Greengrow, Inc.is evaluating three new fertilizer mixes.During a trial period, five similar young plants of equal size were randomly assigned to each fertilizer mix.Plant heights (in centimeters) were measured at the end of the trial period.The tables below summarize (partially) the test results.Use the sample results to test the null hypothesis that average heights for the three populations represented here are the same, using a significance level of 5%.Compute the F value (F_stat) for the test. A)Since the computed F (F_stat) is greater than the critical F value, we cannot reject the "no difference in population means" null hypothesis. B)Since the p-value is greater than .025, we can reject the "no difference in population means" null hypothesis. C)Since the p-value is greater than .05, we cannot reject the "no difference in population means" null hypothesis. D)Since the computed F (F_sta_t) is greater than the critical F value, we can reject the "no difference in population means" null hypothesis.

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The Answer of Greengrow, Inc.is evaluating three new fertilizer mixes.During a trial period, five similar young plants of equal size were randomly assigned to each fertilizer mix.Plant heights (in centimeters) were measured at the end of the trial period.The tables below summarize (partially) the test results.Use the sample results to test the null hypothesis that average heights for the three populations represented here are the same, using a significance level of 5%.Compute the F value (F_stat) for the test. A)Since the computed F (F_stat) is greater than the critical F value, we cannot reject the "no difference in population means" null hypothesis. B)Since the p-value is greater than .025, we can reject the "no difference in population means" null hypothesis. C)Since the p-value is greater than .05, we cannot reject the "no difference in population means" null hypothesis. D)Since the computed F (F_sta_t) is greater than the critical F value, we can reject the "no difference in population means" null hypothesis.