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    Testing for difference in means

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    Name the test to use and the hypotheses first, then:

    1. (18 points) The U.S. Bureau of Prisons publishes data in Statistics Report on the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served in months.

    N M S
    Sample 1: Fraud 10.12 months 4.90 35
    Sample 2:Firearms 18.78 months 4.64 40

    At the 5% level do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses? Identify the appropriate test and test the claim using the traditional method at the 0.05 level of significance.

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    Solution Preview

    Please refer to the attachment.

    This is a test of difference between means, however, the conditions are a little ambiguous. Thus I apply the traditional and general method and take the conditions as: Sample 1: Fraud
    Mean time is M1 = 10.12 months, standard deviation is SD1= 4.90 and size is n1=35
    Sample 2: Firearms
    Mean time is M2 =18.78 months, standard deviation is SD2= 4.64 and size is n2=40

    Using a .05 level of significance, test the hypothesis that the mean time for sample 1 is less than that of sample 2.
    Let Md=µ1 ...

    Solution Summary

    The following exercise outlines the process for conducting a hypothesis test that two means are significantly different.