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    Testing of mean and variance

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    Testing of mean and variance. See attached file for full problem description.

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    Please see the attached file.

    Answer 4
    Here we test the null hypothesis H0:
    i.e the average daily yield =880.
    The test statistic used is . If calculated value of Z is less than , we reject the null hypothesis.

    Here Z =

    One-Sample Z

    Test of mu = 880 vs not = 880
    The assumed standard deviation = 21

    N Mean SE Mean 95% CI Z P
    50 871.000 2.970 (865.179, 876.821) 3.03 0.002

    Since that Z is grater than 1.96 we reject the null hypothesis the daily yield has not changed.

    Answer 5

    Descriptive ...

    Solution Summary

    Testing of mean and variance through students t test and F test are described in the solution.