STAT 425 HW6
Suggested Solution
1.
(a) We can first save the data into a local file and use R command to read the data
into R. We may use the following code and get the summary for the data set.
> football = read.table("Football.txt",header = TRUE);
> summary(football);
Trial
Air
Helium
Min.
: 1.0
Min.
:15.00
Min.
:11.00
1st Qu.:10.5
1st Qu.:23.50
1st Qu.:24.50
Median :20.0
Median :26.00
Median :28.00
Mean
:20.0
Mean
:25.92
Mean
:26.38
3rd Qu.:29.5
3rd Qu.:28.50
3rd Qu.:30.00
Max.
:39.0
Max.
:35.00
Max.
:39.00
(b) It is appropriate to use
paired t-test
here to perform our test.
By using the
following R command we can see the result for our test.
> t.test(football$Air-football$Helium);
One Sample t-test
data:
football$Air - football$Helium
t = -0.4198, df = 38, p-value = 0.677
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
-2.687423
1.764346
sample estimates:
mean of x
-0.4615385
An alternative way to perform paired t-test is to use the following command.
> t.test(football$Air, football$Helium, paired = TRUE);
Paired t-test
data:
football$Air and football$Helium
t = -0.4198, df = 38, p-value = 0.677
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.687423
1.764346
sample estimates:
mean of the differences
-0.4615385
1

From the summary of the test, since the p-value is 0.677 which is larger than 0.05,
we may conclude that the difference between these two balls are not significantly
different from 0. We can NOT reject the null hypothesis.
(c) By the randomization test introduced in the lecture notes, we can randomly mul-
tiply either 1 or -1 to each difference between two balls within one trail, and then
by doing so for lots of times we may get the approximated distribution for the
test statistics. By using the following code, we can implement the randomization
test in R.