Question

Consider the computer output below.

Two-Sample T-Test and CI

Sample	N	Mean	StDev	SE Mean
1	15	54.79	2.13	0.55
2	20	58.60	5.28	1.2

Difference = μ1-μ2

Estimate for difference: –3.91

95% upper bound for difference: ?

T-test of difference = 0 (vs <): T-value = -2.93

P-value = ?

DF = ?

(a) Fill in the missing values. Use lower and upper bounds for the P-value. Suppose that the hypotheses are H0: μ1-μ2=0 versus H1: μ1-μ2<0. Enter your answer; P-value, lower bound <P< Enter your answer; P-value, upper bound DF = Enter your answer; DF Determine a 95% upper bound for difference. Round your answer to four decimal places (e.g. 98.7654). μ1-μ2≤ Enter your answer; 95% upper bound for difference

Is this a one-sided or a two-sided test?

The test is one-sided.

The test is two-sided.

(b) What are your conclusions if α=0.05? What if α=0.01?

We reject the null hypothesis at the 0.05 or the 0.01 level of significance.

We do not reject the null hypothesis at the 0.05 or the 0.01 level of significance.

We do not reject the null hypothesis at the 0.05 and reject it at the 0.01 level of significance.

We reject the null hypothesis at the 0.05 and do not reject it at the 0.01 level of significance.

(c) This test was done assuming that the two population variances were different. Does this seem reasonable?

No.

Yes.

(d) Suppose that the hypotheses had been H0: μ1=μ2 versus H1: μ1≠μ2. What would your conclusions be if α=0.05?

Do not reject the null hypothesis.

Reject the null hypothesis.

Answer 1

a)

Degree of freedom is given by

Putting the values as above we get

df = 26.4499

p-value for t-stat = -2.93 is 0.0035

Upper bound for p-value = 0.0035

Lower bound for p-value = 0.0034

Confidence interval calculation

We need to find standard error

The confidence interval is given by

The critical value of t for 27 degrees of freedom for 95% confidence interval is 2.051

CI = [-6.4824, -1.1376]

Upper bound = -1.1376

Test is one sided (left tail)

b)

P-value is less than 0.05 and 0.01, so at both levels, we reject the null hypothesis

Option A is the right answer

c)

Yes

d)

p-value in that case would be 0.00689 which is still less than 0.05. So we would reject the null