STATISTICS Assignment 5 Solution

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1 STATISTICS 42 - Assignment 5 Solution 1. An experiment will be performed on individuals with elevated serum cholesterol. We want to see if the type of drug (Lipitor or Zocor ) has an effect on the change in serum cholesterol (serum cholesterol level before treatment serum cholesterol level after three months of treatment). A second factor, the amount of drug ( mg/day, 1 mg/day and 2 mg/day), will also be investigated in a two factor completely randomized experiment. a) What is the response? Change in serum cholesterol (serum cholesterol level before treatment serum cholesterol level after three months of treatment). b) What are the experimental units? The experimental units are individuals with elevated serum cholesterol. c) The experimenter will use factorial crossing to create the treatment combinations. How many treatment combinations will there be in the experiment? List all the treatment combinations. There will be 6 treatments. mg Lipitor, mg Zocor, 5 mg Lipitor, 5 mg Zocor, 1 mg Lipitor, 1 mg Zocor. d) The experimenter would like to be able to detect a difference in treatment population means of 1.4 standard deviations with Alpha =.5 and Beta =.1. How many experimental units will the experimenter need? The experimenter will need 18 experimental units for each of the 6 treatments for a total of 18 individuals with elevated serum cholesterol. e) With this number of units, what size difference in factor level population means can be detected when Alpha=.5 and Beta=.1? A difference of between.6 and.7 standard deviations for drug means. A difference of between.8 and.9 standard deviations for amount means. f) Because of budget constraints, only 6 units are available for each treatment combination. How does this choice affect the size of the detectable difference in treatment population means? in factor level population means? Use Alpha=.1 and Beta=.1. With 6 units for each treatment, one can detect a 2.5 standard deviation difference in treatment populations means, a 1 standard deviation difference in drug population means and a 1.4 standard deviation difference in amount population means. 1

2 g) Explain how you would randomly assign treatments to the experimental units. Include a table that indicates the random assignment of treatments for this experiment to experimental units. Remember the budget constraints in f). Number the 36 individuals with elevated serum cholesterol with a unique number between and 36. Use JMP to do the random assignment. Enter Lipitor in 18 rows and Zocor in 18 rows of a column named Drug. In a second column named Amount have six rows with mg, six rows with 1 mg and six rows with 2 mg for each drug. In a third column labeled Individual use Column Formula Random Col Shuffle. For each row, the Drug and Amount are assigned to the individual whose number is in the Individual column. Drug Amount Individual Drug Amount Individual Lipitor mg 1 Zocor mg 22 Lipitor mg 2 Zocor mg 27 Lipitor mg 15 Zocor mg 3 Lipitor mg 31 Zocor mg 34 Lipitor mg 33 Zocor mg 21 Lipitor mg 8 Zocor mg 4 Lipitor 1 mg 6 Zocor 1 mg 9 Lipitor 1 mg 23 Zocor 1 mg 1 Lipitor 1 mg 13 Zocor 1 mg 25 Lipitor 1 mg 26 Zocor 1 mg 3 Lipitor 1 mg 24 Zocor 1 mg 7 Lipitor 1 mg 18 Zocor 1 mg 11 Lipitor 2 mg 32 Zocor 2 mg 12 Lipitor 2 mg 17 Zocor 2 mg 29 Lipitor 2 mg 14 Zocor 2 mg 28 Lipitor 2 mg 19 Zocor 2 mg 35 Lipitor 2 mg 36 Zocor 2 mg 16 Lipitor 2 mg 5 Zocor 2 mg 2 2. An experiment is performed by researchers in kinesiology on the effects of step height and stepping frequency on the change in heart rate. Forty eight individuals are recruited from students majoring in kinesiology. They all give their consent to participate in the experiment. Each individual is asked to step up onto a platform and step down repeatedly. The height of the platform will be either 15 cm or 3 cm. The rate at which a participant steps will be 14, 21, 28 or 35 steps per minute. A combination of platform height and stepping frequency will be assigned to each participant completely at random so that 6 participants are assigned each combination. Participants step up and down off the platform for 5 minutes. The change in heart rate (heart rate after stepping heart rate before stepping) is calculated for each participant. 2

