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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 \(1 \mathrm{mi}\) 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.

Step by step solution

01

Analyzing whether p has a normal distribution

Using the rule of thumb, for approximation of a binomial distribution to a normal distribution, both \( np \) and \( n(1-p) \) should be greater than 5.\n\nHere, \( n = 10 \) and \( p = 0.3 \)\n\nSo, \( np = (10)(0.3) = 3 \) and \( n(1 - p) = (10)(1 - 0.3) = 7 \)\n\nSince \( np \) is not greater than 5, p based on a random sample of only 10 residents will not have approximately a normal distribution.

02

Calculating the mean and standard deviation

For a sample size \( n = 400 \), the mean and standard deviation of the proportion \( p \) are calculated as follows:\n\nThe mean \( µ_p = π \) and here \( π = 0.3\n\)\nSo, \( \mu_p = 0.3 \)\n\nThe standard deviation \( σ_p \) can be calculated as \(\sigma_p = \sqrt{(π(1 − π))/ n}\)\nHere, \( n = 400\n\)\nSo, \( \sigma_p = \sqrt{(0.3×(1-0.3))/400} \)

03

Calculating the probability

p is approximately normally distributed with mean 0.3 and standard deviation calculated in the previous step. To find the probability that p is between 0.25 and 0.35, we first standardize these values using the Z-score formula, which is \(Z = (X – μ) / σ\).\n\nZ1 = (0.25 - 0.3) / σ_p\nZ2 = (0.35 - 0.3) / σ_p\n\nThen find the area underneath the standard normal curve between Z1 and Z2 to get the probability P(.25 ≤ p ≤ .35).

04

Comparing the probabilities for different sample sizes

As the sample size \( n \) increases, the standard deviation decreases and the normal distribution becomes narrower. Therefore, for a fixed range of p, the probability decrease when the sample size increases from 400 to 500. Hence, the probability calculated in Part (c) is larger than it would be if n=500.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal distribution

The normal distribution is a continuous probability distribution that is symmetric around the mean. Most values cluster around the mean, creating a bell-shaped curve when graphed. In statistics, this distribution has key properties such as the mean, median, and mode being equal, and it is completely described by its mean and standard deviation.

For small sample sizes, like the one in our original problem with only 10 people, a binomial distribution cannot be approximated to a normal distribution. This is because the condition of having both parameters \( np \) and \( n(1-p) \) greater than 5 is not satisfied. Here, \( np = 3 \), which is less than 5, leading to the conclusion that this distribution does not adequately approximate to a normal distribution in this scenario.

Binomial distribution

A binomial distribution is a discrete probability distribution describing the number of successes in a fixed number of independent Bernoulli trials. Each trial has exactly two potential outcomes: success or failure.

In the context of our exercise, the binomial distribution accounts for the proportion of residents living close to a hazardous site, where \( \pi = 0.3 \). If you take a random sample of ten individuals, each with a 0.3 probability of living near such a site, your outcome could be modelled using a binomial distribution. However, due to the small sample size, using a normal distribution as an approximation is not ideal. As sample sizes increase, such as to 400 or 500, a binomial distribution more closely approximates a normal distribution, allowing for the use of normal probability calculations.

Standard deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In probability and statistics, the standard deviation provides an idea of how much individual elements deviate from the mean.

In the exercise, when dealing with a sample size \( n = 400 \), the standard deviation of the proportion \( p \) can be determined using the formula:\[ \sigma_p = \sqrt{\frac{\pi(1-\pi)}{n}} \]For \( \pi = 0.3 \), the calculation becomes: \( \sigma_p = \sqrt{\frac{0.3 \times 0.7}{400}} \). This computation reveals how spread out the proportion of people living within a mile of a hazardous waste site is expected to be for that larger sample size.

As the sample size increases, standard deviation decreases, implying more precise estimates of the proportion.

Probability calculation

Probability calculation involves determining the likelihood of a specific outcome or range of outcomes happening by using statistical methods.

In the step-by-step solution, the probability that the proportion \( p \) is between 0.25 and 0.35 was calculated assuming a normal distribution for a large sample size \( n = 400 \). This is done by standardizing these values using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Next, you find the probabilistic result by finding the area under the normal curve between these standardized values.

An important insight given by the exercise is that with larger samples (e.g., \( n = 500 \)), the curve becomes narrower, reducing the probability for a fixed range around the mean, thus, affecting the probability calculations accordingly.