Question

The derailment of a freight train due to the catastrophic failure of a traction motor armature bearing provided the impetus for a study reported in the article "Locomotive Traction Motor Armature Bearing Life Study" (Lubrication Engr., Aug. 1997: 12-19). A sample of high-mileage traction motors was selected, and the amount of cone penetration (mm/ 10) was determined both for the pinion bearing and for the commutator armature bearing, resulting in the following data:

Motor

1 2 3 4 5 6

Commutator 211 273 305 258 270 209

Pinion 266 278 259 244 273 236

Motor

7 8 9 10 11 12

Commutator 223 288 296 233 262 291

Pinion 290 287 315 242 288 242

Motor

13 14 15 16 17

Commutator 278 275 210 272 264

Pinion 278 208 281 274 268

Calculate an estimate of the population mean difference between penetration for the commutator armature bearing and penetration for the pinion bearing, and do so in a way that conveys information about the reliability and precision of the estimate. (Note: A normal probability plot validates the necessary normality assumption.) Would you say that the population mean difference has been precisely estimated? Does it look as though population mean penetration differs for the two types of bearings? Explain.

Short answer

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

Step-by-step solution

To Calculate an estimate of the population mean penetration differs for the two types of bearings

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

\(Motor,i\)	\(Commutator, {x_i}\)	\(Pinion,{y_i}\)	\(Difference, {d_i} = {x_i} - {y_i}\)
1	211	226	-15
2	273	278	-5
3	305	259	46
4	258	244	14
5	270	273	-3
6	209	236	-27
7	223	290	-67
8	288	287	1
9	296	315	-19
10	233	242	-9
11	262	288	-26
12	291	242	49
13	278	278	0
14	275	208	67
15	210	281	-71
16	272	274	-2
17	264	268	-4

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

To Calculate an estimate of the population mean penetration differs for the two types of bearings

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.

To Calculate an estimate of the population mean penetration differs for the two types of bearings

To Calculate an estimate of the population mean penetration differs for the two types of bearings

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

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

To Calculate an estimate of the population mean penetration differs for the two types of bearings

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.