To test whether alcohol has an effect on reaction time, 10 subjects were given reaction-tests on two different days. Each subject drank a glass of liquid containing alcohol before one of the two tests and a glass of liquid not containing alcohol before the other one. The order of presentation was randomised independently for each subject. The reaction times in 1/10 second are given in the table below.
Subject |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
With Alcohol |
4.7 |
5.1 |
5.8 |
4.8 |
6.0 |
4.4 |
5.5 |
6.1 |
5.6 |
4.9 |
Without Alcohol |
3.9 |
4.0 |
4.4 |
3.0 |
5.1 |
4.1 |
5.5 |
6.8 |
4.7 |
4.6 |
(a) Carry out the appropriate test at 5% level of significance and check whether there is significant evidence that alcohol has an effect on reaction time.
(b) Give the p-value of the test. Do you come to the same conclusion by using p-value? Explain. (DETAIL ANSWER PLEASE!)
a)
Difference Scores
Calculations
Treatment 1
N1: 10
df1 = N - 1 = 10 - 1 = 9
M1: 5.29
SS1: 3.13
s21 =
SS1/(N - 1) = 3.13/(10-1) = 0.35
Treatment 2
N2: 10
df2 = N - 1 = 10 - 1 = 9
M2: 4.61
SS2: 9.61
s22 =
SS2/(N - 1) = 9.61/(10-1) = 1.07
T-value
Calculation
s2p =
((df1/(df1 +
df2)) * s21) +
((df2/(df2 +
df2)) * s22) =
((9/18) * 0.35) + ((9/18) * 1.07) = 0.71
s2M1 =
s2p/N1
= 0.71/10 = 0.07
s2M2 =
s2p/N2
= 0.71/10 = 0.07
t = (M1 -
M2)/√(s2M1
+ s2M2) =
0.68/√0.14 = 1.81
The t-value is 1.80751. The p-value is .087426.
b) p > 0.05, null is not rejected. So, no effect
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