Question

Population 1   Population 2
51   54
57   51
65   67
68   60
53   58
58   61
75   66
72   79

Can it be concluded, from this data, that there is a significant difference between the two population means?

Let d=(Population 1 entry)−(Population 2 entry). Use a significance level of α=0.1 for the test. Assume that both populations are normally distributed.

Step 1 of 5: State the null and alternative hypotheses for the test.

Step 2 of 5: Find the value of the standard deviation of the paired differences. Round your answer to one decimal place.

Step 3 of 5: Compute the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to three decimal places.

Step 5 of 5: Make the decision for the hypothesis test. (Reject or Fail to Reject Null Hypothesis)

Answer 1

Solution:

Step 1 of 5: State the null and alternative hypotheses for the test.

Null hypothesis: H₀: there is no any significant difference between the two population means.

Alternative hypothesis: H_a: there is a significant difference between the two population means.

H₀: µ_d = 0 versus H_a: µ_d ≠ 0

Step 2 of 5: Find the value of the standard deviation of the paired differences.

The formula for standard deviation is given as below:

SD = sqrt[∑(Di - DBar)^2/(n – 1)]

The calculation table is given as below:

Pop1	Pop2	Di	(Di - DBar)^2
51	54	-3	11.390625
57	51	6	31.640625
65	67	-2	5.640625
68	60	8	58.140625
53	58	-5	28.890625
58	61	-3	11.390625
75	66	9	74.390625
72	79	-7	54.390625
Total			275.875

∑(Di - DBar)^2 = 275.875

SD = sqrt[275.875/(8 – 1)]

S_d = SD = 6.2778

S_d = SD = 6.3

Step 3 of 5: Compute the value of the test statistic.

Test statistic for paired t test is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = 0.3750

S_d = 6.2778

n = 8

df = n – 1 = 7

α = 0.10

t = (0.3750 – 0)/[6.3/sqrt(8)]

t = 0.1690

Test statistic = t = 0.169

Step 4 of 5: Determine the decision rule for rejecting the null hypothesis H0.

df = n – 1 = 7

α = 0.10

Test is two tailed.

So, critical value by using t-table is given as below:

Critical value = - 1.8946 and 1.8946

Decision rule: Reject H₀ if |t| > 1.895

Step 5 of 5: Make the decision for the hypothesis test.

We have |t| = 0.169 < critical value = 1.895

So, we do not reject the null hypothesis

Fail to reject null hypothesis

There is insufficient evidence to conclude that there is a significant difference between the two population means.