Here is the R code for creating the observed data:
x <- c(159, 190, 204, 206, 222, 223)
y <- c(370, 376, 418, 488, 490, 503, 512, 532, 587, 605, 637)
We can use the function
var.test() to test whether the two treatments
have the same underlying population variance.
var.test(x,y) # P-value = 0.0099; 95% CI for var ratio = (0.017, 0.48)
I would conclude that the variability of response to the two is different.
To obtain a 95% confidence interval for the ratio of the two underlying proportions, we take the square-root of the 95% CI for the ratio of the population variances.
result <- var.test(x,y)
sqrt( result$conf.int ) # 95% CI = (0.13, 0.70)
You might be interested in the ratio of the SD under treatment B to the SD under treatment A (i.e., the reciprocal of that considered above) instead:
result <- var.test(y,x)
sqrt( result$conf.int ) # 95% CI = (1.4, 7.6)
We first download the data file and read it into R. (The file is comma-delimited.)
mydata <- read.csv("data_hw07-1.csv")
The resulting object, mydata has
two columns, "diet" and "gain".
Unfortunately, the "diet" column is not made a
factor, and so the function aov() for
performing the analysis of variance will not work correctly.
Thus we need to do the following.
is.factor(mydata$diet) # Darn! It's not a factor
mydata$diet <- as.factor(mydata$diet)
is.factor(mydata$diet) # Now it is.
To get the ANOVA table and the p-value for the test of whether the average weight gain is the same for the three diets, we do the following.
out <- aov(gain ~ diet, data = mydata) # perform the ANOVA
summary(out) # get table and p-value
The ANOVA table we obtain is as follows:
| Source | SS | df | MS |
| Between | 36 | 2 | 18 |
| Within | 210 | 9 | 23.3 |
| Total | 246 | 11 | |
Because the MSbetween is less than the MSwithin, we're clearly not going to reject the null hypothesis. We get an F statistic of 0.77 and a P-value of 0.49.
Since the P-value is ~50%, we fail to reject the null hypothesis, and conclude that the data are insufficient to conclude that the average weight gains on these three diets are different.
Download and read in the data using something like the following:
mydata <- read.csv("data_hw07-2.csv")
We can calculate the sample means, sample SDs and sample sizes for each group using the following:
tapply(mydata$Length, mydata$Group, mean)
tapply(mydata$Length, mydata$Group, sd)
tapply(mydata$Length, mydata$Group, length)
We can make a dotplot of the data using the following:
stripchart(mydata$Length ~ mydata$Group, method="jitter")
To get the ANOVA table and the p-value for the test of whether the average weight gain is the same for the three diets, we do the following.
out <- aov(Length~Group, data=mydata) # perform the ANOVA
summary(out) # get table and p-value
The ANOVA table we obtain is as follows:
| Source | SS | df | MS |
| Between | 871.4 | 4 | 217.9 |
| Within | 3588.5 | 60 | 59.8 |
| Total | 4459.9 | 64 | |
We get an F statistic of 3.64 and a P-value of 0.01.
Since the P-value is quite small, we conclude that there are differences in the average lengths of daffodils in the different areas.
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