Question

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) Determine whether the sample sizes are large enough for the Central Limit Theorem to apply. Samples of size 300 from population \(A\) with proportion 0.15 and samples of size 300 from population \(B\) with proportion 0.20

Short answer

The standard error for the difference in the two given sample proportions can be calculated by substituting the given values into the standard formula. The Central Limit Theorem applies in this case as both 'successes' and 'failures' exceed 5 in each group.

Step-by-step solution

Calculating Standard error

The standard error for the difference in two proportions is calculated using the formula: \(SE = \sqrt{\frac{{pA(1-pA)}}{nA} + \frac{{pB(1-pB)}}{nB}}\). Here \(pA=0.15, nA=300, pB=0.20, nB=300\). After substitution, this calculation becomes: \(SE = \sqrt{ \frac{{0.15(1-0.15)}}{300} + \frac{{0.20(1-0.20)}}{300} }\)

Evaluation of standard error

Now, we just have to calculate the expression inside the square root and then take the square root of our result to find the standard error. This will give us the final value for the standard error of the distribution of differences in sample proportions.

Check the applicability of the Central Limit Theorem

To determine if the Central Limit Theorem (CLT) applies, one should check that the expected number of 'successes' (np) and 'failures' (n(1-p)) both exceed 5 in each group. For population A, we have \(np_A = 300*0.15=45\) and \(n(1-p)_A = 300*0.85 = 255\). Similarly, for population B, we have \(np_B = 300*0.20 = 60\) and \(n(1-p)_B = 300*0.80 = 240\). In both cases, 'successes' and 'failures' are greater than 5, so the CLT applies.