# Test of hypothesis for difference between means

Exercise #1 A sample of 40 observations is selected from one population. The sample mean is 102 and the sample STDEV is 5. A second sample of 50 observations is selected from a second population. That sample mean is 99 and the STDEV is 6. Conduct the following test of hypotheses using the 0.04 significance level:

H(0): m(1) = m(2)

H(1): m(1) is not equal to m(2)

a. Is this a one-tailed or a two-tailed test?

b. State the decision rule.

c. Compute the value of the z- test statistic.

d. State your decision regarding H(0).

e. What is the p-value?

Exercise #2 The null and alternate hypotheses are:

H(0): m(1) = m(2)

H(1): m(1) is not equal to m(2)

A random sample of 10 observations from one sample revealed a sample mean of 23 and a sample STDEV of 4. A random sample of 8 observations from another population revealed a sample mean of 26 and a sample STDEV of 5. At the 0.05 significance level, is there a difference between the population means? Also,

(a) state the decision rule,

(b) compute the pooled estimate of the population variance,

(c) compute the t-test statistic,

(d) state your decision about the null hypothesis, and

(e) estimate the p-value.

https://brainmass.com/statistics/hypothesis-testing/test-of-hypothesis-for-difference-between-means-167537

#### Solution Preview

Exercise #1: A sample of 40 observations is selected from one population. The sample mean is 102 and the sample STDEV is 5. A second sample of 50 observations is selected from a second population. That sample mean is 99 and the STDEV is 6. Conduct the following test of hypotheses using the 0.04 significance level:

H(0): m(1) = m(2)

H(1): m(1) is not equal to m(2)

Null Hypothesis: Ho: M 1=M 2 There is no difference between the means

Alternative Hypothesis: H1: M 1not equal to M 2 There is difference between the means

At significance level= 0.04

a. Is this a one-tailed or a two-tailed test?

No of tails= 2

This is a 2 tailed test because we are testing M 1not equal to M 2

b. State the decision rule.

Since we are working with large sample size we will use z distribution

The test statistic is z

Significance level= 0.04

Z-value= 2.0537 corresponding to level of significance= 0.04

Thus the z critical value= 2.0537

Decision rule: If

-2.0537 < sample Z value < 2.0537 Accept Null Hypothesis

Else, reject the Null Hypothesis

c. Compute the value of the z- test statistic.

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#### Solution Summary

The solution tests hypotheses for difference between means.