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At university A, 46% of the undergraduate students are female. At university B, 52% of the undergraduate students are female. Suppose a random sample of 100 students from both schools was selected. Assume the sampling distribution of p hat Subscript Upper A Baseline minus p hat Subscript Upper B is approximately Normal.

What is the probability that the proportion of females selected from university A is greater than the proportion of females selected from university B?

If needed, use the z-table to answer the question.

Report your answer to 3 decimal places.

2 Answers

8 votes

Answer: .198

Explanation:

got it right on edg

User Stefan Holdermans
by
8.3k points
5 votes

The probability that the proportion of females selected from University A is greater than from University B is 0.726 to three decimal places.

To calculate the probability that the proportion of females selected from university A (p_hat Subscript Upper A) is greater than the proportion from university B (p_hat Subscript Upper B), we need to determine the difference in sample proportions and use the z-table to find the probability. First, we calculate the standard error of the difference in proportions:

SE = √[(p_A * (1 - p_A) / n_A) + (p_B * (1 - p_B) / n_B)]

Where p_A = 0.46, p_B = 0.52, and both n_A and n_B = 100. Substituting the values gives us:

SE = √[(0.46 * 0.54 / 100) + (0.52 * 0.48 / 100)] = √[(0.2484 / 100) + (0.2496 / 100)] = √[0.004984 + 0.004992] = √[0.009976] = 0.09988

Next, we calculate the z-score:

z = (p_hat Subscript Upper A - p_hat Subscript Upper B) / SE = (0.46 - 0.52) / 0.09988 = -0.06 / 0.09988 = -0.6008

Using the z-table, we look up the probability for z = -0.6008. The z-table gives us the probability that z is less than -0.6008, but we want the probability that z is greater. So, we subtract this probability from 1:

probability = 1 - z-table value for -0.6008

After looking up the value in the z-table, let's assume it is 0.2743. Thus, the probability we're interested in is:

probability = 1 - 0.2743 = 0.7257

To three decimal places, the probability is 0.726.

User CuriousMind
by
8.2k points
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