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    One tailed test for difference between means (small sample)

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    A national manufacturer of ball bearings is experimenting with two different processes for producing precision ball bearings. It is important that the diameters be as close as possible to an industry standard. The output from each process is sampled and the average error from the industry standard is calculated. The results are presented below.

    Process A Process B
    Mean 0.002 mm 0.0026 mm

    Standard Deviation 0.0001 mm 0.00012 mm

    Sample Size 12 14

    The researcher is interested in determining whether there is evidence that the two processes yields different average errors from the industry standard.

    I need to state the null and alternate hypotheses, the decision rule, find the test statistic, state my decision, and estimate the P value for the attached.

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    Please see the attached files for details.

    One tailed test for difference between means (small sample size)

    At significance level= 0.05

    1 2
    Process A Process B
    Mean =?Ê= 0.0020 0.0026 mm Difference of means= 0.0006 = 0.0026 - 0.002
    Standard deviation =s= 0.00010 0.00012 mm
    Sample size=n= 12 14

    pooled estimate of ?Ð^2=sp^2={ (n1-1)s1^2+(n2-1)s2^2}/(n1+n2-2)=
    =(11 * 0.0001 ^2 + 13 * 0.00012 ^2 )/ 24 = 0.000000012383
    pooled estimate of ?Ð=sp= = ...

    Solution Summary

    Solution executes a One tailed test for difference between means (small sample).