Statistics for Maintenance Modelling: Repairable Systems

# Exercise 3¤

Carry out a trend test on the data set from screen 17 of unscheduled maintenance actions for a diesel engine of USS Grampus carried out at times: 860, 1258, 1317, 1442, 1897, 2011, 2122, 2439, 3203, 3298, 3902, 3910, 4000, 4247, 4411, 4456, 4517, 4899, 4910, 5676, 5755, 6137, 6221, 6311, 6613, 6975, 7335, 8158, 8498, 8690, 9042, 9330, 9394, 9426, 9872, 10191, 11511, 11575, 12100, 12126, 12368, 12681, 12795, 13399, 13668, 13780, 13877, 14007, 14028, 14035, 14173, 14176, 14449, 14587, 14610, 15070 at a 2.5% significance level. With the null hypothesis H0: no trend and H1: increasing trend, can we reject the null hypothesis? Select "Yes" or "No" and click the "Check Answer" button when you've finished. Click here for the normal distribution table.

Yes
No

Yes, that's right, the null hypothesis cannot be rejected. t0 = 15070, n = 56, and ti = 461738, which gives
U = (461738 - (56 15070) / 2) / (15070 (56 / 12)1/2 ) = 1.2218. At a 2.5% significance level, the critical value is 1.96. Since this is greater than 1.2218, it suggests that the null hypothesis cannot be rejected.

No, in fact the null hypothesis cannot be rejected. t0 = 15070, n = 56, and ti = 461738, which gives
U = (461738 - (56 15070) / 2) / (15070 (56 / 12)1/2 ) = 1.2218. At a 2.5% significance level, the critical value is 1.96. Since this is greater than 1.2218, it suggests that the null hypothesis cannot be rejected.

Return