Sample 1 |
Sample 2 |
|
Mean |
22.7 |
20.5 |
Variance (s^2) |
5.4 |
3.6 |
Observations (sample size) |
9 |
9 |
This year |
Last Year |
Difference (d) |
|
PDC 1 |
5.1 |
4.1 |
1 |
PDC 2 |
3.9 |
2.9 |
1 |
PDC 3 |
4.8 |
2.8 |
2 |
PDC 4 |
3.4 |
3.4 |
0 |
PDC 5 |
4.6 |
2.6 |
2 |
PDC 6 |
7.7 |
4.7 |
3 |
S.D. of d= 1.048 |
|||
Mean of d=-1.5 |
Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F Statistic |
Between Groups |
210.2788 |
|||
Within Groups |
1483 |
74.15 |
||
Total |
2113.833 |
Level of Significance , α = 0.05
mean of sample 1, x̅1= 22.7000
variance,s² = 5.4
size of sample 1, n1= 9
mean of sample 2, x̅2= 20.5
variance of sample 2 ,s^2 = 3.6
size of sample 2, n2= 9
difference in sample means = x̅1-x̅2 =
2.2000
Degree of freedom, DF= n1+n2-2 =
16
t-critical value = t α/2 =
2.1199 (excel formula =t.inv(α/2,df)
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = 2.1213
std error , SE = Sp*√(1/n1+1/n2) =
1.0000
margin of error, E = t*SE =
2.1199
difference of means = x̅1-x̅2 =
2.2000
confidence interval is
Interval Lower Limit= (x̅1-x̅2) - E =
0.08
Interval Upper Limit= (x̅1-x̅2) + E =
4.32
so, option a) is answer
-------------------------------------------------------------------------------------------------------
sample size , n = 6
mean of sample 1, x̅1= 4.9167
mean of sample 2, x̅2= 3.4167
mean of difference , D̅ = 1.5000
std dev of difference , Sd =
1.0488
std error , SE = Sd / √n = 0.4282
t-statistic = (D̅ - µd)/SE =
3.503
so, answer is option d)
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