1. The data set LM10 in the following table are from an experiment that utilized a completely randomizeddesign.Treatment3.95.41.3SourceTreatmentErrorTotal2.42.00.7df(1)(6)(9)Response4.14.82.2(a) Complete the following ANOVA table (fill in all values in boxes 1 through 10).ANOVA TableSS(2)(7)(10)MS(3)(8)5.53.8F2.3p-Value(5)(b) Test the null hypothesis Ho: #₁ = 2 = 3 versus the alternative hypothesis H₁: at least onepair of means differ. Draw your conclusion using a = 0.1.

Treatment T1 T2 T3 3.9 5.4 1.3 2.4 2.0 0.7 4.1 4.8 2.2 5.5 3.8 2.3 3.5 The data…

Answered: 1. The data set LM10 in the following…

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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Eemaining Time: 1 hour. 37 minutes, 54 seconds. Question Completion Status: QUESTION 10 A normally distributed random variable X with variance o = 400 and P(X> 130) = 0.6554. Then the mean of X is O 122 O 138 O 120 O 290 QUESTION 11 For the given data in stem-and- leaf display form, find the inter-quartile range. 3.2 | 4579 3.3 | 02558 3.4 | 45789 3.5 | 11678 3.6 | 0268 O 2.5 O 0.28 O 0.31
50
Use formulas to solve this and also with explaination
Hi, I'm stuck with question 2. Question 1 given the example for this question
The ANOVA table of the above data is as follows: ANOVA Source of Variation SS df MS 2.083 P-value Factor A J 1 0.725 Marketing startegies 0.750 1 0.750 0.832 Interaction 90.750 K 90.750 0.043 Within 125.333 15.667 Total 218.917 10. Calculate J and L 11. State HO and H1 hypotheses of interaction effect based on the above problem. 12. At a = 0.1, is there any significant effect? What is your decision? Support your answer. 13. What is your conclusion?
Hello! I need answer for letter F only using Anderson-Darling test to support the normal distribution of the data.
Mm1
help with 3
(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 have data drawn from a normal distribution with a known variance of 16. You set up the following NHST: • Ho: data follows a N(2, 4²) • HÃ: data follows a N(µ, 4²) where µ ‡ 2. Test statistic: standardized sample mean z. Significance level set to a = .05. You then collected n = 16 data points with sample mean 1.5. (a) Find the rejection region. Draw a graph indicating the null distribution and the rejection region. (b) Find the z-value and add it to your picture in part (a). (c) Find the p-value for this data and decide whether or not to reject Ho in favor of HA.
Sun Oct 22 webassign.net You may need to use the appropriate technology to answer this question. The following data are from a completely randomized design. In the following calculations, use a = 0.05. xj Treatment Treatment 1 2 63 46 53 46 52 64.67 о но H1 #Hz #Из H₂H₁ = H₂ = 3 83 = 71 87 71 78 68.00 Treatment 3 70 54 61 51 59 (a) Use analysis of variance to test for a significant difference among the means of the three treatments. State the null and alternative hypotheses. H₁ H₁ = H₂ = 13 H: Not all the population means are equal. 71.33 O Ho: Not all the population means are equal. на: H1 = H2 = Из O Ho: At least two of the population means are equal. H: At least two of the population means are different. Ho: H₁ H₂ H3 на: М1 #M2 #Из 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. O Reject Ho. There is not sufficient evidence to conclude that the means of…
Refer to the below table. Using an alpha = 0.05, test the claim that IQ scores are the same for children in three different blood lead level groups: low lead level, medium lead level, and high lead level). One-Way Analysis of Variance Summary Table for IQ Measurements for Children among Three Blood Lead Level Groups: Low Lead Level, Medium Lead Level, and High Lead Level. Source df SS MS F p Between-group (treatment) 2 469.1827 2677.864 2.30 0.104 Within-group (error) 118 203.6918 11745.05 Total 120

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