**Tutorial Exercise**You may need to use the appropriate technology to answer this question.The following data are from a completely randomized design.| | Treatment A | Treatment B | Treatment C ||----------------|-------------|-------------|-------------|| | 33 | 45 | 32 || | 29 | 44 | 37 || | 29 | 44 | 36 || | 27 | 47 | 37 || | 32 | 50 | 33 || **Sample mean**| 30 | 46 | 35 || **Sample variance** | 6.00 | 6.50 | 5.50 |(a) At the \( \alpha = 0.05 \) level of significance, can we reject the null hypothesis that the means of the three treatments are equal?(b) Use Fisher's LSD procedure to test whether there is a significant difference between the means for treatments A and B, treatments A and C, and treatments B and C. Use \( \alpha = 0.05 \).(c) Use Fisher's LSD procedure to develop a 95% confidence interval estimate of the difference between the means of treatments A and B.

Given alpha = 0.05Null hypothesis: Means of the three treatments are equal.After entering the data…

Answered: The following data are from a…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A middle school teacher believes that a reading rewards program results in a higher number of pages read per year after the program is completed. The teacher chooses a simple random sample of students and records the number of pages read per year before starting the program and after the program is completed. The results of the study are below. What are the population parameters? Pick Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level Before 813.25 192250.25 182875.22 100 100 0 198 1.497 0.068 1.653 0.136 1.972 95% After 904.95 What is the level of significance? Ex: 0.12 Is the two-sample hypothesis test a paired or unpaired t-test? Pick The null hypothesis is Ho: after Pick The alternative hypothesis is Ha : Mafter Pick Should Ho be rejected or does Ho fail to be rejected? Pick What conclusion can be drawn from the data? Pick Mbefore before ✓evidence exists to support the claim…
Textbook: Statistics for the Behavioral SciencesGregory J. Privitera Please show your work-Concept and Application Problems (Pg.133) If Sample 1 has a variance of 4 and Sample 2 has variance of 32, can we tell which sample had a larger range? If so, which sample had a greater range?
In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance (to 2 decimals, if necessary). If the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100 Error Total 1,500 At a=.05 level of significance, is there a significant difference between the treatments?
Answer all of the following question, anova table have given
(see attachment) Use technology to conduct a one-factor ANOVA to determine if the group means are equal using α=0.05α=0.05. Group means (report to 2 decimal places): Group 1: Internship: Group 2: Co-op: Group 3: Work Study: ANOVA summary statistics:F-ratio = (report accurate to 3 decimal places)p=p= (report accurate to 4 decimal places)
You may need to use the appropriate technology to answer this question. In a completely randomized design, 15 experimental units were used for the first treatment, 17 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100 Error Total 1,800 At a 0.05 level of significance, is there a significant difference between the treatments? State the null and alternative hypotheses. H0: ?1 = ?2 = ?3Ha: ?1 ≠ ?2 ≠ ?3H0: ?1 ≠ ?2 ≠ ?3Ha: ?1 = ?2 = ?3 H0: At least two of the population means are equal.Ha: At least two of the population means are different.H0: Not all the population means are equal.Ha: ?1 = ?2 = ?3H0: ?1 = ?2 = ?3Ha: Not all the population means are equal. Find the value of the test statistic. (Round your…
Three different treatments and four different methods are tested for their ability to control acne. What is the value of the LSD (0.05) for the comparison among method means? Treatment 1 Treatment 2 Treatment 3 SUMMARY Treatment 1 Count Sum Average Variance Treatment 2 Count Sum Average Variance Treatment 3 Count Sum Average Variance 3 Anova: Two-Factor With Replication Total Count Sum Average Variance Method 1 5 6 7 ANOVA Source of Variation Sample 6 2 Method 1 2 11 5.5 0.5 2 13 6.5 0.5 2 5 2.5 0.5 6 29 4.833 3.767 Method 2 6 SS 36,75 7 8 7 3 4 Method 2 2 13 6.5 0.5 2 15 7.5 0.5 2 7 3.5 0.5 6 35 5.833 3.767 df 2 Method 3 7 8 9 8 4 6 Method 3 2 15 7.5 0.5 2 17 8.5 0.5 2 10 5 2 6 42 7 3.2 MS 18.4 Method 4 8 9 6 5 6 7 Method 4 2 17 8.5 0.5 2 11 5.5 0.5 2 13 6.5 0.5 6 41 6.833 2.167 F Total 8 56 7 1.714 8 56 7 1.714285714 8 35 4.375 3.125 P-value 2:370761-05 F crit 3.89
TRUE OR FALSE
In an industrial experiment, a job was performed by 10 workers using Method I, and by 7 workers using Method II. The experiment yielded the following data. The unit of measurement is minute. A statistician wants to test for the equality of two variances and the difference in the mean times to complete the job using the two methods. (This is a case of two independent samples drawn from two normal populations). Method 1 Method 2 12221915101615222613 61071261012 n1=10 Mean=17 Variance=26 n2=7 Mean=9 Variance=7 (a) Test at the 5% significance level whether the two variances are equal. (Give null and the alternative hypotheses, test statistic, rejection rule, conclusion) (b) Estimate with 95% confidence the ratio of the two population variances. Briefly describe what the interval estimate tells you. (c) Briefly explain how to use the interval estimate in part (b) to test the hypotheses in (a).
When comparing more than two treatment means, why should you use an analysis of variance instead of using multiple t tests? Question 31 options: Using multiple t tests increases the risk of a Type II error The analysis of variance is more likely to detect a treatment effect Using multiple t tests increases the risk of a Type I error There is no advantage to using an analysis of variance instead of multiple t tests
Drug Trial ProblemClinical trials for a drug was conducted using a treatment group and a placebo group. The results are shown in the table below. Treatment Placebo μ μ₁ μ₂ n 33 38 x̄ 2.31 2.87 s 0.85 0.71 Use a 0.05 significance level to test the claim that the two samples are from populations with different means. State the claim, null, and alternative hypothesis. Claim: μ1 μ2 HO: μ1 μ2 H1: μ1 μ2
In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance (to 2 decimals, if necessary). If the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100 Error Total 1,700 At a .05 level of significance, is there a significant difference between the treatments? The -value is - Select your answer - less than .01 between .01 and .025 between .025 and .05 between .05 and .10 greater than .10 What is your conclusion? - Select your answer - Cannot conclude that the means of the three treatments are equal Conclude that the means of the three treatments are equal

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