The null and alternative hypothesis
Test statistic
where
X | (x-13.33)^2 | Y | (y-16.17)^2 | |||
12 | 1.7689 | 16 | 0.0289 | |||
16 | 7.1289 | 19 | 8.0089 | |||
14 | 0.4489 | 14 | 4.7089 | |||
15 | 2.7889 | 20 | 14.6689 | |||
11 | 5.4289 | 13 | 10.0489 | |||
12 | 1.7689 | 15 | 1.3689 | |||
sum/ss | 80 | 19.3334 | sum/ss | 97 | 38.8334 | |
mean =sum/24 | 13.33333 | mean =sum/15 | 16.16667 | |||
sx^2=ss/23 | 3.86668 | sy^2=ss/14 | 7.76668 | |||
sx | 1.9663876 | sy | 2.78687639 |
Therefore
= -2.03
df =6+6-2 =10
P value =0.0698
Since P value > 0.05 , the result is not significant
We fail to reject H0
There is not sufficient evidence to conclude that population means are different .
Note : P value using excel"=T.DIST.2T(2.03,10)"
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