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 value | 2, 4, 6, 8, 9, 10, 13 & 15 |
| population mean | 56 |
| significance level (α) | 0.1 |
| t0 | 30.802 |
| te | 1.89 |
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
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=d̄√s2/n
=8.375 - 56√19.125/8
step 4 To find t0, simplify the above expression
=47.625√19.125/8
=47.625√2.3906
=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.
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=d̄√s2/n
=8.375 - 56√19.125/8
step 4 To find t0, simplify the above expression
=47.625√19.125/8
=47.625√2.3906
=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.
