provide R code as well (i) Chapter 1 Celebration of Lel Chapter +1+Celebratior +

ID: 3041040 • Letter: P

Question

provide R code as well

(i) Chapter 1 Celebration of Lel Chapter +1+Celebratior + X v Packages/Microsoft.MicrosoftEdge 8wekyb3d8bbwe/TempState/Downloads/Chapter+1+Celebration+of+Lear A study was conducted to determine if a medical device is effective for treating a'condition that occurs after an injury. Subjects were randomized to four treatments consisting of a control and three different lengths of time the device is wom each day for a two month period. The response variable is a measure of the ability of the device to moderate the condition. 1 hour/day 2 hours per day 3 hours per day 4.73 5.81 5.69 3.86 4.06 6.56 8.34 Control 7.03 4.65 6.65 49 5.32 6.00 5.12 7.08 5.48 6.52 4.09 6.28 7.77 5.68 8.47 4.58 4.11 5.72 5.91 6.89 6.99 4.98 9.94 6.38 4.51 4.97 3.00 7.97 2.23 4.85 7.26 5.92 5.58 1.91 4.90 4.54 8.18 5.42 6.03 7.04 5.64 9.35 6.52 4.96 6.10 6.71 6.51 1.70 5.89 5.34 5.88 7.50 3.28 5.38 7.30 5.46 4.03 2.72 5.17 5.70 5.85 6.45 7.60 7.90 7.91 a. Write the analysis of variance model for this study, and define each term in the model. b. State the assumptions necessary for an analysis of variance of the data and verify assumptions. c. State the hypotheses and compute the analysis of variance table for the data d. Test the hypothesis of no difference among the means with -005. State your decision with respect to the null hypothesis and give a conclusion in the context of this scenario. The researcher stated before the experiment that the control would be compared to the average of the other three treatments. Give the contrast for this comparison, the hypotheses, and perform the appropriate analysis with = 0.05. State your decision with respect to the null hypothesis and give a conclusion in the context of this scenario. Perform a Duncan's multiple-range test on the four treatments. Give the hypotheses and explain your results e. f.

Explanation / Answer

Part (a)

ANOVA Model

Let xij represent the jth observation in the ith treatment (column), j = 1,2,…,20; i = 1,2,3, 4

Then the ANOVA model is: xij = µ + i + ij, where µ = common effect, i = effect of ith treatment, and ij is the error component. ANSWER

Part (b)

ANOVA Assumption

Error component, ij is assumed to be Normally Distributed with mean 0 and variance 2. ANSWER

Part (c)

Null and Alternate Hypotheses

Null hypothesis: H0: 1 = 2 = 3 = 4 = 0 Vs Alternative: HA: H0 is false. i,e., at least one i is different from other i’s. ANSWER

Part (d)

ANOVA TABLE

Source

Fcrit

p-value

Teatment

10.044

3.3478

1.2979

2.7249

0.28

Error

196.03

2.5793

Total

206.07

2.6085

Since Fcal (1.2979) < Fcrit (2.7249), H0 is accepted. This is also confirmed by the p-value (0.28) being greater than the level of significance (0.05).

Conclusion

There is not enough evidence to suggest that the means of 4 treatments are different. ANSWER

DONE