# 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
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.