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A study reported in the American Journal of Public Health (Science News) - the first to follow lead levels in blood for law-abiding handgun hobbyists using indoor firing ranges- -documents a considerable risk of lead poisoning. \(^{\star}\) Lead exposure measurements were made on 17 members of a law enforcement trainee class before, during, and after a 3 -month period of firearm instruction at a state-owned indoor firing range. No trainees had elevated lead levels in their blood before training, but 15 of the 17 ended training with blood lead levels deemed "elevated" by the Occupational Safety and Health Administration (OSHA). Is there sufficient evidence to claim that indoor firing range use increases blood-level readings? a. Give the associated \(p\) -value. b. What would you conclude at the \(\alpha=.01\) significance level? c. Use the normal approximation to give the approximate \(p\) -value. Does the normal approximation appear to be adequate when \(n=17\) ?

Step by step solution

01

Formulating Hypotheses

We start by formulating the null and alternative hypotheses. Let \( p \) represent the proportion of trainees with elevated blood lead levels after training. The null hypothesis \( H_0 \) states that indoor firing range use does not increase blood lead levels significantly, meaning \( p = 0.5 \). The alternative hypothesis \( H_a \) is that it does increase blood lead levels, or \( p > 0.5 \).

02

Sample Proportion Calculation

Calculate the sample proportion \( \hat{p} \) using the data. Out of the 17 trainees, 15 had elevated levels, so \( \hat{p} = \frac{15}{17} \). This approximates to \( \hat{p} \approx 0.882 \).

03

Test Statistic Calculation

We use the normal approximation for the binomial distribution. The test statistic \( z \) is computed as follows: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Substituting \( \hat{p} = 0.882 \), \( p_0 = 0.5 \), and \( n = 17 \), we get \[ z = \frac{0.882 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{17}}} \approx 3.57 \].

04

Calculating the p-value

For a one-tailed test, we find the p-value by looking up the standard normal \( z \)-table. The p-value for \( z = 3.57 \) is very small, typically less than 0.0002.

05

Conclusion at \(\alpha = 0.01\)

Since the p-value is much smaller than \( \alpha = 0.01 \), we reject the null hypothesis \( H_0 \). This indicates strong evidence that using an indoor firing range increases blood lead levels.

06

Evaluate Normal Approximation Adequacy

The normal approximation requires that both \( np \) and \( n(1-p) \) be greater than 5. Here \( np = 17 \times 0.5 = 8.5 \) and \( n(1-p) = 17 \times 0.5 = 8.5 \), which satisfy this condition. Thus, the normal approximation is adequate for this sample size.

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Key Concepts

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

Normal Approximation

When working with binomial distributions, calculating probabilities directly can be computationally heavy, especially with larger sample sizes or probabilities not directly tabulated. The normal approximation simplifies this by using the standard normal curve to approximate the binomial distribution.

This works particularly well for large sample sizes where the binomial distribution begins to resemble the normal distribution. To employ this approximation, we generally need two conditions to be satisfied:

• Both the product of the sample size and probability of success, (\( np \)), and the product of the sample size and probability of failure, (\( n(1-p) \)), should be greater than 5.

This ensures sufficient data points in both outcomes to resemble a normal distribution.

For the firing range study, we calculated \( np = 8.5 \) and \( n(1-p) = 8.5 \). Both values being greater than 5 confirm this approach is justified, demonstrating the normal approximation’s adequacy for calculating the \( p\)-value.

P-Value Calculation

The \( p\)-value plays a crucial role in hypothesis testing, allowing us to understand how likely the observed data would appear under the null hypothesis.
To compute the \( p\)-value for this hypothesis test, we first transform our sample result into a test statistic, using the normal approximation.

• For a one-tailed test, we need to know how extreme the test statistic is. We compute the standardized \( z \)-score using the formula:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]
• Once calculated, the test statistic \( z \) value (e.g., \( z=3.57 \)) is used to find the corresponding \( p\)-value from a standard normal \( z \)-table.

A very small \( p\)-value, as found in our exercise (\(<0.0002)\), suggests the observed sample proportion is unlikely under the null hypothesis, indicating a significant effect.

Binomial Distribution

Understanding the binomial distribution is central when evaluating cases like the lead exposure study, where outcomes are simply classified as success or failure.
The binomial distribution applies when:

• Each trial in the experiment is independent.
• The probability of success is the same for each trial.
• We are looking at a fixed number of trials.

Here, success refers to having elevated blood lead levels after firing range training.
This distribution helps us to calculate the probability of a given number of successes in a number of trials, in this case, trainees developing elevated lead levels.
As a discrete probability distribution, each possible outcome (number of successes) has a probability, making it invaluable in hypothesis testing to determine significant changes.