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 testwith the sample data. Complete parts (a) through (c).Click the icon to view the data table of the poplar weights and the Bonferroni results.Poplar Weights (kg) and Bonferroni Resultsa. Use a 0.10 significance level to test the claim that the different treatments result in the same mean weight.Determine the null and alternative hypotheses.H₂!H₁:Determine the test statistic.The test statistic is(Round to two decimal places as needed.)Determine the P-value.The P-value is(Round to three decimal places as needed.).What is the conclusion for this hypothesis test at a 0.10 significance level?▼O A. Fail to reject Ho. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poOB. Reject Ho. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weigOC. Reject Ho. There is insufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight.O D. 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.b. What do the displayed Bonferroni results tell us?With a P-value ofWith a P-value oftherethereWith a P-value ofthere(Round to three decimal places as needed.).a significant difference between the No Treatment and Fertilizer groups.a significant difference between the No Treatment and Irrigation groups.H₂:H₂:a significant difference between the No Treatment and Fertilizer and Irrigation groups.The test statistic is(Round to two decimal places as needed.)Find the P-value.GThe P-value is(Round to three decimal places as needed.)What do the results indicate at a 0.10 significance level?No Treatment1.2120.5730.5570.1331.297(1) TREATMENT1.00Fertilizer0.9390.8660.4640.5771.029Bonferroni Results(J) TREATMENT2.003.004.00PrintIrrigation0.0690.6610.0980.8170.935OA. Reject H. There is sufficient evidence to warrant rejection of the claim that the irrigation treatment group and the group treated with both fertilizer and irrigation yield the same mean poplar weight.OB. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that the irrigation treatment group and the group treated with both fertilizer and irrigation yield the same mean poplar weight.OC. Fail to reject Ho. There is insufficient evidence to warrant rejection of the claim that the irrigation treatment group and the group treated with both fertilizer and irrigation yield the same mean poplar weight.D. Reject H. There is insufficient evidence to warrant rejection of the claim that the irrigation treatment group and the group treated with both fertilizer and irrigation yield the same mean poplar weight.MeanDifference (I-J) Std. Error-0.02060.2384-0.8422Fertilizer and Irrigatio0.8521.776Donec. Let P₁, P₂, P3, and H, represent the mean amount of the no treatment, fertilizer, irrigation, and fertilizer and irrigation groups, respectively. Use the Bonferroni test procedure with a 0.10 significance level to test for a significant difference between the mean amountof the irrigation treatment group and the group treated with both fertilizer and irrigation. Identify the test statistic and the P-value. What do the results indicate?Determine the null and alternative hypotheses.0.268980.268980.268981.4662.2521.637Sig.1.0001.0000.039

No treatment Fertilizer Irrigation Fertilizer and irrigation 1.212 0.939 0.069 0.852…

Answered: at a 0.10 significance level?

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 294 people over the age of 55, 78 dream in black and white, and among 291 people under the age of 25, 10 dream in black and white. Use a 0.01 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. Complete parts (a) through (c) below indentify test statistic p value= the 90% confidence interval is
Suppose a study reported that the average persin watched 3.37 hours of television per day. a random sample of 15 people gave the number of hours of television watched per day shown below. at the 1% significance level, do the data provide sufficent evidence to conclude that the amount of television watched per day last year by the average person is greater than the value reported in the study? what is the test statistic value for this problem? a.0.28 b.-38 c.none of the above d. 0.58 e.-4.295 f..04 g. 1.94 . the proper conclusion for this problem is? a. reject the null hypothesis we have sufficent evidence to prove the average number of hours people watch tv is more than 3.37 hours. b. Do not reject the null hypothesis we have sufficient evidence to prove the average number of hours of TV watched is greater than 3.37 hours c. Reject the null hypothesis and claim the average number of hours people watch TV is less than 3.37 because the P value is near zero d. Do not reject the null…
1) If the data is IQ test scores, an individual with a negative z-score what IQ level would he have? 2
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Suppose a study reported that the average person watched 3.34 hours of television per day. A random sample of 15 people gave the number of hours of television watched per day shown. At the 10% significance level, do the data provide sufficient evidence to conclude that the amount of television watched per day last year by the average person differed from the value reported in the study? (Note: x = 4.400 hours and s= 1.262 hours.) Set up the hypotheses for the one-mean t-test. HoiH Ha: H 2.3 4.0 4.5 4.9 5.0 4.5 3.3 2.0 5.7 5.9 6.2 3.8 5.9 3.9 4.1
The average amount of time it takes for couples to further communicate with each other after their first date has ended is 2.35 days. Is this average different for blind dates? A researcher interviewed 70 couples who had recently been on blind dates and found that they averaged 2.5 days to communicate with each other after the date was over. Their standard devlation was 0.655 days. What can be concluded at the the a = 0.10 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: Ho: ?v Select an answerv H: ?vSelect an answerv c. The test statistic ?v (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is ?v a f. Based on this, we should Select an answerv the null hypothesis. g. Thus, the final conclusion is that ... O The data suggest that the population mean is not significantly different from 2.35 at a = 0.10, so there is statistically…
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A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 323 people over the age of 55, 71 dream in black and white, and among 288 people under the age of 25, 14 dream in black and white. Use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. Complete parts (a) through (c) below. Question content area bottom Part 1 a. Test the claim using a hypothesis test. Consider the first sample to be the sample of people over the age of 55 and the second sample to be the sample of people under the age of 25. What are the null and alternative hypotheses for the hypothesis test? A. H0: p1=p2 H1: p1<p2 B. H0: p1≥p2 H1: p1≠p2 C. H0: p1≠p2 H1: p1=p2 D. H0: p1≤p2 H1: p1≠p2 E. H0: p1=p2 H1: p1>p2 Your answer is correct. F. H0: p1=p2 H1:…
Based on these tables, would you accept or reject the hypothesis that individuals in bad neighborhoods commit more crime?
The average number of cavities that thirty-year-old Americans have had in their lifetimes is 4. Do twenty-year-olds have a different number of cavities? The data show the results of a survey of 11 twenty- vear-olds who were asked how many cavities they have had. Assume that the distribution of the population is normal. 2, 5, 3, 2, 5, 3, 5, 2, 2, 2, 4 What can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: ns Ho: ? vSelect an answer v re H: ? v Select an answer c. The test statistic 2 v = (please show your answer to 3 decimal places.) urse d. The p-value = e. The p-value is ? va f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is that... (Please show your answer to 4 decimal places.) O The data suggest that the population mean number of cavities for twenty-year-olds is not significantly different from 4 at a = 0.05, so there is…

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