Two Tailed Distribution

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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.
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<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 ...