Question

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" (Ergonomics, \(1995: 2588 - 2601\)) reported on a test that involved sensing a tiny gap in an otherwise smooth 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\(1.71\;mm\), 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 CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of\(.01\).

Short answer

where the probability can be found in the table in the appendix. Since

\(P = P(T \le - 2.46) = 0.013\)

do not reject null hypothesis at the given significance level. There is not enough evidence to claim that the true average for CTS subjects exceeds the true average for normal subjects.

Step-by-step solution

To find the test statistic value:

The hypotheses of interest are

\({H_0}:{\mu _1} - {\mu _2} = 0{\rm{ vs }}{H_a}:{\mu _1} - {\mu _2} < 0\)

where\({\mu _1}\)is the true average gap for normal subjects, and\({\mu _2}\)is the true average for CTS subjects.

Given two normal distributions, the random variable (standardized)

\(T = \frac{{\bar X - \bar Y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{S_1^2}}{m} + \frac{{S_2^2}}{n}} }}\)

has approximately students\(t\)distribution with degrees of freedom\(\nu \), where\(\nu \)is

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}}\)

has to be rounded down to the nearest integer.

The test statistic value can be computed using the given value in the exercise as follows

\(\begin{array}{l}t = \frac{{\bar x - \bar y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{1.71 - 2.53}}{{\sqrt {\frac{{0.53}}{8} + \frac{{2.53}}{{0.87}}} }}\\ = \frac{{ - 0.82}}{{\sqrt {0.035 + 0.076} }} = \frac{{ - 0.82}}{{0.333}}\\ = - 2.46\end{array}\)

To find the degrees of freedom:

The degrees of freedom can be computed by the given formula

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{(0.035 + 0.076)}^2}}}{{\frac{{{{0.035}^2}}}{7} + \frac{{0.076}}{9}}} = 15.085\)\(P = P(T \le - 2.46) = 0.013\)

and the round down to the nearest integer is the needed degrees of freedom

\(\nu = 15\)

Since the alternative hypothesis is\({H_a}:{\mu _1} - {\mu _2} < 0\), the corresponding\(P\)value is the probability to the left of\(t\)under the students\(T\)statistic with degrees of freedom 15 , therefore

where the probability can be found in the table in the appendix. Since

\(P = P(T \le - 2.46) = 0.013\)

do not reject null hypothesis at the given significance level. There is not enough evidence to claim that the true average for CTS subjects exceeds the true average for normal subjects.