3 a) What is the response? The response is the change in heart rate (heart rate after stepping heart rate before stepping). b) What are the conditions? The conditions are step height (15 cm and 3 cm) and stepping frequency (14, 21, 28 and 35 steps per minute). c) How many treatments are there? There are 2x4 = 8 treatments. d) What are the experimental units? The experimental units are students majoring in kinesiology. e) What is an outside variable that is controlled in this experiment? The amount of time spent stepping is the same. All students step for 5 minutes. f) What is an outside variable that is not controlled in this experiment? The fitness level of each of the students is not controlled. Some may be very fit and some may not be as fit. g) Describe, in some detail, what contributes to chance, random, error variation in this experiment. In general, differences among experimental units that are treated the same contribute to chance, random error variation. For this experiment, differences in the fitness level of students assigned a particular combination of step height and stepping frequency will contribute to chance, random, error variation. h) Give the full model describing the relationship between the response, conditions and random error. Be sure to define all the terms in the model within the context of the problem. Y = μ + τ ij + ε Y = μ + α i + β j + αβ ij + ε Y is the change in heart rate. μ is the overall population mean. τ ij is the treatment effect for step height, i, and stepping frequency, j. α i is the step height effect of step height, i. β j is the stepping frequency effect of stepping frequency, j. αβ ij is the interaction effect for step height, i, and stepping frequency, j. ε is the random error. 3

4 A JMP data table is posted on the course website. Use JMP to analyze the data. i) Estimate the treatment effects. Mean change in heart rate is The means for the 6 treatments are given in the following table. Frequency=14 Frequency=21 Frequency=28 Frequency=35 Height=15 cm Height=3 cm The estimated treatment effects are given in the following table. Frequency=14 Frequency=21 Frequency=28 Frequency=35 Height=15 cm Height=3 cm j) Test the hypothesis that all the treatment effects are zero against the alternative that at least one is not zero. Be sure to give an appropriate null and alternative hypothesis. Report the value of the appropriate test statistic, P-value, decision, reason for the decision and a conclusion within the context of the problem. H : all the τ ij = H A : some τ ij F = , P-value <.1 Because the P-value is so small (<.5) we reject the null hypothesis and conclude that some of the combinations of step height and stepping frequency affect the change in heart rate. k) Estimate the step height effects. Estimated Effect Height=15 cm = Height=3 cm = l) Test the hypothesis that the effects of step height are zero against the alternative that at least one is not zero. Be sure to give an appropriate null and alternative hypothesis using the notation from your answer to h). Report the value of the appropriate test statistic, P-value, decision, reason for the decision and a conclusion within the context of the problem. H : all the α i = H A : some α i F = , P-value <.1 Because the P-value is so small (<.5) we reject the null hypothesis and conclude that the two step heights affect the change in heart rate differently. 4

5 m) Where are the statistically significant differences in step height sample means? Support you answer statistically. Because there are only two step heights, 15 cm has a statistically different mean change in heart rate than 3 cm. The F test in l) supports this conclusion. n) Estimate the stepping frequency effects. Estimated Effect Frequency= = Frequency= = Frequency= = Frequency= = o) Test the hypothesis that all the stepping frequency effects are zero against the alternative that at least one is not zero. Be sure to give an appropriate null and alternative hypothesis using the notation from your answer to h). Report the value of the appropriate test statistic, P-value, decision, reason for the decision and a conclusion within the context of the problem. H : all the β j = H A : some β j F = , P-value <.1 Because the P-value is so small (<.5) we reject the null hypothesis and conclude that some of the stepping frequencies affect the change in heart rate. p) Where are the statistically significant differences in stepping frequency sample means? Support you answer statistically. HSD = ( ) = 9. 39, so any differences in stepping frequency sample means greater than 9.39 are deemed statistically significant. Level Mean 35 A B C D Levels not connected with the same letter are significantly different. Therefore each stepping frequency has a sample mean change in heart rate that is different from all of the other stepping frequencies. 5

