Given | |||
X1 bar | 4.7 | X2 bar | 6.8 |
S1 | 2 | S2 | 2.5 |
n1 | 13 | n2 | 10 |
b)
Hypothesis : | |
Ho: | μ1 = μ2 |
Ha: | μ1 not = μ2 |
i)
Test : | |||
Sp^2 | 4.964285714 | ((n1-1)S1^2+(n2-1)S2^2)/(n1+n2-2) | |
t | -2.240777156 | (X1 bar-X2 bar )/SQRT(Sp^2*(1/n1 + 1/n2)) | Equal vriance |
ii)
t Critical Value : | ||||
tc | 2.079613845 | T.INV.2T(alpha,df) | TWO | |
Rejction region: | ||||
ts | < for - | tc | TWO | To reject |
ts | > for + | tc | TWO | To reject |
iii)
Decision : | |||
If | |||
P value | < | α = 0.05 | Reject H0 |
P value | > | α = 0.05 | Do not reject |
iv)
P value : | |||
P | 0.035978961 | T.DIST.2T(ts,df) | TWO |
P value = 0.036
P value < 0.05, rejcet H0
v)
There is enough evidence to conclude that population means are different at 5% significance level
c)
95% CI
α= | 0.05 | |
df | 21 | n1+n2-2 |
CI | Equal variance | |
tc | 2.079613845 | T.INV.2T(alpha,df) |
Upper | -0.151038203 | (X1 bar-X2 bar )+tc*Sp*SQRT(1/n1 + 1/n2) |
Lower | -4.048961797 | (X1 bar-X2 bar )-tc*Sp*SQRT(1/n1 + 1/n2) |
CI = (-4.0490, -0.1510)
The confidence intarval does not include null hypothesis value (0), so, we reject H0 and there is enough evidence to conclude that claim is true
1. Consider the hypothesis test Ho:ui= u2 against Hi: 417 42. Suppose that sample sizes are...
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