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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). The article "A Gap Detection Tactility Test for Sensory Deficits Associated with Carpal Tunnel Syndrome" (Erganamics, 1995: 2588-2601) reported on a test that involved sensing a tiny gap in an otherwise smoosh surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for \(m-8\) normal subjects was \(1.71 \mathrm{~mm}\), and the sample standard deviation was \(.53\); for \(n=10\) CTS subjects, the sample mean and sample standard deviation were \(2.53\) and \(.87\), respectively. Does this data suggest that the true average gap detection threshold for CIS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of .01

Step by step solution

01

State the Hypotheses

We start by defining our null and alternative hypotheses. The null hypothesis, \( H_0 \), is that the true average gap detection threshold for CTS subjects is less than or equal to that for normal subjects: \( \mu_{CTS} \leq \mu_{normal} \). The alternative hypothesis, \( H_a \), is that the true average gap detection threshold for CTS subjects exceeds that for normal subjects: \( \mu_{CTS} > \mu_{normal} \).

02

Set the Significance Level

The problem states that we use a significance level of \( \alpha = 0.01 \). This is the probability of rejecting the null hypothesis when it is actually true, and it is a measure of the risk we are willing to take to make a Type I error.

03

Calculate the Test Statistic

We will use a two-sample t-test to compare the means. The test statistic \( t \) can be calculated using the formula: \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \( \bar{x}_1 = 2.53 \), \( s_1 = 0.87 \), \( n_1 = 10 \) for the CTS group, and \( \bar{x}_2 = 1.71 \), \( s_2 = 0.53 \), \( n_2 = m-8 \) (we assume \( m = 8 \) here for normal subjects). Substitute the values into the formula.

04

Compute Pooled Standard Error

First, compute the pooled standard error:\[ SE = \sqrt{\frac{0.87^2}{10} + \frac{0.53^2}{8}} \]Calculate this to obtain the value used in the t-statistic calculation.

05

Calculate the Test Statistic Again

Insert the pooled standard error into the test statistic formula:\[ t = \frac{2.53 - 1.71}{SE} \]Calculate the value of \( t \).

06

Compare the Test Statistic to the Critical Value

Determine the degrees of freedom using the formula for unequal variances:\[ u \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}} \]Using the calculated \( u \), find the critical t-value for a one-tailed test with \( \alpha = 0.01 \). Compare this against the calculated test statistic from Step 5.

07

Decision and Conclusion

If the calculated t-value is greater than the critical t-value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Draw a conclusion about whether there is sufficient evidence to suggest that the average gap detection threshold for CTS subjects exceeds that for normal subjects.

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Key Concepts

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

Two-sample t-test

The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two groups. This test is particularly useful when you want to compare the averages of two independent samples, such as the gap detection thresholds of CTS subjects versus normal subjects.

To perform a two-sample t-test, you'll need to collect data from both groups and calculate the sample mean and sample standard deviation for each. The formula for the test statistic (t) is:

• \[t = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]

Here,

• \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means,
• \( s_1 \) and \( s_2 \) are the standard deviations,
• \( n_1 \) and \( n_2 \) are the sample sizes of the two groups.

This test provides a way to infer if the difference between the means is not due to random chance, thus helping you to draw conclusions about your hypothesis.

Significance Level

The significance level, often denoted by \( \alpha \), is a critical part of hypothesis testing. It represents the probability of making a Type I error, which occurs when you reject the null hypothesis even though it is true. Common significance levels are 0.05, 0.01, or 0.10, depending on how rigorous you want your test to be.

In this exercise, a significance level of 0.01 was chosen. This means you are willing to accept a 1% risk of incorrectly rejecting the null hypothesis. A lower significance level like this indicates a stricter criterion for proving a statistical difference exists, making it less likely to conclude there is a difference when there actually isn't.

Choosing the right significance level is crucial. It involves balancing the need to minimize false positives while being sensitive enough to detect an actual effect when it exists. For this example, the researchers wanted a high confidence in their findings, leading to the choice of 0.01.

Pooled Standard Error

The pooled standard error is used when you assume that two samples come from populations having the same variance but different means. It combines the variability of the two samples into a single measure and helps calculate the t-statistic.

For the exercise at hand, calculate the pooled standard error as follows:

• \[SE = \sqrt{\frac{0.87^2}{10} + \frac{0.53^2}{8}}\]

This value reflects the average amount by which the sample mean differs from the true population mean. By using the pooled standard error, the calculation accounts for the sample sizes and variances of both groups, providing a robust basis for the t-statistic.

In the context of the problem, the pooled standard error is critical in determining whether the observed difference in gap detection thresholds is statistically significant.

Degrees of Freedom Calculation

Degrees of freedom (df) in the context of a two-sample t-test indicate the number of values in the final calculation of a statistic that are free to vary. It's a complex topic but crucial for accurately determining the critical value of the t-distribution.

When calculating degrees of freedom for a two-sample t-test with unequal variances, you can use the following formula:

• \[u \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}}\]

This is known as the Welch-Satterthwaite equation. It approximates the degrees of freedom and is particularly useful when the variances of the groups are not equal, as is often the case in real-world data.

Accurate degrees of freedom ensure that the critical t-value is correct, which is fundamental in making your hypothesis testing decision robust and reliable.