6 q) Are there any interaction effects that are different from zero? Be sure to give an appropriate null and alternative hypothesis using the notation from your answer to h). Report the value of the appropriate test statistic, P-value, decision, reason for the decision and a conclusion within the context of the problem. H : all the αβ ij = H A : some αβ ij F = , P-value <.16 Because the P-value is small (<.5) we reject the null hypothesis and conclude that there is statistically significant interaction between step height and stepping frequency. r) Construct an interaction plot. Comment on the plot and what it tells you about the interaction between the two factors. Be specific and be sure your answer deals with the context of this experiment. As stepping frequency increases, the average change in heart rate tends to increase. As height increases, the average change in heart rate tends to increase. However, the effect of height is greater when the stepping frequency is 35 steps per minute compared to 14 steps per minute. This is an indication of interaction. For 21 steps per minute and 28 steps per minute the effect of height is about the same. s) Based on the analysis of these data, what combination of step height and stepping frequency would you recommend if someone wanted to increase their heart rate the most? Using the 35 steps per minute frequency and 3 cm step height produced a mean increase in heart rate of beats per minute. The HSD for comparing treatment combination means is () = Because the mean for 35 steps per minute frequency and 3 cm step height is more than larger than the any of the other treatment combination means the differences between this treatment combination and all others are statistically significant. 6

7 Therefore I would recommend 35 steps per minute frequency and a 3 cm height in order for kinesiology majors similar to those who participated in the experiment to increase their heart rate the most after 5 minutes of stepping. t) Construct plots of residuals versus the levels of the two factors. Describe the plots and indicate what this tells you about the Fisher conditions necessary for the analysis of variance. Level n Mean Std Dev Level n Mean Std Dev The amount of variation within each step height is not too different. The ratio of the largest standard deviation to the smallest is less than 2. The amount of variation for each stepping frequency differs. The ratio of the largest standard deviation to the smallest is greater than 2. The equal standard deviation condition may not be met. 7

8 u) Construct plots of the distribution of residuals. Describe each of the plots in the distribution of residuals. Indicate what this tells you about the Fisher conditions necessary for the analysis of variance. Residual The histogram looks fairly symmetric (there may be a slight skew to the right). The box plot looks fairly symmetric (there may be a slight skew to the right, with the mean slightly bigger than the median). The points on the normal quantile plot follow the diagonal (normal model) line fairly well. It appears that the errors could be normally distributed. 8

9 Response: Change Turn in the JMP output that you have used to answer the questions. Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 48 Analysis of Variance Source DF Sum of Squares Mean Square F Ratio Model Error Prob > F C. Total <.1* Effect Tests Source DF Sum of Squares F Ratio Prob > F Frequency <.1* Height <.1* Frequency*Height * Effect Details Frequency Least Squares Means Table Level Least Sq Mean Std Error Mean

10 LSMeans Differences Tukey HSD α=.5 Q= HSD = (3.5327) = 9.39 LSMean[i] By LSMean[j] Mean[i]-Mean[j] Std Err Dif Lower CL Dif Upper CL Dif Level Least Sq Mean 35 A B C D Levels not connected by same letter are significantly different. Height Least Squares Means Table Level Least Sq Mean Std Error Mean

11 Frequency*Height Least Squares Means Table Level Least Sq Mean Std Error 14, , , , , , , , LS Means Plot Level Least Sq Mean 35,3 A ,3 B ,15 B C ,3 B C ,15 C D ,15 D E ,3 D E ,15 E 12. Levels not connected by same letter are significantly different. 11

12 LSMeans Differences Tukey HSD α=.5 Q= HSD = () = LSMean[i] By LSMean[j] Mean[i]-Mean[j] 14,15 14,3 21,15 21,3 28,15 28,3 35,15 35,3 Std Err Dif Lower CL Dif Upper CL Dif 14, , , , , , , , e e