Question

Suppose that two groups are being compared, one with a sample size, n = 10, and the other with a sample size, n = 15. The standard deviation of the measurements within the first group is 1.5, and the standard deviation of the measurements within the second group is 2.5. The sample mean within the first group is 10 and the sample mean within the second group is 15. Find the standard error of (round your answer to three decimal places).

Answer 1

The Estimated standard error = 0.8010

Explanation:

Step(I):-

Given first sample size n₁ = 10

Given second sample size n₂ = 15

Mean of the first sample x₁⁻ = 10

Mean of the second sample x₂⁻ = 15

The standard deviation of the first sample 'S₁'= 1.5

The standard deviation of the second sample 'S₂'= 2.5

step(ii):-

The standard error of two groups is determined by

`S.E( X^(-) _(1) - X^(-) _(2) ) = \sqrt{(S^(2) _(1) )/(n_(1) ) +(S^(2) _(2) )/(n_(2) ) }`

`S.E( X^(-) _(1) - X^(-) _(2) ) = \sqrt{((1.5)^(2) )/(10) +((2.5)^(2) )/(15) }`

`S.E( X^(-) _(1) - X^(-) _(2) ) = √(0.64166) = 0.8010`

Final answer:-

The Estimated standard error = 0.8010