knitr::opts_chunk$set(echo = TRUE)

IE 5342 Flipped Assignment 9

Meghan Cephus

09/29/2024

Problem 1

  1. Suppose we wish to design a new experiment that tests for a significant difference between the mean effective life of four insulating fluids at an accelerated load of 35kV.  The variance of fluid life is estimated to be 3.5 hrs based on preliminary data.  We would like this test to have a type 1 error probability of 0.05, and for this test to have an 80% probability of rejecting the assumption that the mean life of all the fluids are the same if there is a difference greater than 2 hours between the mean lives of the fluids, with a min of 18 hrs and max of 20 hrs.  How many samples of each fluid will need to be collected to achieve this design criterion in the case of Min, Intermediate, and Max variability?

The variability determines the the effect size. For each variability the effect size, f, will change.

Min Variability

f = \[ d\sqrt{\frac{1}{2k}}\]

sigma <- 3.5
range <- 2
d <- range/sigma
library(pwr)
pwr.anova.test(k=4,n=NULL,f=d*sqrt(1/(2*4)),sig.level=0.05,power=0.8)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 67.76303
##               f = 0.2020305
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group

Intermediate Variability

f = \[ \frac{d}{2}\sqrt{\frac{k+1}{3(k-1)}}\]

pwr.anova.test(k=4,n=NULL,f=(d/2)*sqrt(5/(3*3)),sig.level=0.05,power=0.8)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 61.08609
##               f = 0.2129589
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group

Max Variability

f = even\[ \frac{d}{2}\]

pwr.anova.test(k=4,n=NULL,f=d/2,sig.level=0.05,power=0.8)
## 
##      Balanced one-way analysis of variance power calculation 
## 
##               k = 4
##               n = 34.38178
##               f = 0.2857143
##       sig.level = 0.05
##           power = 0.8
## 
## NOTE: n is number in each group

Conclusion

I believe there might be an error in my calculations since as the variability increases the sample size decreases. The sample sizes are also quite large. I suspect there is a problem with the effect size, f, calculation. Either with the formula or the calculation of d.

Problem 2

The experimenter decided to collect six observations for each type of fluid, resulting in the following test data.

a. Test the hypothesis that the life of fluids is the same against the alternative that they differ at an α=0.10 level of significance (Remember to enter the data in a tidy format when using R, or to pivot_longer to a tidy format using tidyr)

F1 <- c(17.6,18.9,16.3,17.4,20.1,21.6)
F2 <- c(16.9,15.3,18.6,17.1,19.5,20.3)
F3 <- c(21.4,23.6,19.4,18.5,20.5,22.3)
F4 <- c(19.3,21.1,16.9,17.5,18.3,19.8)

dat <- data.frame(F1,F2,F3,F4)

library(tidyr)

dat <- pivot_longer(dat,c(F1,F2,F3,F4))

aov <- aov(value~name, data=dat)
summary(aov)
##             Df Sum Sq Mean Sq F value Pr(>F)  
## name         3  30.17   10.05   3.047 0.0525 .
## Residuals   20  65.99    3.30                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
qf(0.9, 3, 20)
## [1] 2.380087

Conclusion:

F = 3.047 and F critical = 2.38 for α=0.10. F > F critical 3.047 > 2.38. We CAN reject the Null Hypothesis. We can conclude that the fluid type DOES make a difference in life hours.

b. Is the model adequate? (show plots and comment)

plot(aov)

Conclusion:

The residuals vs. fitted graph shows the variance is reasonably consistent. The Q-Q residuals show that the model is reasonably normal.

c. Assuming the null hypothesis in question 1 is rejected, which fluids significantly differ using a family-wise error rate of α=0.10 (use Tukey’s test).  Include the plot of confidence intervals.

TukeyHSD(aov,conf.level = 0.90)
##   Tukey multiple comparisons of means
##     90% family-wise confidence level
## 
## Fit: aov(formula = value ~ name, data = dat)
## 
## $name
##             diff        lwr       upr     p adj
## F2-F1 -0.7000000 -3.2670196 1.8670196 0.9080815
## F3-F1  2.3000000 -0.2670196 4.8670196 0.1593262
## F4-F1  0.1666667 -2.4003529 2.7336862 0.9985213
## F3-F2  3.0000000  0.4329804 5.5670196 0.0440578
## F4-F2  0.8666667 -1.7003529 3.4336862 0.8413288
## F4-F3 -2.1333333 -4.7003529 0.4336862 0.2090635
plot(TukeyHSD(aov,conf.level = 0.90))

Conclusion:

Only the means of Fluid 2 and Fluid 3 significantly differed. This is because their range of differences does not include the number 0 unlike the rest of the comparisons. This is also seen in the graph where the graph of Fluid 2 vs Fluid 3 is the only one that does not cross the 0 line.