91Ó°ÊÓ

The hypotheses of interest here are\({H_0}:{\mu _1} - {\mu _2} = 0\)versus\({H_1}:{\mu _1} - {\mu _2} \ne 0\), where\({\mu _1},{\mu _2}\)are the true averages of the stress limits.

Unpooled test:

Given two normal distributions, the random variable (standardized)

\(t = \frac{{\overline x - \overline y - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

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

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

has to be rounded down to the nearest integer.

The two-sample t test for testing\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)uses the following value of test statistic

For the adequate alternative hypothesis the adequate area under the\({t_\nu }\)curve is the P value.

The test statistic t value is

\(\begin{array}{c}t = \frac{{8.48 - 9.36 - 0}}{{\sqrt {\frac{{{{0.79}^2}}}{{14}} + \frac{{{{1.52}^2}}}{{12}}} }}\\ = \frac{{ - 0.88}}{{0.3869}}\\ = - 1.81\end{array}\)

Compute the degrees of freedom now in order to find the P value. From the formula, the following is true

\(\begin{array}{c}\nu = \frac{{{{\left( {\frac{{{{0.79}^2}}}{{14}} + \frac{{{{1.52}^2}}}{{12}}} \right)}^2}}}{{\frac{{{{\left( {{{0.79}^2}/14} \right)}^2}}}{{14 - 1}} + \frac{{{{\left( {{{1.52}^2}/12} \right)}^2}}}{{12 - 1}}}}\\ = 15.95\end{array}\)

and when you round down to the nearest integer you get 15 which is the degrees of freedom. Thus, \(\nu = 15\)

The test is two sided which means that the P value is two times the area under the\({t_{15}}\)curve to the right of\(|t|\). The P value is

\(\begin{array}{c}P = 2 \times P(T > 1.81)\\ = 2 \times 0.046\\ = 0.092\end{array}\)

The value was computed using a software, you can find it or estimate it from the table in the appendix. Random variable T has students distribution with 15 degrees of freedom. It is not needed to draw conclusions about the hypothesis, you just need to compare the pooled and unpooled tests.

Pooled test:

The pooled t test for testing\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)uses the following t value

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{{s_p} \times \sqrt {\frac{1}{m} + \frac{1}{n}} }}\)

where the corresponding T statistic has students distribution with m+n-2degrees of freedom, and\({s_p}\)is

\({s_p} = \sqrt {\frac{{(m - 1)s_1^2 + (n - 1)s_2^2}}{{m + n - 2}}} \)

The pooled standard deviation\({s_p}\)is

\(\begin{array}{c}{s_p} = \sqrt {\frac{{(m - 1)s_1^2 + (n - 1)s_2^2}}{{m + n - 2}}} \\ = \sqrt {\frac{{(14 - 1) \times {{0.79}^2} + (12 - 1) \times {{1.52}^2}}}{{14 + 12 - 2}}} \\ = \sqrt {1.3970} \\ = 1.181\end{array}\)

The t statistic value for the pooled test is

\(\begin{array}{c}t = \frac{{8.48 - 9.36 - 0}}{{1.181 \times \sqrt {1/14 + 1/12} }}\\ = \frac{{ - 0.88}}{{0.456}}\\ = - 1.89\end{array}\)

The degrees of freedom\(\nu \)are

\(\begin{array}{c}\nu = m + n - 2\\ = 14 + 12 - 2\\ = 24\end{array}\)

The alternative test hypothesis is\({H_a}:{\mu _1} - {\mu _2} \ne 0\); thus the P value is corresponding area under the\({t_{24}}\)curve (the same way as for the pooled test)

\(\begin{array}{c}P = 2 \times P(T > 1.89)\\ = 2 \times 0.035\\ = 0.07\end{array}\)

The value was computed using a software, you can find it or estimate it from the table in the appendix. Random variable T has students distribution with 24 degrees of freedom.

Comparison:

The P value for the pooled test is smaller than the P value for the unpooled test Also, the pooled test has more degrees of freedom (24) than the unpooled test (15).