MS Mod 7 Difference of means for small samples practice exam questions


MS Mod 7 Difference of means for small samples practice exam questions...

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MS Module 7 Difference of means for small samples practice exam questions

(The attached PDF file has better formatting.)

Samples from two groups have the following samples sizes, means, and standard deviations:

Group    Sample Size    Sample Mean    Sample SD
Group 1    5    12    1.1
Group 2    7    15    1.4


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 7.1: Variance of estimated mean

What is the variance of the estimated mean of each group?

Answer 7.1: variance of the estimated mean = (standard deviation)2 / number of observations

●    Group 1: 1.12 / 5 = 0.242
●    Group 2: 1.42 / 7 = 0.280


Question 7.2: Variance of estimated difference of group means

What is the variance of the estimated difference of the group means?

Answer 7.2: 0.242 + 0.280 = 0.522

(variance of the estimated difference of the group means = sum of variances of estimated group means)


Question 7.3: Standard deviation of estimated difference of group means

What is the standard deviation of the estimated difference of the group means?

Answer 7.3: 0.5220.5 = 0.7225

(standard deviation = square root of the variance)


Question 7.4: Degrees of freedom

What are the degrees of freedom for a t test of each group’s mean?

Answer 7.4: Degrees of freedom = number of observations – 1

●    Group 1: 5 – 1 = 4
●    Group 2: 7 – 1 = 6


Question 7.5: Degrees of freedom

What are the degrees of freedom for a t test of the difference of the group means?

Answer 7.5: The approximate degrees of freedom is

(s21/m + s22/n)2 / (s21/m)2/(m-1) + (s22/n)2/(n-1) =

(variance of group 1 mean + variance of group 1 mean)2 /
    (square of variance of group 1 mean / (group 1 observations – 1)
+    square of variance of group 2 mean / (group 2 observations – 1) ) =

    (0.242 + 0.280)2 / (0.2422 / (5 – 1) + 0.282 / (7 – 1) ) = 9.834

We truncate the degrees of freedom to 9.


Question 7.6: t value for difference of group means

What is the t value for a 90% two-sided confidence interval of the difference of the group means?

Answer 7.6: 1.833

(the t value for a 90% two-sided confidence interval is the t value for a 5% one-tailed test with 9 degrees of freedom)


Question 7.7: Confidence interval

What is the 90% two-sided confidence interval of the difference of the group means (μ1 – μ2)?

Answer 7.7: Confidence interval = (μ1 – μ2) ± t value × standard deviation of the difference of the group means

●    lower bound: (12 – 15) – 1.833 × 0.7225 = -4.324
●    upper bound: (12 – 15) + 1.833 × 0.7225 = -1.676


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