Question

To compare three brands of computer keyboards, four data entry specialists were randomly selected. Each specialist used all three keyboards to enter the same kind of text material for 10 minutes, and the number of words entered per minute was recorded. The data obtained are given in the following table:

Keyboard Brand

Data Entry Specialist A B C

1 77 67 63

2 71 62 59

3 74 63 59

4 67 57 54

a. Test the null hypothesis H0 that no differences exist between the effects of the keyboard brands on the mean number of words entered per minute. Set α = 0.025

b. Test the null hypothesis H0 that no differences exist between the effects of the data entry specialists on the mean number of words entered per minute. Set α= 0.025. c. Use Tukey simultaneous 95 percent confidence intervals to make pairwise comparisons of the keyboard brand effects on the mean number of words entered per minute. Which keyboard brand maximizes the mean number of words entered per minute?

Answer 1

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ANSWER:

we can do this quickly in the open source statisitcal package R

The R snippet is as follows

A<- c(77,71,74,67)
B<- c(67,62,63,57)
C<- c(63,59,59,54)
  
df <- data.frame(A,B,C)

library(reshape2)

df <- melt(df)

library(ggpubr)

# Global test
compare_means(value ~ variable, data = df, method = "anova")

# Change method to anova
ggboxplot(df, x = "variable", y = "value",
color = "variable", palette = "dark2")+
stat_compare_means(method = "anova")

# perform anova analysis
a<- aov(lm(value~ variable,data=df))

TukeyHSD(a)
plot(TukeyHSD(a))

The result is

> compare_means(value ~ variable, data = df, method = "anova")
# A tibble: 1 x 6
.y. p p.adj p.format p.signif method
<chr> <dbl> <dbl> <chr> <chr> <chr>
1 value 0.002825428 0.002825428 0.0028 ** Anova

The p value of the anova is 0.0028 , which is less than 0.05 hence we can say that there is a difference between atleast 2 groups

The tukeyhsd is as follows

> TukeyHSD(a)
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = lm(value ~ variable, data = df))

$variable
diff lwr upr p adj
B-A -10.0 -17.95844 -2.04156 0.0164748 ## not signficant
C-A -13.5 -21.45844 -5.54156 0.0027409 ## signficant difference
C-B -3.5 -11.45844 4.45844 0.4674238 ## not signficant

only p values that are less than 0.05 are considered signficant

also , as evident from the boxplot above , brand A of keyboard has the highest median for average typing speed , hence brand A maximises the mean number of words entered per minute

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