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 100 from Population 1 with mean 87 and standard deviation 12 and samples of size 80 from Population 2 with mean 81 and standard deviation 15

Short answer

The standard error of the distribution of differences in sample means is approximately 2.06

Step-by-step solution

Identify Values for Calculations

Identify the given values from the problem: the sizes of the samples (n_1 = 100, n_2 = 80), means of the samples (\(\mu_1 = 87, \mu_2 = 81\)) and standard deviations of the samples (\(s_1 = 12, s_2 = 15\)).

Substitute Values into Standard Error Formula

Insert the identified values into the standard error formula. It translates into: \(\sqrt{{\frac{12^2}{100}} + {\frac{15^2}{80}}}\)

Calculation

Carry out the operations inside the square root first, then the square root operation. This translates into: \(\sqrt{1.44+2.8125} = \sqrt{4.2525}\). The value of the square root of 4.2525 is approximately 2.06 when rounded to two decimal places.

Interpret the result

The computed standard error, approximately 2.06, is the standard deviation of the sampling distribution of the difference in means. It shows the randomness of the difference in means and helps understand the closeness of the sample mean to the population mean.