SE of 2 Sample Means Difference for n1 = 6 & n2 = 5, σ1 = 1.8257 & σ2 = 3.562
Calculation summary & work with steps for n1 = 6 & n2 = 5 with population A 94, 95, 94, 95, 97, . . . . , 93 & 99 and population B 97, 86, 89, 91, 92, 95, 87 & 89 to estimate the standard error of difference between two sample means. The below is the calculation summary for SE of (x̄1 - x̄2) for sample size n1 = 6 and n2 = 5 with polulation standard deviations σ1 = 1.8257 and σ2 = 3.562.
| Calculation Summary | |
|---|---|
| Population A | 94, 95, 94, 95, 97, . . . . , 93 & 99 |
| Sample Size n1 | 6 |
| Population B | 97, 86, 89, 91, 92, 95, 87 & 89 |
| Sample Size n2 | 5 |
| population standard deviation (σ1) | 1.8257 |
| Population standard deviation (σ2) | 3.562 |
| SE of difference between 2 means (x̄1-x̄2) | 1.7587 |
SE of x̄1 - x̄2 Work with Steps for n1 = 6, n2 = 5, σ1 = 1.8257 & σ2 = 3.562
The below is the example work with step by step calculation shows how to estimate the standard error of difference between two means for sample size n1 = 6, sample size n2 = 5 with population dataset A 94, 95, 94, 95, 97, . . . . , 93 & 99 and dataset B 97, 86, 89, 91, 92, 95, 87 & 89 to help grade school students to solve the similar SE of (x̄1-x̄2 worksheet problems efficiently.
step 1 Address the input parameters and values
Population Dataset A = 94, 95, 94, 95, 97, . . . . , 93 & 99
Population Dataset B = 97, 86, 89, 91, 92, 95, 87 & 89
Sample size (n1) = 6
Sample size (n2) = 5
step 2 Find the mean for dataset A 94, 95, 94, 95, 97, 93, 95, 93 and 99
µ = n ∑ i = 1 Xin
=(94 + 95 + 94 + . . . . + 99)/9
Mean A (μA) = 95
Find the mean for dataset B 97, 86, 89, 91, 92, 95, 87 and 89
=(97 + 86 + 89 + . . . . + 89)/8
Mean B (μB) = 90.75
step 3 Apply the values of μA and dataset A in the below population standard deviation formula
Find the population standard deviation for 94, 95, 94, 95, 97, 93, 95, 93 & 99
=√{ (94 - 95)² + (95 - 95)² + (94 - 95)² + . . . . + (99 - 95)²}/9
= √(-1)² + (0)² + (-1)² + . . . . + (4)²/9
= √(1 + 0 + 1 + . . . . + 16)/9
= √30/9
= √3.3333
σ1 = 1.8257
Find the population standard deviation for 97, 86, 89, 91, 92, 95, 87 & 89
=√{ (97 - 90.75)² + (86 - 90.75)² + (89 - 90.75)² + . . . . + (89 - 90.75)²}/7
= √(6.25)² + (-4.75)² + (-1.75)² + . . . . + (-1.75)²/8
= √(39.0625 + 22.5625 + 3.0625 + . . . . + 3.0625)/8
= √101.5/8
= √12.6875
σ2 = 3.562
step 4 Substitute σ1, σ2, n1, n1 the values in below formula
SE(x̄1-x̄2) = √
= √
= √
= √0.5555 + 2.5376
= √3.0931
= 1.7587
1.7587 is the standard error of difference between two sample means.
