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.