91Ó°ÊÓ

Persons having Reynaud's syndrome are apt to sufter a sudden impoirment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal)cm² \(/\) min) was measured. For \(m=10\) subjects with the syndrome, the average heat output was \(\bar{x}=.64\), and for \(n=10\) nonsufferers, the average output was \(2.05\). Let \(\mu_{1}\) and \(\mu_{2}\) denote the true average heat outputs for the two types of subjects. Assume that the two distributions of heat output are normal with \(\sigma_{1}=2\) and \(\sigma_{2}=4\) a. Consider testing \(H_{0}: \mu_{t}-\mu_{2}=-1.0\) versus \(H_{2}: \mu_{1}-\) \(\mu_{t}<-1.0\) at level .01. Describe in words what \(H_{a}\) says, and then carry out the test. b. Compute the \(P\)-value for the value of \(Z\) obtained in part (a). c. What is the probability of a type II error when the actual difference between \(\mu_{1}\) and \(\mu_{2}\) is \(\mu_{1}-\mu_{2}=-1.2\) ? d. Assuming that \(m=n\), what sample sizes are required to ensure that \(\beta=.1\) when \(\mu_{1}-\mu_{2}=-1.2\) ?

Step by step solution

01

Define Hypotheses for Test

For part (a), we are testing the null hypothesis \( H_0: \mu_1 - \mu_2 = -1.0 \) against the alternative hypothesis \( H_a: \mu_1 - \mu_2 < -1.0 \). The alternative hypothesis \( H_a \) states that the difference in average heat output between subjects with and without the syndrome is less than -1.0, indicating that subjects with the syndrome have a significantly lower heat output.

02

Calculate Test Statistic

To test the hypothesis, we use a z-test for the difference of means. The test statistic \( Z \) is calculated as follows:\[ Z = \frac{(\bar{x}_1 - \bar{x}_2) - (-1)}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}} \]Substitute the given values:\( \bar{x}_1 = 0.64 \), \( \bar{x}_2 = 2.05 \), \( \sigma_1 = 2 \), \( \sigma_2 = 4 \), \( m = 10 \), and \( n = 10 \):\[ Z = \frac{(0.64 - 2.05) - (-1)}{\sqrt{\frac{2^2}{10} + \frac{4^2}{10}}} = \frac{-0.41}{\sqrt{0.4 + 1.6}} = \frac{-0.41}{\sqrt{2}} \approx -0.29 \]

03

Determine the Critical Value and Decision

The significance level is \( \alpha = 0.01 \). For a lower-tailed test, the critical z-value corresponding to \( \alpha = 0.01 \) is \(-2.33\). Since our calculated \( Z \approx -0.29 \) does not fall into the critical region (i.e., it is not less than -2.33), we do not reject the null hypothesis \( H_0 \).

04

Compute the P-value

The P-value corresponds to the probability that the test statistic \( Z \) is less than the observed value when the null hypothesis is true. Given \( Z \approx -0.29 \), the P-value can be found using standard normal distribution tables or software:The P-value \( P(Z < -0.29) \approx 0.3859 \).

05

Calculate Type II Error Probability

For part (c), where \( \mu_1 - \mu_2 = -1.2 \), calculate the new test statistic:\[ Z = \frac{(0.64 - 2.05) - (-1.2)}{\sqrt{0.4 + 1.6}} = \frac{-0.21}{\sqrt{2}} \approx -0.15 \]To find \( \beta \), the probability of a type II error, we compute:\( \beta = P(Z > -2.33 + (\text{\text{New Z-value} based on} -1.2 \text{ difference})) \approx 0.9357 \).

06

Calculate Required Sample Size for Beta = 0.1

For part (d), we want \( \beta = 0.1 \) with \( \mu_1 - \mu_2 = -1.2 \). This requires:\[ n = \left(\frac{(Z_{\alpha} + Z_{\beta}) \sqrt{\sigma_1^2 + \sigma_2^2}}{\Delta}\right)^2 \]Substituting \( \Delta = 0.2 \), \( Z_{\alpha} = 2.33 \), and \( Z_{\beta} = 1.28 \):\[ n = \left(\frac{(2.33 + 1.28) \sqrt{2^2 + 4^2}}{1.2}\right)^2 \approx 45.5 \text{ per group} \] So, a sample size of \( n = 46 \) is required for each group.

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Key Concepts

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

Type II Error

In hypothesis testing, a Type II error occurs when we fail to reject a false null hypothesis. This means that even though there is a real effect or difference in the data, the test does not detect it. This error is inversely related to the test's power, which measures its ability to detect an effect when there is one. A common notation for the probability of committing a Type II error is \( \beta \).
For instance, in our example where the difference between means \( \mu_1 - \mu_2 \) is actually \(-1.2\), not detecting this difference would be a Type II error if we failed to reject the null hypothesis \( H_0: \mu_1 - \mu_2 = -1.0 \).
The calculated probability of this error in the exercise was approximately 0.9357. This high probability suggests a low power for the test under these specific conditions, indicating the test is not very effective in this situation. To reduce \( \beta \) and thus decrease the chance of a Type II error, we can increase the sample size or the significance level, or use a more sensitive test.

Sample Size Calculation

Calculating the appropriate sample size is crucial in hypothesis testing because it affects the test's reliability and validity. If the sample size is too small, the probability of a Type II error can be high, which means we might miss detecting a true effect. Conversely, too large a sample size may be unnecessary and resource-consuming.
In the exercise, the sample size required for sufficient power with \( \beta = 0.1 \) when \( \mu_1 - \mu_2 = -1.2 \) was calculated using the formula:

• \[ n = \left(\frac{(Z_{\alpha} + Z_{\beta}) \sqrt{\sigma_1^2 + \sigma_2^2}}{\Delta}\right)^2 \]
• \( Z_{\alpha} \) and \( Z_{\beta} \) are critical values from the standard normal distribution, corresponding to the desired levels of significance \( \alpha \) and power \( 1 - \beta \).
• \( \Delta \) is the expected difference in means.

Substituting the appropriate values yielded a sample size of approximately 46 per group. This ensures the test is effective enough to detect the effect with high confidence.

Z-test

The \( Z \)-test is a statistical test utilized to determine if there is a significant difference between the means of two groups, assuming the data follows a normal distribution. It is especially suitable when the population variances are known and the sample size is large, though it can also be used for smaller samples under certain conditions.
For the hypothesis test in the exercise, the \( Z \)-test was employed to test the difference between means \( \mu_1 \) and \( \mu_2 \). The test statistic was calculated using the formula:

• \[ Z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}} \]
• \( \bar{x}_1 \) and \( \bar{x}_2 \) are sample means, \( \sigma_1 \) and \( \sigma_2 \) are population standard deviations, and \( m \) and \( n \) are sample sizes.

Through substitution of the given values, the calculated \( Z \)-value was approximately -0.29.
Since this did not exceed the critical \(-2.33\) value at \( \alpha = 0.01 \), it led to the decision not to reject the null hypothesis. Though apparently straightforward, understanding the foundations of a \( Z \)-test is crucial for interpreting its results correctly.