In: Statistics and Probability
Given | |||
X1 bar | 2597.333333 (AVERAGE()) | X2 bar | 2491.167 |
S1 | 57.91603117(STADEV()) | S2 | 44.81257 |
n1 | 6 | n2 | 6 |
Let alpha = 0.05
Equal variance test:
F = S1^2/S2^2 = 1.6703
F critical
FL=0.14
FU=7.146
FL=0.14 < F=1.67 < FU=7.146 ,DO not reject equal variance of null hypothesis
T test:
Hypothesis : | α= | 0.05 | ||
df | 10 | n1+n2-2 | ||
Ho: | μ1 = μ2 | |||
Ha: | μ1 > μ2 | |||
t Critical Value : | ||||
tc | 1.812461123 | T.INV(1-α,df) | RIGHT | |
ts | >= | tc | RIGHT | To reject |
Test : | ||||
Sp^2 | 2681.216667 | ((n1-1)S1^2+(n2-1)S2^2)/(n1+n2-2) | ||
t stat | 3.551263144 | (X1 bar-X2 bar )/SQRT(Sp*(1/n1 + 1/n2)) | Equal vriance | |
P value : | ||||
P value | 0.002628586 | T.DIST.RT(ts,df) | RIGHT | |
Decision : | ||||
P value | < | α | Reject H0 |
P value < 0.05, Reject H0
There is enough evidence to conclude that the mean birth weight for newborns from mothers who complete the program is higher than the birth weight for newborns from mothers who do not complete the program.