91Ó°ÊÓ

The level of monoamine oxidase (MAO) activity in blood platelets (mm/mg protein/h) was determined for each individual in a sample of 43 chronic schizophrenics, resulting in \(\bar{x}-2.69\) and \(s_{i}=2.30\), as well as for 45 normal subjects, resulting in \(\bar{y}-6.35\) and \(x_{2}=4.03\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha=.01\). [Hint: \(H_{0}\) and \(H_{a}\) here have a different form from the three standard cases. Let \(\mu_{1}\) and \(\mu_{2}\) refer to true averagse MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta-2 \mu_{1}-\mu_{2}\). Write \(H_{0}\) and \(H_{a}\) in terms of \(\theta\), estimate \(\theta\), and derive \(\hat{\omega}_{d}\) ("Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients" Nature, July 28, 1972: 225-226).

Step by step solution

01

Understand the Problem

We want to determine if the true average monoamine oxidase (MAO) activity for normal subjects is more than twice that of chronic schizophrenics. This involves a hypothesis test between two sample means.

02

Define the Hypotheses

Let \(\mu_1\) be the true average MAO activity for schizophrenics and \(\mu_2\) be for normal subjects. Define \(\theta = 2\mu_1 - \mu_2\). We wish to test \(H_0: \theta = 0\) against \(H_a: \theta < 0\), reflecting that \(2\mu_1 < \mu_2\).

03

Calculate the Estimate for \(\theta\)

From the sample data, we have \(\bar{x} = 2.69\) for schizophrenics and \(\bar{y} = 6.35\) for normals. The estimator for \(\theta\) is \(\hat{\theta} = 2\bar{x} - \bar{y}\). Substitute the values to get:\[\hat{\theta} = 2(2.69) - 6.35 = 5.38 - 6.35 = -0.97\]

04

Calculate the Standard Error \(\hat{\omega}_d\)

We compute the standard error \(\hat{\omega}_d\). With sample sizes \(n_1 = 43\) (schizophrenics) and \(n_2 = 45\) (normals), and standard deviations \(s_1 = 2.30\) and \(s_2 = 4.03\), the variance is estimated by:\[S^2_d = 4s_1^2/n_1 + s_2^2/n_2 = \frac{4(2.30)^2}{43} + \frac{4.03^2}{45}\]Calculate numerically: \[S^2_d = \frac{21.16}{43} + \frac{16.2409}{45}\approx 0.492 + 0.361\approx 0.853\]Then \(\hat{\omega}_d = \sqrt{S^2_d} \approx \sqrt{0.853} \approx 0.923\).

05

Compute the Test Statistic

Calculate the test statistic \(Z = \frac{\hat{\theta}}{\hat{\omega}_d}\) where \(\hat{\theta} = -0.97\) and \(\hat{\omega}_d \approx 0.923\).\[Z = \frac{-0.97}{0.923} \approx -1.051\]

06

Determine the Critical Value and Conclusion

For \(\alpha = 0.01\), the critical value is \(-2.33\) using a standard normal distribution table for a one-tailed test. Since \(Z = -1.051 > -2.33\), we do not reject \(H_0\).

07

State the Conclusion

There is not enough statistical evidence at the \(\alpha = 0.01\) level to conclude that the true average MAO activity for normal subjects is more than twice that of schizophrenics.

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Key Concepts

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

Sample Sizes

Sample size is a critical component in hypothesis testing as it impacts the reliability and accuracy of the results. In this exercise, two distinct groups provide data:

• 43 individuals with chronic schizophrenia
• 45 normal subjects

The sample size of each group determines the degree of confidence in the findings. Larger sample sizes generally lead to more stable and dependable estimates of population parameters. When comparing two groups, as in this study, the sample sizes are similar, which helps maintain balance and fairness in the analysis. A balanced sample size minimizes bias and enhances the credibility of the statistical test outcomes. In general, more subjects lead to better estimates of the population means and consequently a more powerful statistical test that can potentially detect any real differences between groups.

Test Statistic

The test statistic is a standardized value that reflects how far your sample statistic is from the null hypothesis. It helps determine whether to reject the null hypothesis. This value in hypothesis testing synthesizes information from your data by incorporating estimated parameters and sample sizes. In the given task, the test statistic is denoted by:

• \[ Z = \frac{\hat{\theta}}{\hat{\omega}_d} \]

The test statistic calculation involves dividing the estimate of the parameter (\( \hat{\theta} \)) by the standard error (\( \hat{\omega}_d \)). The resultant value can then be compared against a critical value to evaluate our hypotheses. In this exercise, the computed test statistic is approximately -1.051, which indicates how significant the estimated difference is when viewed through the lens of the assumed statistical distribution (usually assumed to be normal in large sample sizes). The sign and magnitude describe how the sample estimate deviates from what is expected under the null hypothesis.

Critical Value

Critical values are threshold values that dictate whether to reject or fail to reject the null hypothesis. They derive from the statistical distribution that fits the data, typically the standard normal distribution in large sample size cases. In this exercise, given a significance level of \( \alpha = 0.01 \) and a one-tailed test, the critical value is found to be -2.33.

• A critical value helps frame the decision-making criterion to determine hypothesis validity.
• For a one-tailed test, it checks the extreme possibility of either end of the distribution, depending on the direction of the alternative hypothesis.

Comparing the test statistic (-1.051) to this critical value enables statisticians to decide if there is enough evidence against the null hypothesis. If the test statistic falls in a more extreme region than the critical value, the null hypothesis gets refuted. However, since \( -1.051 > -2.33 \), we fail to reject the null hypothesis.

Hypotheses

In hypothesis testing, formulating the right hypotheses is vital. In the present case, the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)) take a unique form:

• \( H_0: \theta = 0 \)
• \( H_a: \theta < 0 \)

Where \( \theta \) represents \( 2\mu_1 - \mu_2 \), with \( \mu_1 \) as the average MAO level for schizophrenics and \( \mu_2 \) for normal subjects. The null hypothesis states there is no difference in terms of double the average MAO activity between groups, while the alternative hypothesis seeks evidence that normal subjects have more than twice the MAO activity of schizophrenics. The set hypotheses elegantly capture the research question and help guide the testing procedure to find meaningful conclusions.

Standard Error

Standard error provides insight into the variability of the sample mean estimates. It helps convey how much the sample statistic represents the population parameter. In this context, the standard error (\( \hat{\omega}_d \)) is indispensable in determining the test statistic. Here, calculation of the standard error involves both sample sizes and their respective standard deviations:

• Sample standard deviation for schizophrenic subjects (\( 2.30 \))
• Sample standard deviation for normal subjects (\( 4.03 \))

Combining this information involves estimating the variance: \[ S^2_d = \frac{4(2.30)^2}{43} + \frac{4.03^2}{45} \]Resulting in a standard error of approximately 0.923. Having a low standard error indicates a closer estimate of the population parameter. A critical element of constructing confidence intervals or testing hypotheses, the standard error mitigates the influence of random sampling errors.