Weights (kg) of poplar trees were obtained from trees planted in a rich and mois

Question

Weights (kg) of poplar trees were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most effective? Click the icon to view the data table of the poplar weights. Determine the null and alternative hypotheses. H1: At least one of the four population means is different from the others. Find the F test statistic. F(Round to four decimal places as needed.) Find the P-value using the F test statistic. P-value = 10.61 56 (Round to four decimal places as needed.) What is the conclusion for this hypothesis test? OA. Fail to reject Ho- There is insufficient evidence to warrant rejection of the claimthat the four different treatments yield the same mean poplar weight O B. Reject Ho. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight c. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight D Reject o There is insufficient evidence to warrant rejection of the claim that the four different treatments yeld the same mean p ar weight Is there a treatment that appears to be most effective?

Explanation / Answer

From th above out put we got F = 4.0873

p value = 0.0248

Decision rule : If p value < singnificance level we reject the null hypothsis

                        If p value > singnificance level we fail to reject the null hypothsis

So, here p value < significance level

So we reject the null hypothesis.

Reject H0 : there is sufficient evidence to warrent rejection of the claim that the four different treatments yield the same mean popular weight

Anova: Single Factor SUMMARY Groups Count Sum Average Variance no treatment 5 3.75 0.75 0.24615 Fertilizer 5 3.68 0.736 0.04498 Irrigation 5 2.36 0.472 0.11752 Fertilizer and Irrigation 5 6.07 1.214 0.05683 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 1.4269 3 0.475633 4.08725 0.024794 3.238872 Within Groups 1.86192 16 0.11637 Total 3.28882 19

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