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    Two Tailed Distribution

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    Regression analysis

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    P3. 1. a) Ho: U=5 vs H1: U≠5
    <br>b) since we are to test whether it's "different from" 5 sec. This is a two-tailed distribution. And on either end of the t-distribution, a=0.025 are the rejection region.
    <br>C) please refer to the attached Excel file for calculation steps:
    <br>now we have: Xm=4.5, U=5, S=4.119 df=n-1=13
    <br>Then t^=(Xm-U)/(S/SQRT(N))= (4.5-5)/(4.119/SQRT(13))= -0.438
    <br>d) from (c) we get |t^|=0.438
    <br>while from t-table, t*(0.975,13)=2.160
    <br>since |t^|<t*, we can accept the H0 with 95% level of confidence.
    <br>e) the above decision means that the average washing time is not different from 5 seconds for level of significance of 0.05.
    <br>P4. ANOVA
    <br>1) The null hypothesis will be that all population means are equal, the alternative hypothesis is that at least one mean is ...

    Solution Summary

    The solution addresses a two tailed distribution, f-statistic and mean square within groups.