A random sample of 24 items from the first population showed a mean of 111 and a

Question

A random sample of 24 items from the first population showed a mean of 111 and a standard deviation of 13. A sample of 19 items for the second population showed a mean of 101 and a standard deviation of 9. Assume the sample populations do not have equal standard deviations.

Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.)

The null and alternate hypotheses are:    H0: 1 2 H1: 1 > 2

A random sample of 24 items from the first population showed a mean of 111 and a standard deviation of 13. A sample of 19 items for the second population showed a mean of 101 and a standard deviation of 9. Assume the sample populations do not have equal standard deviations.

Explanation / Answer

Statcrunch output for the test is:

Two sample T hypothesis test:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 > 0
(without pooled variances)

Hypothesis test results:

So,

a) Degree of freedom = 40 (Rounded off to the nearest whole number)

b) For 40 degrees of freedom and right tailed test,

critical t - value = 1.303

So,

Reject Ho if t > 1.303

c) From Statcrunch output,

Value of test statistic = 2.974

d) This lies in the rejection region so reject Ho.

Difference Sample Diff. Std. Err. DF T-Stat P-value 1 - 2 10 3.3622648 40.371606 2.9741857 0.0025

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