Question

The article "Fatigue Testing of Condoms" cited in Exercise 7.32 reported that for a sample of 20 natural latex condoms of a certain type, the sample mean and sample standard deviation of the number of cycles to break were 4358 and 2218 , respectively, whereas a sample of 20 polyisoprene condoms gave a sample mean and sample standard deviation of 5805 and 3990 , respectively. Is there strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles? Carry out a test using a significance level of 01 . (Note: The cited paper reported P-values of t tests for comparing means of the various types considered.)

Short answer

Do not reject null hypothesis

Step-by-step solution

To find the strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles

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.

To find the strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles

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

\(\nu = 29.\)

To find the strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles

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