Question

Ch. 11, 2. Given two dependent random samples with the following results:

Population 1	71	68	50	84	76	76	80	79
Population 2	76	63	54	80	79	82	75	82

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)d=(Population 1 entry)−(Population 2 entry). Use a significance level of α=0.2 for the test. Assume that both populations are normally distributed.

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

Ho: μd(=,≠,<,>,≤,≥) 0

Ha:μd (=,≠,<,>,≤,≥) 0

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.

Reject Ho if (t, I t I) (<,>) _____

Step 5 of 5:

Make the decision for the hypothesis testTop of Form

Reject Null Hypothesis Fail to Reject Null Hypothesis

Answer 1

Population 1	71	68	50	84	76	76	80	79
Population 2	76	63	54	80	79	82	75	82

The Hypotheses are

The Paired sample Test is computed in Excel by addons shown below:

t-Test: Paired Two Sample for Means
	Population 1	Population 2
Mean	73	73.875
Variance	111.7142857	102.125
Observations	8	8
Pearson Correlation	0.897437244
Hypothesized Mean Difference	0
df	7
t Stat	-0.52615222
P(T<=t) one-tail	0.307517568
t Critical one-tail	0.896029644
P(T<=t) two-tail	0.615035135
t Critical two-tail	1.414923928

Step 2:

the standard deviation of the paired differences Calculated as 4.7

Step3:

the value of the test statistic cumputed as 0.526

Step 4:

Rejection region:

Rejcet Ho if t:∣t∣>1.415

Step5:

decision for the hypothesis test

Since it is observed that |t| = 0.526 ≤tc​=1.415, it is then concluded that the null hypothesis is not rejected.