Neas-Seminars

MS Mod 5 Hypothesis testing of proportions practice exam questions


http://33771.hs2.instantasp.net/Topic15985.aspx

By NEAS - 8/5/2018 10:30:43 PM


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.
By NEAS - 10/16/2019 1:47:17 PM

 
tud55886 - 10/16/2019 10:13:44 AM
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. 


See the variance of the z value in Example 9.11 (page 452).