Question

In: Statistics and Probability

A coach times 6 randomly selected runners to measure their times early in the week and...

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

Solutions

Expert Solution

(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


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