Reaction times were measured on five volunteers before and after taking a night time cold medicine. The hypothesis being tested is whether there is a change in reaction time after taking the cold medicine. A 5% level of significance is used.
Before:
After:
Difference, d: 2, 2, 1, 5, 1
p*p: 4, 17, 13, 19, 15
4, 1, 11, 12, 12
17, 25, 1, 11, 12
SUM:
35
To conduct a Paired t-test, the following steps are taken:
1. Null hypothesis (H0): There is no change in reaction time after taking the cold medicine.
Alternative hypothesis (Ha): Reaction time changed after taking the cold medicine.
2. Calculate the mean difference (d̄):
d̄ = (Σd) / n
d̄ = (2 + 2 + 1 + 5 + 1) / 5
d̄ = 11 / 5
d̄ = 2.2
3. Calculate the standard deviation of the differences (s):
s = √((Σ(d - d̄)^2) / (n - 1))
s = √((4 + 1 + 11 + 12 + 12) / (5 - 1))
s = √(40 / 4)
s = √10
s ≈ 3.16
4. Calculate the standard error of the mean difference (SE):
SE = s / √n
SE = 3.16 / √5
SE ≈ 1.41
5. Calculate the t-value:
t = (d̄ - μ) / (SE / √n)
μ = 0 (since the null hypothesis assumes no change)
t = (2.2 - 0) / (1.41 / √5)
t ≈ 2.48
6. Determine the critical value at a 5% level of significance (α = 0.05) with 4 degrees of freedom (n - 1):
The critical value can be obtained from a t-distribution table or a statistical software.
7. Compare the calculated t-value with the critical value:
If the calculated t-value is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
8. Conclusion:
Based on the calculated t-value and the critical value, we either reject or fail to reject the null hypothesis. The conclusion should be stated in terms of the original problem and the significance level.
9. Interpretation:
The interpretation of the conclusion should explain the implications of the results in the context of the problem.