91Ó°ÊÓ

The following table represents data in a way where you can see everything needed to analyze it.

The differences in the table are given because the paired CI has to be used.

The Paired t Test:

When \(D = X - Y\) (difference between observations within a pair), \({\mu _D} = {\mu _1} - {\mu _2}\) and null hypothesis

\({H_0}:{\mu _D} = {\Delta _0}\)

the test statistic value for testing the hypotheses is

\(t = \frac{{\bar d - {\Delta _0}}}{{{s_D}/\sqrt n }}\)

where \(\bar d and {s_D}\) are the sample mean and the sample standard deviation of differences\({d_i}\) , respectively. In order to use this test, assume that the differences \({D_i}\) are from a normal population. Depending on alternative hypothesis, the P value can be determined as the corresponding area under the \({t_{n - 1}}\) curve.

The paired t confidence interval for \({\mu _D}\) is

\(\left( {\bar d - {t_{\alpha /2,n - 1}} \times \frac{{{s_D}}}{{\sqrt n }},\bar d + {t_{\alpha /2,n - 1}} \times \frac{{{s_D}}}{{\sqrt n }}} \right)\)

A one-sided confidence bound can be obtained by retaining the relevant sing \(( + or - ),and by replacing {t_{\alpha /2,n - 1}} by {t_{\alpha ,n - 1}}\)

Normal probability plot suggest that the confidence interval is appropriate to use.

The sample mean and the sample standard deviation of the differences has to be computed.

The Sample Mean \(\bar x\) of observations \({x_1},{x_2}, \ldots ,{x_n}\) is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean $\bar{d}$ for the differences is

\(\bar d = \frac{1}{{17}} \times ( - 15 - 5 + \ldots - 4) = - 4.18\)

The Sample Variance \({s^2}\) is

\({s^2} = \frac{1}{{n - 1}} \times {S_{xx}}\)

Where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \times {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation s is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \times {S_{xx}}} \)

The sample variance is

\(\begin{array}{l}s_D^2 = \frac{1}{{17 - 1}} \times \left( {{{( - 15 - ( - 4.18))}^2} + {{( - 5 - ( - 4.18))}^2} + \ldots + {{( - 4 - ( - 4.18))}^2}} \right)\\ = 1285.154\end{array}\)

and the sample standard deviation

\({s_D} = \sqrt {1285.154} = 35.85\)

For example, \(95\% \) confidence interval \((\alpha = 0.05),\) where \({t_{\alpha /2,n - 1}} = {t_{0.025,16}} = 2.12\) which was obtained from the table in the appendix, is

\(\begin{array}{l}\left( {\bar d - {t_{\alpha /2,n - 1}} \times \frac{{{s_D}}}{{\sqrt n }},\bar d + {t_{\alpha /2,n - 1}} \times \frac{{{s_D}}}{{\sqrt n }}} \right)\\\\ = \left( { - 4.18 - 2.12 \times \frac{{35.85}}{{\sqrt {17} }}, - 4.18 + 2.12 \times \frac{{35.85}}{{\sqrt {17} }}} \right)\end{array}\)

\( = ( - 22.61,14.25).\)

This indicates that the population mean difference has not been precisely estimated because the limits are far apart (big difference in the limits). Because zero belong to the interval the conclusion is that the population mean penetration differs for the two types of bearings.