Two different brand of latex (water-based) paint are being considered for use in
ID: 3265712 • Letter: T
Question
Two different brand of latex (water-based) paint are being considered for use in a large construction project. To choose between the brands, one of the key factors is the time it takes for the paint to dry. Engineers sampled 24 specimens of each brand and measured the drying times (in hours) for each specimen. The data collected is given below. (a) Treat these sample as independent samples, one taken from the Brand A population and one taken from the Brand B population. The goal is to learn how the population mean drying times for the two brands compare. Perform a thorough analysis that addresses this question. A thorough analysis will contain a complete description of the statistical assumptions as well as checking these assumptions. showing all calculations (carried out "by hand" or preferably by R) (if helpful/needed) some well-constructed, informative graphs which are relevant to the problem at hand a well-written paragraph that summarizes the entire analysis (which should include the final main conclusions). (b) If you performed a matched-pairs analysis in part (a), then you did the analysis incorrectly because a matched-pairs analysis assumes that the samples are dependent. However, would it be possible to learn about the p pairs design? If so, explain how you could use a matched-pairs experiment to accomplish this: make sure you explicitly state who you are accomplishing the "matching." If you don't think it is possible, then explain why.Explanation / Answer
a)
> Brand_A=c(4.6,4.9,4.5,6.4,5.9,4.8,4.2,6.1,5.4,4.6,5.9,4.8,4.3,3.8,4.2,
+ 4.9,6.4,6.7,4.3,4.4,3.8,6.1,5.3,4.3)
> Brand_B=c(3.8,4.4,3.7,4.2,4.1,4.5,4,4.5,4,4,3.4,3.5,4,3.5,4.5,3.6,4.3,
+ 4.3,4.3,4.6,4.7,4.9,4.1,3.5)
>
> ## Descriptive test
> summary(Brand_A)
Min. 1st Qu. Median Mean 3rd Qu. Max.
3.800 4.300 4.800 5.025 5.900 6.700
> summary(Brand_B)
Min. 1st Qu. Median Mean 3rd Qu. Max.
3.400 3.775 4.100 4.100 4.425 4.900
# Normality test
> shapiro.test(Brand_A)
Shapiro-Wilk normality test
data: Brand_A
W = 0.91641, p-value = 0.04867
> shapiro.test(Brand_B)
Shapiro-Wilk normality test
data: Brand_B
W = 0.96103, p-value = 0.4595
Conclusion: The Brand A violates the assumption of normality at 0.05 level of significance. Whereas, Brand B accepts the assumption of normality at 0.05 level of significance.
# T test
> t.test(Brand_A,Brand_B)
Welch Two Sample t-test
data: Brand_A and Brand_B
t = 4.6541, df = 33.001, p-value = 5.093e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.5206393 1.3293607
sample estimates:
mean of x mean of y
5.025 4.100
Conclusion: The estimated p-value of the test is 5.093e-05. Hence, we can conclude that there is a significant difference in mean between the two brand at 0.05 level of significance.
b) Paire T test
> t.test(Brand_A, Brand_B, paired=TRUE)
Paired t-test
data: Brand_A and Brand_B
t = 5.0493, df = 23, p-value = 4.128e-05
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.5460358 1.3039642
sample estimates:
mean of the differences
0.925
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