The accompanying data are the weights​ (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results from using the Bonferroni test with the sample data. Complete parts​ (a) through​ (c). Determine the test statistic. The test statistic is enter your response here. ​(Round to two decimal places as​ needed.) Determine the​ P-value. The​ P-value is enter your response here. ​(Round to three decimal places as​ needed.) What is the conclusion for this hypothesis test at a 0.10 significance​ level? A. Fail to reject H0. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. B. Reject H0. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. C. Fail to reject H0. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. D. Reject H0. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.
Added by Jorge O.
Step 1
However, since the data is not provided, we will assume some values for the test statistic and P-value to demonstrate the process. Let's assume the test statistic is 3.25 and the P-value is 0.012. Part 1: Test Statistic The test statistic is 3.25. (This value is Show more…
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The accompanying data are the weights (kg) of poplar trees that were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Also shown are partial results from using the Bonferroni test with the sample data. Complete parts (a) through (c). No Treatment Fertilizer Irrigation Fertilizer and Irrigation 1.22 1.02 0.09 0.82 0.63 0.83 0.67 1.83 0.47 0.44 0.07 1.46 0.08 0.69 0.72 2.21 1.32 1.09 0.92 1.69 Bonferroni Results Mean (I) TREATMENT (J) TREATMENT Difference (I-J) Std. Error Sig. 1.00 2.00 -0.07 0.27183 1.000 3.00 0.25 0.27183 1.000 4.00 -0.8580 0.27183 0.040 a. Use a significance level to test the claim that the different treatments result in the same mean weight. Determine the null and alternative hypotheses. H0: ÎĽ1 = ÎĽ2 = ÎĽ3 = ÎĽ4 H1: At least one of the four population means is different from the others. b. Find the test statistic. (Round to two decimal places as needed.)
Sri K.
An experiment has been conducted for four treatments with eight blocks. Complete the following analysis of variance table. (Round your values for mean squares and F to two decimal places, and your p-value to three decimal places.) Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p-value --------------------- | -------------- | ------------------ | ----------- | - | ------- Treatments | 300 | | | | Blocks | 400 | | | | Error | | | | | Total | 900 | | | | Use α = 0.05 to test for any significant differences. State the null and alternative hypotheses. H0: μ1 = μ2 = μ3 = μ4 Ha: Not all the population means are equal. Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to three decimal places.) p-value = State your conclusion. Reject H0. There is sufficient evidence to conclude that the means of the four treatments are not all equal.
Sheryl E.
reject the null hypothesis? Since the P-value is greater than the level of significance at α = 0.05, we do not reject H0. Since the P-value is less than or equal to the level of significance at α = 0.05, we reject H0. Since the P-value is greater than the level of significance at α = 0.05, we reject H0. Since the P-value is less than or equal to the level of significance at α = 0.05, we do not reject H0. (e) Interpret your conclusion in the context of the application. At the 5% level of significance there is insufficient evidence to conclude that the means are not all equal. At the 5% level of significance there is sufficient evidence to conclude that the means are not all equal. At the 5% level of significance there is insufficient evidence to conclude that the means are all equal. At the 5% level of significance there is sufficient evidence to conclude that the means are all equal. (f) Make a summary table for your ANOVA test.
Madhur L.
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