No_Treatment Fertilizer Irrigation Fertilizer_and_Irrigation 1.14 0.84 0.09 0.92

Question

No_Treatment Fertilizer Irrigation Fertilizer_and_Irrigation

1.14 0.84 0.09 0.92

0.42 0.75 0.68 1.46

0.48 0.48 0.14 1.16

0.05 0.59 0.81 1.48

1.35 0.93 0.83 , 1.14

Weights (g) of poplar trees were obtained from trees planted in a ich and moiest region The trees were given different treatments identiod in the accompanying table Use a 0.05 signilicance leved to tesa the claikn that the lor reatmont categories yiekd poplar frees with the same mean weight Is there a treatment that appears to be most effective? ? ick the icon to view the data table ofthe poplar weights Determine the null and aternative hypotheses H1 Find the F test statistic FRound to four decimal places as needed) Find the Pvalue using the F test statistic P value-(Round to four decimal places as needed ) What is the conclusion for this hypothesis test? There is insufficient evidence to warraot rejection ef the clalm that 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 yleild the same mean poplar wght Click to select your answer(s) O Type here to search

Explanation / Answer

H0: mu1 = mu2 = mu3 = mu4

H1: At least one of the means is different from the others.

The p-value is .032568. The result is significant at p < .05.

Reject null hypothesis.

There is sufficient evidence to support the claim that the four population means are not all the same. We conclude that those weights come from populations having means that are not all the same.

1.14 0.84 0.09 0.92 0.42 0.75 0.68 1.46 0.48 0.48 0.14 1.16 0.05 0.59 0.81 1.48 1.35 0.93 0.83 1.14 n 5 5 5 5 x-bar 0.688 0.718 0.51 1.232 s 0.539602 0.182948 0.365582 0.236897

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