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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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    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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    The solution tests hypotheses for difference between means.

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