Question

A geology teacher takes her class on an annual field trip to analyze rock samples. They use randomized sets of

coordinates to select samples from two regions.
Suppose they examine a sample of 70 rocks from region A, which is composed of 15% igneous rock. Then they
examine a sample of 60 rocks from region B, which is composed of 20% igneous rock. Then they look at the
difference between the sample proportions (PA - B).
What are the mean and standard deviation of the sampling distribution of pA - PB?

Answer 1

The mean of the sampling distribution of pA - pB is -0.05 and the standard deviation is approximately 0.0670.

Step-by-step explanation:

The question asks us to find both the mean and the standard deviation of the sampling distribution of the difference between two sample proportions, pA - pB, where pA is the sample proportion of igneous rocks in region A and pB is the sample proportion in region B. The mean of this sampling distribution, assuming independence between the two samples, is simply the difference between the two population proportions, which in this case is 0.15 - 0.20 = -0.05. The standard deviation of the sampling distribution can be found using the formula:

√[(pA * (1 - pA) / nA) + (pB * (1 - pB) / nB)]

where pA = 0.15, nA = 70, pB = 0.20, and nB = 60. Substituting these values into the formula and calculating gives:

√[(0.15 * 0.85 / 70) + (0.20 * 0.80 / 60)] = √[(0.1275 / 70) + (0.1600 / 60)] = √[0.0018214 + 0.0026667] = √[0.0044881] ≈ 0.0670

Answer 2

Second option would be the answer