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With domestic sources of building supplies running low several years ago, roughly 60,000 homes were built with imported Chinese drywall. According to the article "Report Links Chinese Drywall to Home Problems" (New York Times, Nov. 24, 2009), federal investigators identified a strong association between chemicals in the drywall and electrical problems, and there is also strong evidence of respiratory difficulties due to the emission of hydrogen sulfide gas. An extensive examination of 51 homes found that 41 had such problems. Suppose these 51 were randomly sampled from the population of all homes having Chinese drywall. a. Does the data provide strong evidence for concluding that more than \(50 \%\) of all homes with Chinese drywall have electrical/environmental problems? Carry out a test of hypotheses using \(\alpha=.01\). b. Calculate a lower confidence bound using a confidence level of \(99 \%\) for the percentage of all such homes that have electrical/environmental problems. c. If it is actually the case that \(80 \%\) of all such homes have problems, how likely is it that the test of (a) would not conclude that more than \(50 \%\) do?

Step by step solution

01

Formulate Hypotheses for Part (a)

We need to test whether more than 50% of homes with Chinese drywall have electrical/environmental problems. Our null hypothesis is that the true proportion of such homes, \( p \), is 0.5, and the alternative hypothesis is that \( p > 0.5 \). Thus, we have:\[H_0: p = 0.5\]\[H_a: p > 0.5\]We will use a significance level of \( \alpha = 0.01 \).

02

Calculate Test Statistic for Part (a)

We will use the z-test for proportions. Calculate the sample proportion \( \hat{p} \) as follows: \( \hat{p} = \frac{41}{51} \approx 0.8039 \). The test statistic \( z \) is given by:\[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]Substituting the values, we have:\[z = \frac{0.8039 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{51}}} \approx \frac{0.3039}{0.070}\approx 4.34\]

03

Make a Decision for Part (a)

Compare the calculated z-value to the critical value from the standard normal distribution at \( \alpha = 0.01 \). The critical z-value is 2.33. Since 4.34 > 2.33, we reject the null hypothesis. This provides strong evidence that more than 50% of homes have problems.

04

Calculate Lower Confidence Bound for Part (b)

We want a 99% lower confidence bound for \( p \). For a 99% confidence level, the z-value is 2.33 (one-tailed). The formula for the confidence interval lower bound is:\[\hat{p} - z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Substitute \( \hat{p} = 0.8039 \), \( n = 51 \), and \( z = 2.33 \):\[0.8039 - 2.33 \times \sqrt{\frac{0.8039 \times 0.1961}{51}} \approx 0.8039 - 0.086 \approx 0.7179\]Thus, we are 99% confident that at least 71.79% of the homes have problems.

05

Calculate Probability Not to Conclude (c)

If the actual proportion is 0.8, we want to find the probability of failing to reject \( H_0: p = 0.5 \) for \( \alpha = 0.01 \). Using the z-score formula, the z-value when \( p = 0.8 \) is:\[z = \frac{0.8 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{51}}} \approx \frac{0.3}{0.070} \approx 4.29\]We need to find the probability of observing such a high z-value under \( H_0 \), which is practically zero (probability of a Type II error is very small). Therefore, the likelihood of not concluding more than 50% is virtually 0.

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Key Concepts

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

Z-test for proportions

In hypothesis testing, when we want to compare a sample proportion to a known population proportion, we use the z-test for proportions. This is particularly useful when analyzing binary data, where observations fall into two categories. For example, in our drywall problem, we need to decide if more than 50% of homes are affected, labeling them as "problematic" or "not problematic."

The formula for the test statistic in a z-test is:

• Calculate the sample proportion \( \hat{p} \) which is the number of successes divided by the total sample size \( n \).
• The test statistic \( z \) formula is given by: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \] Here, \( p_0 \) is the population proportion under the null hypothesis.

Once we have the z-value, we can compare it to critical values from the standard normal distribution to decide if we should reject the null hypothesis.

Confidence interval calculation

A confidence interval gives us a range of possible values for a population parameter. When we calculate a confidence interval for a proportion, we get a span of values within which the true population proportion is likely to fall.

For calculating the confidence interval, the sample proportion \( \hat{p} \) is central. The formula for a lower confidence bound on a proportion, particularly when we are interested in a one-tailed interval, is:

• The formula is given by: \[ \hat{p} - z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \( z \) is the z-value corresponding to the desired confidence level, and \( n \) is the sample size.
• This calculation gives us the minimum plausible proportion, meaning we are sure that the actual proportion is above this value to a specific degree of confidence.

In the drywall example, we calculated a 99% lower bound. This means we can be very confident that at least this percentage of homes have problems.

Null and alternative hypotheses

Setting up the null and alternative hypotheses is the first critical step in hypothesis testing. These hypotheses help us determine what we're testing and what conclusions we can draw from our statistical analysis.

The null hypothesis \( H_0 \) usually represents the status quo or a statement of no effect or difference. In our drywall scenario:

• You state \( H_0: p = 0.5 \), suggesting that exactly 50% of the homes have issues.

The alternative hypothesis \( H_a \) is what you want to prove:

• We set \( H_a: p > 0.5 \), meaning we believe more than 50% of the homes face problems.

By using the test statistic, we assess these hypotheses to decide whether the alternative hypothesis provides enough evidence to reject the null hypothesis.

Significance level alpha

The significance level, denoted as \( \alpha \), is a crucial threshold in hypothesis testing that determines the risk of concluding that an effect exists when it doesn't—a Type I error.

It represents the probability of rejecting the null hypothesis when it's actually true. Common choices for \( \alpha \) are 0.05 or 0.01.

• This value helps decide the critical z-value you compare against the test statistic.
• If the computed test statistic surpasses the critical z-value, the result is statistically significant.

For our exercise, an \( \alpha \) of 0.01 was selected, offering strict criteria for evidence, thus making rejecting the null hypothesis more challenging. This level of significance ensures greater confidence in the conclusion that more than 50% of homes have the problems indicated.