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 means x̄1 = 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.
|sample mean (x̄1)||25|
|sample mean (x̄2)||18|
|sample size (n1)||16|
|sample size (n2)||20|
|Significance Level (α)||0.2|
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.
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
x̄1 - x̄2
S2 =n1s1² + n2s2²n1 + n2 - 2
S2 =(16 x 7) + (20 x 11)16 + 20 - 2
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
step 4 To find t0, simplify the above expression
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
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.