(a) do a paired t-test. (b) verify your assumptions for the paired t-test (make
ID: 2933532 • Letter: #
Question
(a) do a paired t-test.
(b) verify your assumptions for the paired t-test (make sure you do this correctly).
(c) now do a regular t test; don't worry about verifying your assumptions but do assume equal variances!)
(d) finally, do a Wilcoxon signed rank test on the dat
Collagen shrinkage temperature (°C) Carcass Treated side Control side Difference 69.50 67.00 70.75 68.50 66.75 68.50 69.50 69,00 66.75 69.00 69.50 69.00 70.50 68.00 69.00 68.750 70.00 69.00 69.50 69.25 67.75 66.50 68.75 70.00 66.75 68.50 69.00 69.75 70.25 66.25 68.25 68.633 -0.50 -2.00 1.25 -0.75 2.00 0.75 0.00 0.50 0.50 -0.75 0.25 1.75 0.75 10 12 13 Mean SD 1.217 1.302 1.118Explanation / Answer
Here,I used R language for the calculation.
Null Hypothesis:True difference in means equal to Zero.
Alternative Hypothesis:True difference in means is not equal to Zero.
Steps:
1.Have to create two vectors
a-Treated side
b-control side
> a<-c(69.50,67,70.75,68.5,66.75,68.50,69.50,69,66.75,69,69.5,69,70.5,68,69)
> b<-c(70,69,69.5,69.25,67.75,66.5,68.75,70,66.75,68.5,69,69.75,70.25,66.25,68.25)
2. Apply paired t-test on the given data
> t.test(a,b,paired=T)
Output:
Paired t-test
data: a and b
t = 0.40434, df = 14, p-value = 0.6921
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.5021850 0.7355184
sample estimates:
mean of the differences
0.1166667
p-value >0.05 we may accept the null hypothesis.
b)Checking the assumptions
1.Are the two samples are paired?
Yes.
2.Is this a large sample?
No. Since, n < 30 we have check the normality.
plot gives approximately a straught line so,normality is checked.
3.There are no outliers in the data.
c)Regular T-test with equal variances
Null Hypothesis:There is no significance difference in the two means.
Alternative Hypothesis:There is a significance difference in the two means.
> t.test(a,b,var.eaual=T)
Output:
Welch Two Sample t-test
data: a and b
t = 0.25349, df = 27.875, p-value = 0.8018
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.8262934 1.0596267
sample estimates:
mean of x mean of y
68.75000 68.63333
P-value=0.8018
P-value >0.05
We may accept the null hypothesis.
d) > wilcox.test(a,b)
Wilcoxon rank sum test with continuity correction
data: a and b
W = 116.5, p-value = 0.8839
null hypothesis:True location shift equal to 0
alternative hypothesis: true location shift is not equal to 0
p-value>0.05,Hence we may accept the null hypothesis.
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