11 (page 452).
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