Consider the experimental results for the following randomized block design. Make the calculations necessary to set up the analysis of variance table.TreatmentB96315418587Use a = 0.05 to test for any significant differences (to 2 decimals but p-value to 4 decimals, if necessary). If your answer is zero, enter "0".Source of Variation Sum of Squares Degrees of Freedom Mean Square Fp-value6.60TreatmentsBlocksError26.53312.3216.08Total354.93The p-value is between 0.01 and 0.025 ✓What is your conclusion?Conclude not all treatment means are equal ✓2481413.2778.082.01BlocksA1210121820с8514188

Given data: Treatment A B C Blocks 1 10 9 8 2 12 6 5 3 18 15 14 4 20 18 18 5 8 7 8…

Answered: Consider the experimental results for…

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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The manager knows from previous years that the company will provide an annual bonus if theaverage sales per customer is greater than $80.00. Based on this sample – with a 5% level ofconfidence – should they expect an annual bonus? t-Test: Paired Two Sample for Means Variable 1 Variable 2 Mean 77.6005 0 Variance 3098.585 0 Observations 100 100 Pearson Correlation #DIV/0! Hypothesized Mean Difference 80 df 99 t Stat -0.43106 P(T<=t) one-tail 0.33368 t Critical one-tail 1.660391 P(T<=t) two-tail 0.667361 t Critical two-tail 1.984217 Here is the data that I ran through excel. I am having a hard time with the hypothesis, decision, and summary.
The accompanying data table contains chest deceleration measurements (in g, whereg is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below. Click the icon to view the table of chest deceleration measurements A Click the icon to view the table of analysis of variance results. ..... a. What characteristic of the data specifically indicates that one-way analysis of variance should be used? O A. There are three samples of measurements. O B. The population means are approximately normal. O C. The measurements are categorized according to the one characteristic of size. O D. Nothing specifically indicates that one-way analysis of variance should be used. b. If the objective is to test…
A researcher takes sample temperatures in Fahrenheit of 17 days from New York City and 18 days from Phoenix. Test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix. Use a significance level of α=0.05. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. New York City Phoenix 99 94.2 95.5 72 93.2 86.8 102 122.1 85.4 114.4 80 94.7 86.4 89.7 75.4 104.7 79.5 77.6 83.4 106.8 64.3 98.6 65.5 91.5 87.7 82 104 97.7 74.3 64.9 59.5 82 82.8 72 116.2 The Hypotheses for this problem are: H0: μ1 = μ2 H1: μ1μ2 Find the p-value. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value =
An experiment has been conducted for four treatments with seven blocks. Complete the following analysis of variance table. (Round your values for mean squares and F to two your p-value to three decimal places.) Source of Variation Treatments Blocks Error Total Sum of Squares 1,500 400 2,600 Degrees of Freedom Use a 0.05 to test for any significant differences. Mean Square p-value
Heights The heights (cm) in the following table are from Data Set 1 “Body Data” in Appendix B. Results from two-way analysis of variance are also shown. Use the displayed results and use a 0.05 significance level. What do you conclude? XLSTAT
A medical researcher wants to test whether the mean cholesterol levels are the same for 3 treatment groups of subjects (with 18 total randomly selected subjects): those who take a new medication, those who follow a diet, and those who make no changes to medication or diet. They plan to use a 5% level of significance for their decision. Below is a partial Analysis of Variance table. Use it to answer the following questions deg of freedom Sum of Squares Mean Square F statistic Treatment Group 3.121 1.5607 Error 1.517 0.1011 Total 17 What are the degrees of freedom for the treatment groups? What are the degrees of freedom for the error? What is the F-statistic for this data (3 decimal places)
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
Please use ANOVA
A researcher wants to measure average cardiovascular health of university students and compare those scores to the average scores in the general population. Assuming that population variance is known, what statistical test is most appropriate for this study? independent-samples t-test single-sample t-test z-test for sample mean related-samples t-test
Need help with missing boxes
The article cited in Exercise 19 provides measures of electrical conductivity (in uS/cm). The results are presented in the following table. Sample Size 49 Location Mean SD Upstream 463 208 Midstream 363 98 31 Downstream 647 878 30 Can a one-way analysis of variance be used to determine whether conductivity varies with location? Or is one of the necessary assumptions violated? Explain.

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