In: Statistics and Probability
A coach times 6 randomly selected runners to measure their times early in the week and at the end of the week (in minutes). Using the parts below, help the coach run a paired t-test to determine if the runners were faster (had lower times) at the end of the week. Use α = 0.05.
Runner | Monday time | Fridays time |
1 |
26.2 | 25.8 |
2 | 25.2 | 24.9 |
3 | 24.8 | 24.9 |
4 | 24.1 | 24.6 |
5 | 25.9 | 25.2 |
6 | 25.1 | 24.5 |
A. State the null and alternative hypotheses in terms of µD.
B. Find the critical value, test statistic, p-value.
C. What is the decision. Reject / fail to Reject
D. Based on the results, would you tell the coach that the runners were faster at the end of the week? yes/no
(A)
H0: Null Hypothesis: ( The runners were not faster (had lower times) at the end of the week)
HA: Alternative Hypothesis: ( The runners were faster (had lower times) at the end of the week) (claim)
(B)
(i)
= 0.05
ndf = n - 1 = 6 - 1 = 5
One Tail - Right Side Test
From Table, critical value of t = 2.0150
(ii)
From the given data, the following Table is calculated:
Runner | Monday time | Fridays time | d = Monday time - Fridays time |
1 |
26.2 | 25.8 | 0.4 |
2 | 25.2 | 24.9 | 0.3 |
3 | 24.8 | 24.9 | - 0.1 |
4 | 24.1 | 24.6 | - 0.5 |
5 | 25.9 | 25.2 | 0.7 |
6 | 25.1 | 24.5 | 0.6 |
From d values, the following Table is calculated:
n = 6
d | (d - ) | (d - )2 |
0.4 | 0.1667 | 0.0278 |
0.3 | 0.0667 | 0.0044 |
-0.1 | -0.3333 | 0.1111 |
-0.5 | 0.7333 | 0.5378 |
0.7 | 0.4667 | 0.2178 |
0.6 | 0.3667 | 0.1374 |
Total = | 1.0333 |
SE=sd/
= 0.4546/
= 0.1856
Test Statistic is given by:
t = 0.2333/0.1856
= 1.2572
So,
test statistic = 1.2572
(iii) By Technology, p - value = 0.1321
So,
p-value = 0.1321
(C)
Since p - value = 0.1321 is greater than = 0.05:
the decision. Fail to Reject
(D)
Correct option:
No