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The level of monoamine oxidase (MAO) activity in blood platelets (nm/mg protein/h) was determined for each individual in a sample of 43 chronic schizophrenics, resulting in \(\bar{x}=2.69\) and \(s_{1}=2.30\), as well as for 45 normal subjects, resulting in \(\bar{y}=6.35\) and \(s_{2}=4.03\). 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 average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta=2 \mu_{1}-\mu_{2}\). Write \(H_{0}\) and \(H_{2}\) in terms of \(\theta\), estimate \(\theta\), and derive \(\overrightarrow{\sigma_{\theta}}\) ("Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients, "Nature, July 28, 1972: \(225-226\) ).]

Step by step solution

01

Define Hypotheses

We need to test if the true average MAO activity for normal subjects is more than twice that of schizophrenics. Define the parameter \(\theta = 2\mu_1 - \mu_2\), where \(\mu_1\) is the average MAO of schizophrenics, and \(\mu_2\) is that of normal subjects. Thus, the null hypothesis \(H_0\) is \(\theta = 0\), and the alternative hypothesis \(H_a\) is \(\theta < 0\). This means we're testing if \( 2\mu_1 < \mu_2 \).

02

Estimate θ

Estimate \(\theta\) using the sample means \(\bar{x}\) and \(\bar{y}\): \(\hat{\theta} = 2\bar{x} - \bar{y} = 2(2.69) - 6.35 = 5.38 - 6.35 = -0.97\).

03

Calculate Standard Error of θ

The standard error \(\sigma_{\theta}\) of \(\hat{\theta}\) can be derived from the standard deviations \(s_1\) and \(s_2\) given: \[ \sigma_{\theta} = \sqrt{4 \left(\frac{s_1^2}{n_1}\right) + \frac{s_2^2}{n_2}} = \sqrt{4 \left(\frac{2.30^2}{43}\right) + \frac{4.03^2}{45}} \].

04

Calculate the Test Statistic

Compute the test statistic \(z\) as \[ z = \frac{\hat{\theta} - 0}{\sigma_{\theta}} \]. Substitute \(\hat{\theta} = -0.97\) and the \(\sigma_{\theta}\) calculated in Step 3. This will provide the test statistic value for our hypothesis test.

05

Determine Critical Value and Decision

For \(\alpha = 0.01\), determine the critical \(z\) value for a left-tailed test from standard normal distribution tables. If the calculated \(z\) statistic is less than the critical value, reject \(H_0\). Otherwise, do not reject \(H_0\).

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Key Concepts

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

Statistical Inference

Statistical inference is the process of using data analysis to infer properties of an underlying probability distribution. It is a methodology that allows statisticians to draw conclusions about populations based on samples taken from these populations. In the context of the exercise, we use statistical inference to determine whether the average monoamine oxidase (MAO) activity in normal subjects is more than twice that in schizophrenics. This involves analyzing sample data and making decisions about the population at large.

Key techniques in statistical inference include:

• Estimation: Determining the approximate value of a population parameter.
• Hypothesis Testing: Assessing the plausibility of a hypothesis using sample data.
• Prediction: Making forecasts about future observations based on the sample data.

These techniques help us make educated guesses about the world and are an essential part of many scientific studies and research.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is a fundamental aspect of hypothesis testing. It is a statement that there is no effect or no difference, and it serves as the starting point for testing. In this exercise, the null hypothesis is set as \(\theta = 0\). This means that the parameter \(\theta\), defined as \(2\mu_1 - \mu_2\), equals zero. Essentially, this states that the true average MAO activity for normal subjects is not more than twice that of schizophrenics.

The null hypothesis is always a statement of equality. We only reject the null hypothesis if there's strong evidence against it, based on our sample data. Importantly, it allows us to quantify the strength of evidence by using statistical tests.

Alternative Hypothesis

In contrast to the null hypothesis, the alternative hypothesis, denoted as \(H_a\), proposes that a substantial effect or difference does exist. It is what we seek to provide evidence for. In this scenario, the alternative hypothesis is \(\theta < 0\), meaning that \(2\mu_1 < \mu_2\).

This hypothesis suggests that the MAO activity in normal subjects is indeed more than double the level found in schizophrenics. By conducting a statistical test, we assess whether this claim holds true based on the sample data. If the evidence strongly supports \(H_a\), we may reject \(H_0\) in favor of \(H_a\), implying a significant difference.

Test Statistic

A test statistic is a standardized value that is calculated from sample data during a hypothesis test. The purpose of the test statistic is to assess whether the observed data deviate from what was expected under the null hypothesis. In the exercise, the test statistic \(z\) is calculated as \(z = \frac{\hat{\theta} - 0}{\sigma_{\theta}}\).

The test statistic converts observed data into a number that can be evaluated in terms of probability. If this value is extreme in the distribution under \(H_0\), it indicates evidence against the null hypothesis. Describing this transformation allows for comparisons to known distributions, like the normal distribution, which inform decision-making in hypothesis testing.

Standard Error

The standard error measures the variability or standard deviation of a sampling distribution. It quantifies the variability of the parameter estimate from different possible samples. In the given problem, the standard error of \(\theta\) is calculated using the standard deviations of the individual groups \(s_1\) and \(s_2\), along with their sample sizes.

Specifically, the formula used is \(\sigma_{\theta} = \sqrt{4 \left(\frac{s_1^2}{n_1}\right) + \frac{s_2^2}{n_2}}\). Calculating the standard error gives us a picture of potential estimation error in \(\hat{\theta}\) and helps in creating a margin of error in hypothesis testing.

Critical Value

The critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis. It is determined by the significance level \(\alpha\) set for the test, and for one-tailed tests, such as this, corresponds to a particular percentile of the distribution. In the task, we use \(\alpha = 0.01\), which conveys a 1% significance level.

For a left-tailed test in a standard normal distribution, this means looking up the critical value that corresponds to the left 1% of the distribution. If the test statistic \(z\) calculated is less than this critical point, we reject \(H_0\). Understanding and determining the critical value is essential for making well-informed decisions in hypothesis testing.