In: Statistics and Probability
Patient |
Placebo |
Valproate |
1 |
37 |
5 |
2 |
52 |
22 |
3 |
63 |
41 |
4 |
2 |
4 |
5 |
25 |
32 |
6 |
29 |
20 |
7 |
15 |
10 |
8 |
52 |
25 |
9 |
19 |
17 |
10 |
12 |
14 |
The number of seizures during the placebo period come from a normal distribution with mean µ1 and variance .
The number of seizures during the Valproate period come from a normal distribution with mean µ2 and variance .
Perform a hypothesis test at the 5% significance level to determine if the mean number of seizures is lower with Valproate than with the placebo. Give hypotheses, test statistic, P-value, conclusion, and interpretation.
paired t test
SAMPLE 1 | SAMPLE 2 | difference , Di =sample1-sample2 | (Di - Dbar)² |
37 | 5 | 32 | 416.160 |
52 | 22 | 30 | 338.560 |
63 | 41 | 22 | 108.160 |
2 | 4 | -2 | 184.960 |
25 | 32 | -7 | 345.960 |
29 | 20 | 9 | 6.760 |
15 | 10 | 5 | 43.560 |
52 | 25 | 27 | 237.160 |
19 | 17 | 2 | 92.160 |
12 | 14 | -2 | 184.960 |
Ho : µd= 0
Ha : µd > 0
mean of difference , D̅ =ΣDi / n =
11.600
std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) =
14.751
std error , SE = Sd / √n = 14.7513 /
√ 10 = 4.6648
t-statistic = (D̅ - µd)/SE = ( 11.6
- 0 ) / 4.6648
= 2.4867
Degree of freedom, DF= n - 1 =
9
t-critical value , t* =
1.8331 [excel function: =t.inv(α,df) ]
p-value =
0.0173 [excel function: =t.dist.rt(t-stat,df)
]
Decision: p-value <α , Reject null
hypothesis
there is enough evidence to conlcude that the mean number of seizures is lower with Valproate than with the placebo.