Example 1: The paper tensile strength experiment value is smaller than o.01. The

Question

Example 1: The paper tensile strength experiment value is smaller than o.01. The actual P - value is 3.59 x 10-6. However, since the P-value is considerably smaller than a - 0.01, we have strong concludethatH is not true. Note that Minitab also provides some summary information about each concentration in the pulp significantly level of hardwood concentration, SSE = SST-SSTreatments 512.96-382.79 130.17 F 19.61 (see ANOVA table next slide), evidence to and since F0.01,320-4.94. we reject H0 and conclude that hardwood affects the strength of the paper. Note that the computer output reports a P-value for the test statistic F = 19.61 in Table 4.7 of zero. This is a truncated value. Appendix Table V reports that including the confidence interval on each mean. 001320-4.94, so clearly the P-

Explanation / Answer

here Null hypothesis H0: all four group means are same

alternate hypothesis Ha: atleast one group mean is different from another

critical value approach:since observed or calculated F=19.61 (p-value=0.000) is more than critical F(0.01,3,20)=4.94 ( alpha=0.01), so we fail to accept Null hypothesis and conclude that alteast one mean is different from another.

using p-value approach :also p-value is less than alpha=0.01 this means we should not accept null hypothesis (same interpretation)

any other information you want from this question please respond back

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