Question 1
a <- c(15,26,13,28,17,20,7,36,12,18)
b <- c(13,20,10,21,17,22,5,30,7,11)
if the mean of the two samples is the same, then the difference in the two means is zero. Null hypothesis.meanA-meanB=0. The alternate hypothesis is that the difference in means is not zero. meanA-meanB!=0
t.test(a,b,paired=TRUE)
##
## Paired t-test
##
## data: a and b
## t = 3.6742, df = 9, p-value = 0.005121
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 1.383548 5.816452
## sample estimates:
## mean of the differences
## 3.6
the p-value = 0.005121, this is much less than 0.05, so we can reject the null hypothesis. So, the alternate hypothesis of the two means not being equal is true.
t.test(a,b)
##
## Welch Two Sample t-test
##
## data: a and b
## t = 0.9802, df = 17.811, p-value = 0.3401
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -4.12199 11.32199
## sample estimates:
## mean of x mean of y
## 19.2 15.6
the p-value = 0.3401, this is much larger than 0.05, so we fail to reject the null hypothesis.
Question 2
ae <- c(9.50, 10.00, 9.75, 9.75, 9.00, 13.0)
ne <- c(11.50, 12.00, 13.25, 11.50, 13.00, 9.00)
if the mean of the two samples is the same, then the difference in the two means is zero. Null hypothesis.meanA-meanB=0. The alternate hypothesis is that the difference in means is not zero. meanae-meanne!=0
qqnorm(ae,main="Active Exercise")
qqnorm(ne,main="No Exercise")
After creating normal probability plots of both data sets, it was determined that there was not enough samples to confidently assume normality. For this reason, the non-parametric method could be used to analyze this data.
wilcox.test(ae,ne)
## Warning in wilcox.test.default(ae, ne): cannot compute exact p-value with ties
##
## Wilcoxon rank sum test with continuity correction
##
## data: ae and ne
## W = 9, p-value = 0.1705
## alternative hypothesis: true location shift is not equal to 0
The p-value = 0.1705, which is greater than 0.05, so we cannot reject the null hypothesis.
a <- c(15,26,13,28,17,20,7,36,12,18)
b <- c(13,20,10,21,17,22,5,30,7,11)
t.test(a,b,paired=TRUE)
t.test(a,b)
ae <- c(9.50, 10.00, 9.75, 9.75, 9.00, 13.0)
ne <- c(11.50, 12.00, 13.25, 11.50, 13.00, 9.00)
qqnorm(ae,main="Active Exercise")
qqnorm(ne,main="No Exercise")
wilcox.test(ae,ne)