Question

In an experiment to compare bearing strengths of pegs inserted in two different types of mounts, a sample of 14 observations on stress limit for red oak mounts resulted in a sample mean and sample standard deviation of \(8.48MPa\) and .79 MPa, respectively, whereas a sample of 12 observations when Douglas fir mounts were used gave a mean of \(9.36\) and a standard deviation of \(1.52\) ('Bearing Strength of White Oak Pegs in Red Oak and Douglas Fir Timbers," J. of Testing and Evaluation, 1998, 109-114). Consider testing whether or not true average stress limits are identical for the two types of mounts. Compare df's and P-values for the unpooled and pooled t tests.

Short answer

\(\begin{array}{l}{\rm{Un pooled:df}} = 15{\rm{,P - value}} \simeq 0.092\\{\rm{Pooled : df}} = 24{\rm{,P - value; = }} \simeq 0.070\\{\rm{Un pooled(df) < Pooled(df)}}\\{\rm{Un pooled(P - value) > Pooled(P - value)}}\end{array}\)

Step-by-step solution

To Find the Unpooled test

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

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

Find the t value

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.

To Find the pooled test

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

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

To Find the P value

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.

Final proof

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).