91Ó°ÊÓ

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

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

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.