MS Mod 6 Difference of means practice exam questions


MS Mod 6 Difference of means practice exam questions

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NEAS
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MS Module 6 Difference of means practice exam questions

We test if the mean values in two groups differ. The number of observations in each group (sample size), the sample means, and the standard deviation of the sample values (sample SD) are shown below.

Group    Sample Size    Sample Mean    Sample SD
Group #1    118    23.99    25.74
Group #2    155    25.27    28.72


μ1 = the mean of Group #1; μ2 = the mean of Group #2.
The null hypothesis is H0: μ1 = μ2; the alternative hypothesis is Ha: μ1 ≠ μ2.


Question 6.2: Variance of difference of sample means

What is the variance of the difference (group #1 – group #2) of the two sample means?

Answer 6.2: 25.742 / 118 + 28.722 / 155 = 10.936348

(variance of the difference of independent samples = sum of the variances of each sample)


Question 6.3: Standard deviation of difference of sample means

What is the standard deviation of the difference (group #1 – group #2) of the two sample means?

Answer 6.3: 10.9363480.5 = 3.307015

(standard deviation = square root of variance)


Question 6.4: z value

What is the z value for the difference (group #1 – group #2) in the two sample means?

Answer 6.4: (23.99 – 25.27) / 3.307015 = -0.3871

(z value = difference in means / standard deviation of this difference)


Question 6.5: p value

What is the p value for a two-tailed test of the null hypothesis?

Answer 6.5: 2 × (1 – Φ(0.3871) ) = 2 × (1 – 0.65066) = 0.6987

Interpolating in the statistical tables:

Φ(0.38) = 0.6480
Φ(0.39) = 0.6517
Φ(0.3871) = ( (0.3871 – 0.38) × 0.6517 + (0.39 – 0.3871) × 0.6480) / (0.39 – 0.38) = 0.6506



Question 6.6: Confidence interval for difference in means

What is the lower bound of the two-sided 90% confidence interval for μ1 – μ2 (the mean of Sample #1 minus the mean of Sample #2)?

Answer 6.6: (23.99 – 25.27) – 1.645 × 3.307015 = -6.720

(difference in sample means – critical t value × standard deviation of the difference)


Question 6.7: Probability of a Type II error

If we use an α of 10% to test the null hypothesis, and the true difference in the means μ2 – μ1 (the mean of Sample #2 minus the mean of Sample #1) = 4.32, what is β, the probability of a Type II error?

Answer 6.7: The probability of a Type II error for the difference in means =

Φ(zα/2 – (Δʹ – Δ0) / σ) – Φ( –zα/2 – (Δʹ – Δ0) / σ) =
Φ(1.645 – 4.32 / 3.307015) – Φ( –1.645 – 4.32 / 3.307015) =
Φ(0.3387) – Φ(-2.9513) = 0.631

●    1.645 – 4.32 / 3.307015 = 0.3387
●    –1.645 – 4.32 / 3.307015 = -2.9513

Φ(0.3387) = 0.6326

Interpolating in the statistical tables:

Φ(0.33) = 0.6293
Φ(0.34) = 0.6331
Φ(0.3387) = ( (0.3387 – 0.33) × 0.6331 + (0.34 – 0.3387) × 0.6293) / (0.34 – 0.33) = 0.6326

Φ(-2.9513) = 0.0016

Interpolating in the statistical tables:

Φ(2.95) = 0.9984
Φ(2.96) = 0.9985
Φ(-2.9513) = 1 – ( (2.9513 – 2.95) × 0.9985 + (2.96 – 2.9513) × 0.9984) / (2.96 – 2.95) = 0.0016



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Why on Question 6.6 you talk about "t critical value" but you are using a z value in the calculation? Are we allowed to do this in the exam?
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tud55886 - 10/16/2019 10:41:37 AM
Why on Question 6.6 you talk about "t critical value" but you are using a z value in the calculation? Are we allowed to do this in the exam?


If the sample size is large enough, the critical t value is the same as the critical z value. We approximate the critical t value by the critical z value.

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