# t-Test Example for 2, 4, 6, 8, 9, 10, 13 & 15 & x̄ = 56

Student's two tailed t-test summary & example for test of significance (hypothesis) for mean of small samples from population with unknown variance by using t-statistic (t0) for dataset {2, 4, 6, 8, 9, 10, 13 & 15}, population mean μ = 56 and critical value (te) from t-distribution table at 10% significance level for the degrees of freedom ν = 7.
Calculation Summary
data set value2, 4, 6, 8, 9, 10, 13 & 15
population mean56
significance level (α)0.1
t030.802
te1.89

## Solved example for dataset {2, 4, 6, 8, 9, 10, 13 & 15} & x̄ = 56

The t-test work with steps for sample data set values {2, 4, 6, 8, 9, 10, 13 & 15}, population mean = 56 to conduct the test of significance for mean 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
data set value = 2, 4, 6, 8, 9, 10, 13 & 15
Significance level (α) = 0.1
population mean = 56

step 2 To find sample mean x̄ & variance s2

 x d = x-x̄ (x - 8.375) d2 2 -6.375 40.6406 4 -4.375 19.1406 6 -2.375 5.6406 8 -0.375 0.1406 9 0.625 0.3906 10 1.625 2.6406 13 4.625 21.3906 15 6.625 43.8906 ∑x = 67 ∑d = 0 ∑d2 = 133.8748

x̄ = xn

x̄ =678
x̄ = 8.375

s2 = 1/n - 1[ ∑d2 - {(∑d)2/n} ]
s2 = 1/8 - 1[ 133.8748 - (02/8)]
s2 = 1/7 x 133.8748
s2 = 19.125

step 3 Substitute sample mean, population mean, standard deviation, sample size value in the below formula for test of significance for sample mean.

d̄ = x̄ - μ

t0=s2/n

=8.375 - 5619.125/8

step 4 To find t0, simplify the above expression
=47.62519.125/8

=47.6252.3906

=47.6251.5462

t0 = 30.802

step 5 Find expected te with level of significance α = 0.1 , degrees of freedom ν = n - 1 = 7 from two tailed t-distribution table

t e = 1.89

Inference
t0 > te
The null hypothesis H0 is rejected since t0 = 30.802 is greater than the critical value for degrees of freedom te(6) = 1.89. Therefore, there is significant difference between the sample & population means.