In: Statistics and Probability
= 195.6 ms
Sd = 230.53 ms
Np = 10
= 195.6 ms
Sd = 230.53 ms
Np = 10
= 195.6 ms
Sd = 230.53 ms
Np = 10
sample size , n = 10
Degree of freedom, DF= n - 1 =
9 and α = 0.05
t-critical value = t α/2,df =
2.2622 [excel function: =t.inv.2t(α/2,df) ]
std dev of difference , Sd =
230.5300
std error , SE = Sd / √n = 230.5300 /
√ 10 = 72.9000
margin of error, E = t*SE = 2.2622
* 72.9000 = 164.9112
mean of difference , D̅ =
195.600
confidence interval is
Interval Lower Limit= D̅ - E = 195.600
- 164.9112 = 30.6888
Interval Upper Limit= D̅ + E = 195.600
+ 164.9112 = 360.5112
so, confidence interval is ( 31
< µd < 361 )
---------------------------
Ho : µd= 0
Ha : µd ╪ 0
Level of Significance , α =
0.05
sample size , n = 10
mean of sample 1, x̅1= 0.000
mean of sample 2, x̅2= 0.000
mean of difference , D̅ =
195.6000
std dev of difference , Sd =
230.5300
std error , SE = Sd / √n = 230.5300 /
√ 10 = 72.9000
t-statistic = (D̅ - µd)/SE = ( 195.6
- 0 ) / 72.9000
= 2.683
Degree of freedom, DF= n - 1 =
9
p-value = 0.0251 [excel
function: =t.dist.2t(t-stat,df) ]
Decision: p-value <α , Reject null
hypothesis
Hence alchohol effect reaction time.
Please let me know in case of any doubt.
Thanks in advance!
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