Question

Use the formula to find the standard error of the distribution of differences in sample means, .

Samples of size 105 from Population 1 with mean 88 and standard deviation 12 and samples of size 70 from Population 2 with mean 78 and standard deviation 16
Round your answer for the standard error to two decimal places.

Answer 1

`SE= \sqrt{(\sigma^2_1)/(n_1)+(\sigma^2_2)/(n_2)}`

And replacing we got:

`SE= \sqrt{(12^2_1)/(105)+(16^2_2)/(70)}= 2.24`

Explanation:

We have the following info given:

`n_1= 105, \mu_1 = 88, \sigma_1 =12`

`n_2= 70, \mu_2 = 78, \sigma_2 =16`

And for this case we want to find the standard error for the following variable
`\bar X_1 -\bar X_2` and the standard error for this distribution is given by:

`SE= \sqrt{(\sigma^2_1)/(n_1)+(\sigma^2_2)/(n_2)}`

And replacing we got:

`SE= \sqrt{(12^2_1)/(105)+(16^2_2)/(70)}= 2.24`