t-Test for x̄1 = 156.75, x̄2 = 167, n1 = 4, n2 = 5 & α = 10%
Solved example work with steps, calculation summary to estimate the t-statistic (t0), critical (table) value (te) for degrees of freedom ν & hypothesis test (H0) at a stated level of significance α = 0.1 for difference between two small samples x = {150, 160, 152 & 165} & y = {173, 154, 164, 179 & 165}. The below is the calculation summary for the test of significance or hypothesis using student's t-test for difference between two small samples with means x̄ = 156.75 & ȳ = 167.
| Calculation Summary | |
|---|---|
| Dataset (x) | {150, 160, 152 & 165} |
| Dataset (y) | {173, 154, 164, 179 & 165} |
| Significance level (α) | 0.1 or 10% |
| t-statistic (t0) | 1.7923 |
| Critical value (te) | 1.89 |
Work with Steps for x̄ = 156.75, ȳ = 167 at α = 0.1
The t-test work with steps to find t-statistic (t0), t-critical (te) and hypothesis (H0) for dataset x = {150, 160, 152 & 165} & y = {173, 154, 164, 179 & 165} to estimate the test of significance for difference between two small sample means.
Workout :
step 1 Address the formula, input parameters and values
Dataset (x) = 150, 160, 152 & 165
Dataset (y) = 173, 154, 164, 179 & 165
Level of significance (α) = 0.1
t0 =
x̄ - ȳ
√
step 2 To find x̄, ȳ, & s2
x̄ =∑xn1
x̄ =627/4 = 156.75
ȳ =∑yn2
ȳ =835/5 = 167
s2 = 1/n1 + n2 - 2[ ∑(x - x̄)2 + ∑(y - ȳ)2 ] s2 = 1/4 + 5 - 2[ 146.75 + 362 ] s2 = 1/7 x 508.75 s2 = 72.6786
step 3 Substitute x̄ , ȳ, s2, n1 & n2 value in the formula for test significance for difference between two means.
=156.75 - 167
√
step 4 To find t0, simplify the above expression
=10.25
√
=10.25
√
=10.25
√
=10.255.7189
t0= 1.7923
step 5 Find expected te using α , degrees of freedom from t-distribution table
d.f = n1 + n2 - 2
d.f = 4 + 5 - 2
d.f = 7
The critical value te from the two tailed t-distribution table for the degrees of freedom ν = 7 at 0.1 significance level.
te = 1.89
Inference
t0 < te
The null hypothesis H0 is accepted since t0 = 1.7923 is smaller than the critical value for degrees of freedom te(6) = 1.89. Therefore, there is no significance difference between the sample & population means.
step 1 Address the formula, input parameters and values
Dataset (x) = 150, 160, 152 & 165
Dataset (y) = 173, 154, 164, 179 & 165
Level of significance (α) = 0.1
t0 =
x̄ - ȳ
√
S2 x {(1 / n1) + (1 / n2)}
step 2 To find x̄, ȳ, & s2
| x | x-x̄ (x-156.75) | (x-x̄)2 |
| 150 | -6.75 | 45.5625 |
| 160 | 3.25 | 10.5625 |
| 152 | -4.75 | 22.5625 |
| 165 | 8.25 | 68.0625 |
| ∑x = 627 | x-x̄ = 0 | (x-x̄)2 = 146.75 |
| y | y-ȳ (y-167) | (y-ȳ)2 |
| 173 | 6 | 36 |
| 154 | -13 | 169 |
| 164 | -3 | 9 |
| 179 | 12 | 144 |
| 165 | -2 | 4 |
| ∑y = 835 | y-ȳ = 0 | (y-ȳ)2 = 362 |
x̄ =∑xn1
x̄ =627/4 = 156.75
ȳ =∑yn2
ȳ =835/5 = 167
s2 = 1/n1 + n2 - 2[ ∑(x - x̄)2 + ∑(y - ȳ)2 ] s2 = 1/4 + 5 - 2[ 146.75 + 362 ] s2 = 1/7 x 508.75 s2 = 72.6786
step 3 Substitute x̄ , ȳ, s2, n1 & n2 value in the formula for test significance for difference between two means.
=156.75 - 167
√
72.6786 x {(1 / 4) + (1 / 5)}
step 4 To find t0, simplify the above expression
=10.25
√
72.6786 x {(1 / 4) + (1 / 5)}
=10.25
√
72.6786 x 0.45
=10.25
√
32.7054
=10.255.7189
t0= 1.7923
step 5 Find expected te using α , degrees of freedom from t-distribution table
d.f = n1 + n2 - 2
d.f = 4 + 5 - 2
d.f = 7
The critical value te from the two tailed t-distribution table for the degrees of freedom ν = 7 at 0.1 significance level.
te = 1.89
Inference
t0 < te
The null hypothesis H0 is accepted since t0 = 1.7923 is smaller than the critical value for degrees of freedom te(6) = 1.89. Therefore, there is no significance difference between the sample & population means.
