91Ó°ÊÓ

The level of monoamine oxidase (MAO) activity in blood platelets ( \(\mathrm{nm} / \mathrm{mg}\) 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 \(H_{2}\) 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 \(H_{a}\) in terms of \(\theta\), estimate \(\theta\), and derive \(\hat{\sigma}_{\dot{\theta}}\) ("Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients," Nature, July 28, 1972: 225-226).]

Step by step solution

01

Define Hypotheses

Start by defining the null hypothesis \(H_0\) and alternative hypothesis \(H_a\) in terms of \(\theta\). We have \(\theta = 2\mu_1 - \mu_2\). The condition that normal subjects have more than twice the activity level for schizophrenics means \(\mu_2 > 2\mu_1\), which translates to \(\theta < 0\). Therefore, \(H_0: \theta = 0\) and \(H_a: \theta < 0\).

02

Estimate the Parameter \(\theta\)

We estimated \(\theta\) using the sample means \(\bar{x}\) and \(\bar{y}\): \(\hat{\theta} = 2\bar{x} - \bar{y}\). Substituting the values, we get \(\hat{\theta} = 2\times2.69 - 6.35 = 5.38 - 6.35 = -0.97\).

03

Calculate the Standard Error \(\hat{\sigma}_{\dot{\theta}}\)

The estimated standard deviation of \(\hat{\theta}\) is computed as \(\hat{\sigma}_{\dot{\theta}} = \sqrt{4\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)}\). Substituting \(s_1 = 2.30\), \(s_2 = 4.03\), \(n_1 = 43\), and \(n_2 = 45\), we calculate: \(\hat{\sigma}_{\dot{\theta}} = \sqrt{4 \left(\frac{2.30^2}{43}\right) + \left(\frac{4.03^2}{45}\right)} = \sqrt{4 \left(\frac{5.29}{43}\right) + \left(\frac{16.24}{45}\right)} = \sqrt{0.49 + 0.36} = \sqrt{0.85} \approx 0.922\).

04

Determine Test Statistic

The test statistic \(T\) is calculated by \(T = \frac{\hat{\theta}}{\hat{\sigma}_{\dot{\theta}}}\). Substituting our values, \(T = \frac{-0.97}{0.922} \approx -1.05\).

05

Compare with Critical Value

For a one-sided test with \(\alpha = 0.01\), the critical value from a standard normal distribution is approximately \(-2.33\). The decision rule is to reject \(H_0\) if \(T < -2.33\).

06

Make a Decision

Since \(-1.05 > -2.33\), we fail to reject \(H_0\). 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 for schizophrenics.

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Key Concepts

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

Null and Alternative Hypotheses

When conducting a statistical hypothesis test, you start by clearly defining the null and alternative hypotheses. These hypotheses are crucial because they establish what you are testing and the possible conclusions. In the given exercise, the null hypothesis (\(H_0\)) is defined as \(\theta = 0\). Here, \(\theta\) is calculated as \(2\mu_1 - \mu_2\), where \(\mu_1\) represents the true average MAO activity for schizophrenics and \(\mu_2\) for normal subjects.

The alternative hypothesis (\(H_a\)) is what you aim to gather evidence for, and in this case, it is \(\theta < 0\). This signifies that the average MAO activity in normal subjects is more than twice that of schizophrenics. Essentially, you are testing if \(\mu_2 > 2\mu_1\).

• **Null Hypothesis**: No difference or effect (\(\theta = 0\)).
• **Alternative Hypothesis**: Evidence contrary to the null (\(\theta < 0\)).

Stating these hypotheses accurately provides clarity on the focus of the test, guiding your subsequent calculations and decision-making.

Test Statistic Calculation

Calculating the test statistic is a vital part of hypothesis testing. It allows you to express how far the sample statistic is from the null hypothesis, in terms of standard errors. In this problem, the test statistic \(T\) is calculated using the formula:\[T = \frac{\hat{\theta}}{\hat{\sigma}_{\dot{\theta}}}\]This formula enables you to standardize your estimate of \(\theta\).

Substitute the estimated \(\theta\) value, \(\hat{\theta} = -0.97,\) and the standard error, \(\hat{\sigma}_{\dot{\theta}} = 0.922\). Plug them into your formula:\[T = \frac{-0.97}{0.922} \approx -1.05\]

• **Test statistic**: This calculation shows you how many standard errors your estimate is from the hypothesized value under \(H_0\).

With the test statistic calculated, you move to compare this value against critical values to make decisions about the hypotheses.

Standard Error Estimation

Estimating the standard error is crucial for understanding the variability of the sample statistic. It shows how much \(\hat{\theta}\) would differ from \(\theta\) just by chance in different samples. In this exercise, the formula used for the standard error of \(\hat{\theta}\) is:\[\hat{\sigma}_{\dot{\theta}} = \sqrt{4\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)}\]

Here:

• \(s_1 = 2.30\),\(s_2 = 4.03\) are the sample standard deviations.
• \(n_1 = 43\),\(n_2 = 45\) are the sample sizes.

Use these to compute the standard error as:\[\hat{\sigma}_{\dot{\theta}} = \sqrt{4\left(\frac{2.30^2}{43}\right) + \left(\frac{4.03^2}{45}\right)} = \sqrt{0.49 + 0.36} = \sqrt{0.85} \approx 0.922\]This value offers insight into the precision of \(\hat{\theta}\). A smaller standard error indicates a more precise estimate, suggesting less variability among samples.

One-Sided Test

In hypothesis testing, the choice between a one-sided and a two-sided test is significant. A one-sided test examines if a parameter is greater or smaller than a particular value, whereas a two-sided test checks for any significant difference from the specified value. In this exercise, the one-sided test assesses if \(\theta\) is less than zero.

To conduct the one-sided test, determine the critical value from a standard normal distribution relevant to your alpha level, \(\alpha = 0.01\). For a one-sided test, the critical value is approximately \(-2.33\).

• **Decision Rule**: Reject \(H_0\) if \(T < -2.33\).

With the test statistic \(T \approx -1.05\), it fails to meet the condition for rejecting \(H_0\). This means there is insufficient evidence to prove the mean MAO activity for normal subjects is more than twice that of schizophrenic subjects, at a \(\alpha = 0.01\) significance level.