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A better drug? In a pilot study, a company's new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood \((\mathrm{mg} / \mathrm{dl})\) greater than with the current drug, the company will begin the expensive process of mass- producing the new drug. For the 14 subjects who were assigned at random to the current drug. the mean cholesterol reduction was \(54.1 \mathrm{mg} / \mathrm{dl}\) with a standard deviation of \(11.93 \mathrm{mg} / \mathrm{dl}\). For the \(15 \mathrm{sub}\) jects who were randomly assigned to the new drug, the mean cholesterol reduction was \(68.7 \mathrm{mg} / \mathrm{dl}\) with a standard deviation of \(13.3 \mathrm{mg} / \mathrm{dl}\). Graphs of the data reveal no outliers or strong skewness. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not \(\left.H_{0}: \mu_{1}-\mu_{2}=0 .\right)\) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

Step by step solution

01

Define the Hypotheses

To solve this problem, we first define our null and alternative hypotheses. The null hypothesis, denoted as \( H_0 \), claims there is no difference or not a significant difference (greater than 10 mg/dl) in mean cholesterol reduction. Therefore, \( H_0: \mu_2 - \mu_1 = 10 \) mg/dl. The alternative hypothesis \( H_a \) claims that the new drug significantly reduces cholesterol by more than 10 mg/dl compared to the old. So, \( H_a: \mu_2 - \mu_1 > 10 \) mg/dl. Here, \( \mu_1 \) and \( \mu_2 \) represent the mean cholesterol reductions for the current and new drugs, respectively.

02

Calculate the Test Statistic

To test the hypothesis, we use the formula for the test statistic for two independent sample means: \[ t = \frac{(\bar{x}_2 - \bar{x}_1) - D}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] \( \bar{x}_1 = 54.1 \), \( \bar{x}_2 = 68.7 \), \( D = 10 \), \( s_1 = 11.93 \), \( s_2 = 13.3 \), \( n_1 = 14 \), \( n_2 = 15 \). Substituting these values gives: \[ t = \frac{(68.7 - 54.1) - 10}{\sqrt{\frac{11.93^2}{14} + \frac{13.3^2}{15}}} \] Calculating, we find the test statistic \( t \approx 2.62 \).

03

Determine the Critical Value or P-value

Using a t-distribution table or calculator for degrees of freedom calculated using the approximation formula, we determine the critical value or P-value. With degrees of freedom \( \approx 27 \), if we assume a significance level \( \alpha = 0.05 \), the critical value for a one-tailed test is around 1.703. Since \( t = 2.62 \) exceeds 1.703, we find our result is significant.

04

Make a Conclusion

Since the calculated \( t \)-value (2.62) exceeds the critical \( t \)-value (1.703), we reject the null hypothesis. We conclude that there is significant evidence that the new drug reduces mean cholesterol by more than 10 mg/dl compared to the current drug, supporting the company's case to consider mass production of the new drug.

05

Analyze Type I and Type II Errors

Based on our conclusion of rejecting the null hypothesis, there is a possibility of making a Type I error, which occurs if we wrongly reject a true null hypothesis. Since we have sufficient evidence to reject \( H_0 \) based on the data, a Type II error is not possible in this context (failing to reject a false \( H_0 \)). Hence, if our conclusion was incorrect, it would imply a Type I error.

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Key Concepts

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

Significance Test

A significance test is essentially a tool that helps us decide if the observed data can be explained by random chance or if there is evidence of a genuine effect. In our example, we're assessing whether the new drug has a greater impact on cholesterol reduction than the current one by more than 10 mg/dl.

The procedure begins by stating two competing hypotheses. The null hypothesis (\(H_0\)) suggests no real difference, specifically, that the difference in mean reduction is exactly 10 mg/dl (\(\mu_2 - \mu_1 = 10\)). The alternative hypothesis (\(H_a\)) posits a mean reduction exceeding 10 mg/dl (\(\mu_2 - \mu_1 > 10\)).

Using a t-test allows us to compare these averages knowing both the sample mean and standard deviation for each drug group. We calculate a test statistic based on sample data. This statistic indicates how far the observed data are from the null hypothesis expectation, measured in units of standard error.

• If this statistic falls within a critical region, or gives a P-value lower than our threshold (\(\alpha\)), we have grounds to reject \(H_0\).
• On the other hand, if our test statistic does not meet these thresholds, we would "fail to reject" \(H_0\), meaning our data don't provide sufficient proof of a significant effect.

Type I Error

A Type I error occurs when we incorrectly reject a true null hypothesis. It represents a false positive result, suggesting an effect that isn't there.

Imagine this in the context of the drug test. If we decided the new drug was significantly better than the current one (rejecting \(H_0\)) when, in fact, there was no such difference, we would commit a Type I error. Our confidence in saying "Yes, the drug works better than thought" would be misplaced.

The probability of making a Type I error is denoted by \(\alpha\), the significance level, which researchers set before exploring. Common levels include 0.05 or 0.01, which represent a 5% or 1% risk of incorrectly rejecting the true \(H_0\), respectively.

• In practice, controlling the significance level is crucial since too high a risk means researchers may make unreliable conclusions.

Type II Error

A Type II error happens when we fail to reject a null hypothesis that is actually false. It's essentially a missed opportunity to capture the true effect.

Returning to our scenario, a Type II error would mean concluding that the new drug isn't significantly better when it is. In essence, our inability to prove its better performance equates to continued usage of a possibly less effective treatment.

Type II errors are denoted by \(\beta\), and their probability depends on several factors such as sample size, effect size, and the chosen significance level. Generally, researchers seek to minimize this error by designing studies with enough power—a concept tied directly to sample size and the magnitude of effect they're looking for.

• By adequately powering a study, researchers reduce the chances of overlooking a genuine effect, boosting the credibility of their conclusive findings.

Statistical Conclusion

A statistical conclusion is essentially our logical endpoint where we decide whether our data provide enough evidence to support or refute a hypothesis. It stems from the results of our significance tests.

In the drug scenario, after calculating the test statistic and comparing it to the critical value, we rejected the null hypothesis. Thus, our statistical conclusion would be that the new drug results in a cholesterol reduction that significantly exceeds 10 mg/dl more than the current treatment.

However, it is crucial to understand that while statistical conclusions provide strong evidence-based signals, they always carry inherent uncertainty. Researchers must weigh the risks of Type I and II errors when making decisions and interpreting results.

• Significant results can prompt further research, lead to medical advancements, or guide policy changes. But responsible interpretation is key, ensuring claims are made confidently but cautiously.