91Ó°ÊÓ

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.

By replacing\({z_{\alpha /2}} by {z_\alpha }\) and\( \pm \) with only\( + and - \)an upper/lower bound is obtained.

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

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

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

The hypotheses of interest are\({H_0}:{\mu _1} - {\mu _2} = 1000\) versus\({H_a}:{\mu _1} - {\mu _2} > 1000\), where\({\mu _1},and {\mu _2}\) are the true mean of cycles to break for polyisprene and the true mean for natural latex condoms, respectively. For given values

\(\begin{array}{l}\bar x = 5805;{s_1} = 3990;m = 20 \\\bar y = 4358;{s_2} = 2218;n = 20.\end{array}\)

The t statistic value is

\(t = \frac{{\bar x - \bar y - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{5805 - 4358 - 1000}}{{\sqrt {\frac{{399{0^2}}}{{20}} + \frac{{221{8^2}}}{{20}}} }} = 0.438\)

To compute the P value, you need to compute the degrees of freedom first

\(\nu = \frac{{{{\left( {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/m} \right)}^2}}}{{m - 1}} + \frac{{{{\left( {s_2^2/n} \right)}^2}}}{{n - 1}}}} = \frac{{{{\left( {\frac{{399{0^2}}}{{20}} + \frac{{221{8^2}}}{{20}}} \right)}^2}}}{{\frac{{{{\left( {399{0^2}/20} \right)}^2}}}{{19}} + \frac{{{{\left( {221{8^2}/20} \right)}^2}}}{{19}}}} = 29.72\)

and the nearest round down integer is 29 , thus

\(\nu = 29.\)

The alternative hypothesis is\({H_a}:{\mu _1} - {\mu _2} > 0\), this the P value is the area under the\({t_{29}}\)curve to the right of t. Thus

\(P = P(T > 0.438) = 0.332\)

where T is the students statistic with 29 degrees of freedom, and the value was computed by a software (you could have taken the approximate value from the table - estimate it).

The P value is

\(P = 0.332 > 0.01 = \alpha \)

Thus

do not reject null hypothesis