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Chapter 8: Problem 28

The article "Should Pregnant Women Move? Linking Risks for Birth Defects with Proximity to Toxic Waste Sites" (Chance [1992]: 40-45) reported that in a large study carried out in the state of New York, approximately \(30 \%\) of the study subjects lived within 1 mi of a hazardous waste site. Let \(\pi\) denote the proportion of all New York residents who live within 1 mi of such a site, and suppose that \(\pi=.3\). a. Would \(p\) based on a random sample of only 10 residents have approximately a normal distribution? Explain why or why not. b. What are the mean value and standard deviation of \(p\) based on a random sample of size \(400 ?\) c. When \(n=400\), what is \(P(.25 \leq p \leq .35)\) ? d. Is the probability calculated in Part (c) larger or smaller than would be the case if \(n=500 ?\) Answer without actually calculating this probability.

Short Answer

Expert verified
a. No, it will not follow an approximately normal distribution. b. Mean \(\mu=0.3\) and Standard Deviation would be calculated using the formula. c. The probability \(P(.25 \leq p \leq .35)\) will be the area under the normal curve between the calculated Z-scores. d. The probability is smaller when n=500.

Step by step solution

01

Check for Normal Distribution

The sample proportion \(p\) will follow an approximately normal distribution if both \(n \pi\) and \(n (1-\pi)\) are greater than or equal to 10 where \(\pi\) is the true proportion and \(n\) is the sample size. In this case, \(\pi = 0.3\) and sample size \(n = 10\). Now calculate \(n \pi = 10 * 0.3 = 3\) and \(n (1-\pi) = 10 * (1-0.3) = 7\). Both values are less than 10, so the sample proportion will not approximately follow a normal distribution.
02

Calculate Mean and Standard Deviation

The mean and standard deviation of \(p\) satisfying \(n \pi\) or \(n (1-\pi)\) > 10 are given by, Mean: \(\mu =\pi\) and Standard Deviation: \(\sigma =\sqrt{\frac{\pi * (1 - \pi)}{n}}\). Inputting the given values, mean is \(\mu = 0.3\) and the standard deviation is \(\sigma = \sqrt{\frac{0.3 * (1 - 0.3)}{400}}\).
03

Calculate Probability Range

In order to calculate the probability \(P(.25 \leq p \leq .35)\), we need to standardize \(p\) by subtracting the mean and dividing by the standard deviation to compute the Z-scores, correspondingly. Z_score_1 = \((0.25 - \mu) / \sigma\) and Z_score_2 = \((0.35 - \mu) / \sigma\). The required probability is the area under the standard normal curve between these two Z scores.
04

Determine Probability for Different Sample Sizes

If we increase the sample size from 400 to 500, the standard deviation decreases, because the standard deviation of \(p\) is inversely proportional to the square root of the size of the sample (i.e., \(\sigma = \sqrt{\pi * (1 - \pi) / n}\)). As a result, the interval between 0.25 and 0.35 around the mean represents a wider range, therefore, the probability for this range \(P(.25 ≤ p ≤ .35)\) would be smaller when \(n=500\) compared to \(n=400\).

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Most popular questions from this chapter

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