Test of hypothesis for difference between means
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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.
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The solution tests hypotheses for difference between means.
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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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