Question

General Electric recently conducted a study to evaluate filaments in their industrial high intensity bulbs. Investigators recorded the number of weeks each high-intensity bulb would last before failure for three test filaments (Groups 1, 2, and 3) and the standard filament (Group 4). The results are as follows. Using ? = 0.01,

Group       1          2       3        4

                 15       14     25     28

                 18       18     19     31

                 21       20     22     27

                 16       16     20     32

                 17       15     18     23

                 20       16     24     25

                18         22     27     30

                              14    18     27     

                                      24     25

                                               26

1. Write an appropriate ANOVA hypothesis to test the difference in means of the four groups (null and alternative).
2. Read the data into R or R-studio, run an ANOVA model in R and paste the code used as well as the output here. What is the decision based on the ANOVA test? You need to explain what part of the output led you to the conclusion you made.
3. Continue using R: Use the Tukey method to test all pairwise contrasts. Show the R code, output, and explain the results of all the comparisons in complete sentences while referencing the parts/numbers on the output that support your conclusions.

Answer 1

we have given 4 groups and ask to do a ANOVA test of them, so

H0: means of all groups are equal

vs

H1: not H0

we get R output as below

Analysis of Variance Table

Response: bulbs
Df Sum Sq Mean Sq F value Pr(>F)   
group 3 613.48 204.493 24.835 2.836e-08 ***
Residuals 30 247.02 8.234   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

here p-value < 0.05 we can say that the group effect is significantly effective and we are rejecting H0

since we get the group means are different we search for which are different by Tukey HSD

the R output is

Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = bulbs ~ group)

$`group`
diff lwr upr p adj
2-1 -0.9821429 -5.02031317 3.056027 0.9107047
3-1 4.0317460 0.09966253 7.963830 0.0428315
4-1 9.5428571 5.69774991 13.387964 0.0000010
3-2 5.0138889 1.22256203 8.805216 0.0059628
4-2 10.5250000 6.82395578 14.226044 0.0000001
4-3 5.5111111 1.92611606 9.096106 0.0012635

here p-value for 1 and 2 difference is >0.05 so we can say that group 1 and 2 are same mean where as all other combinations are significantly effective..

R-code for this problem is

data=read.csv("Book1.csv")
bulbs=data$bulbs
group=as.factor(data$group)
anova(lm(bulbs~group))
TukeyHSD(aov(bulbs~group))