Question

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 300 from Population 1 with mean 75 and standard deviation 18 and samples of size 500 from Population 2 with mean 83 and standard deviation 22

Short answer

The standard error of the distribution of differences in sample means is approximately 3.59

Step-by-step solution

Identify the values

Identify the values from the problem: \(\sigma_{1} = 18\), \(n_{1} = 300\), \(\sigma_{2} = 22\), \(n_{2} = 500\)

Plug into the equation

Plug these values into the equation for the standard error of the distribution of differences in sample means: \(\sqrt{(18^{2}/300 + 22^{2}/500)}\)

Calculate

Preliminary calculation results in \(\sqrt{(3.24 + 9.68)}\)

Compute the square root

Finally, compute the square root: \(\sqrt{12.92}\)