# t-Test for x̄1 = 25, x̄2 = 18, n1 = 16, n2 = 20 & α = 0.2 Solved example work with steps, formula & 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 = 0.2 for difference between two small sample means1 = 25 & x̄2 = 18, s1 = 7 & s2 = 11 for sample size n1 = 16 & n2= 20. The below is the calculation summary for the test of significance or hypothesis using student's t-test for difference between two small sample means x̄1 = 25 & x̄2 = 18.

Calculation Summary
sample mean (x̄1)25
sample mean (x̄2)18
variance (s1²) 7
variance (s2²) 11
sample size (n1)16
sample size (n2)20
Significance Level (α)0.2
t06.6787
te1.31

## Work with Steps for x̄1 = 25, x̄2 = 18 at α = 20%

Solved example work with steps to find t-statistic (t0), critical (table) value (te) for the test of significance at α = 0.2 using student's two tailed t-test for degrees of freedom ν = 34, sample means x̄1 = 25 & x̄2 = 18, s1 = 7 & s2 = 11, sample size n1 = 16 & n2= 20 in statistical experiment may helpful for beginners or grade school students to learn how to solve t-test worksheet or homework problems efficiently by using t-statistic & critical value of t from t-distribution table.

Workout :
step 1 Address the formula, input parameters and values
Sample mean (x̄1)= 25
Sample mean (x̄2)= 18
Sample variance (s1²) = 7
Sample variance (s2²) = 11
Sample size (n1) = 16
Sample size (n2) = 20
Level of significance (α) = 0.2

t0 =
1 - x̄2
S2 x {(1 / n1) + (1 / n2)}
step 2 To find S2

S2 =n1s1² + n2s2²n1 + n2 - 2

S2 =(16 x 7) + (20 x 11)16 + 20 - 2

S2 =33234

= 9.7647

step 3 Substitute x̄1, x̄2, variance 1,variance 2, n1, n2 value in the below formula for test of significance for difference between two means.

=25 - 18
9.7647 x {(1 / 16) + (1 / 20)}

step 4 To find t0, simplify the above expression
=7
9.7647 x {(1 / 16) + (1 / 20)}

=7
9.7647 x 0.1125

=7
1.0985

=71.0481

t0= 6.6787

step 4 Find expected te using α , degrees of freedom from t-distribution table
d.f = n1 + n2 - 2
d.f = 16 + 20 - 2
d.f = 34
The critical value of te from the t-distribution table for the degrees of freedom (n1 + n2 - 2) at 0.2 significance level.
The critical value te(34) from t-distribution table at 0.2 of significance level is 1.31
t e = 1.31

Inference
t0 > te
The null hypothesis H0 is rejected since t0 = 6.6787 is greater than the critical value for degrees of freedom te(33) = 1.31. Therefore, there is significant difference between the sample & population means. 