Homework 5 – Solutions by Lan Lan

 

This homework is based on the following example:

 

Bearings for high-speed turbine engines can be made of many different types of compounds.  The times to fatigue failure (in units of millions of cycles) for high-speed bearings were measured for five different types of compounds.  Compounds I and II are made of an alloy containing lead.  Compounds III, IV, and V are made of an alloy containing iron.  Compounds I and III were made by US companies and compounds II, IV and V were made by foreign companies.  The data are listed below:

Compound	Failure Time	Compound	Failure Time	Compound	Failure Time	Compound	Failure Time	Compound	Failure Time
I	3.03	II	3.19	III	3.46	IV	5.88	V	6.43
I	5.53	II	4.26	III	5.22	IV	6.74	V	9.97
I	5.6	II	4.47	III	5.69	IV	6.9	V	10.39
I	9.3	II	4.53	III	6.54	IV	6.98	V	13.55
I	9.92	II	4.67	III	9.16	IV	7.21	V	14.45
I	12.51	II	4.69	III	9.4	IV	8.14	V	14.72
I	12.95	II	5.78	III	10.19	IV	8.59	V	16.81
I	15.21	II	6.79	III	10.71	IV	9.8	V	18.39
I	16.04	II	9.37	III	12.58	IV	12.28	V	20.84
I	16.84	II	12.75	III	13.41	IV	25.46	V	21.51

 

 

 

 

 

 

 

 

 

 

The goal of this homework is to compare the average failure times for the bearings made from the 5 different compounds.

 

1.          Use PROC MEANS to calculate the average times to fatigue failure for each of the 5 compounds.  Which compound had the longest average time to fatigue failure?  The shortest time?

 

The output of PROC MEANS shows compound V has the highest time to fatigue failure (mean = 14.7) while compound II has the lowest fatigue failure time (mean = 6.05).

 

2.          Use PROC UNIVARIATE to construct side-by-side box plots.  Based on these plots, do you notice any obvious differences between the 5 compounds?

 

By checking the box plot, compound V has higher times and compound II has lower times.  Also, the range between the 25^th percentile and the 75^th percentile are different for each compound.  Compounds I and V have bigger ranges and compound II has a smaller range.

 

3.          Use PROC GLM to compute the analysis of variance for these data.  Based on the overall F-test, is there a significant difference between the treatment means for the 5 different compounds?  State the null and alternative hypotheses for the test, and report the F-statistic and p-value.

 

Ho:  mI = mII = mIII = mIV = mV

Ha:  not all means are equal

 

F = 5.02 and p-value = 0.002

 

At the 5% significance level, reject the null hypothesis and conclude that there IS a significant difference between the average failure times for the 5 compounds.

 

4.          Based on what you know about the compounds, come up with 2 pre-planned comparisons for the means.  Define the contrasts associated with these comparisons and use the contrast statement in PROC GLM to test your hypotheses.  Be sure to state the null and alternative hypotheses for your test, report the p-value, and state your conclusion in non-statistical terms.

 

Are the average failure times for alloys containing lead significantly different than for alloys containing iron?

 

Ho:  (mI + mII)/2 = (mIII + mIV + mV)/3

Ha: (mI + mII)/2 not equal (mIII + mIV + mV)/3

 

F = 4.3 and p-value = 0.0439

At the 5% significance level, reject the null hypothesis and conclude that there is a significant difference between the average failure times between lead alloys and iron alloys.

 

Are the average failure times for compounds made by US companies significantly different than for compounds made by foreign companies?

 

Ho:  (mI + mIII)/2 = (mII + mIV + mV)/3

Ha: (mI + mIII)/2 not equal (mII + mIV + mV)/3

 

F = 0.16 and p-value = 0.689

Fail to reject the null hypothesis and conclude that there is no difference between the average failure times for US made and foreign made compounds.

 

5.          Now, use multiple comparison methods to compare all possible pairs of treatment means.  Using PROC GLM, do the multiple comparisons using the F-protected LSD, Tukey, and Duncan methods.  For each method, which means are declared to be the same, and which methods are declared to be significantly different.

 

LSD:  Same - I&V  /   I, III, & IV  /   II, III, & IV

          Different -  I&II   /  II & V  /  III & V   /   IV & V

 

Tukey:  Same – I, IV, & V   /   I, II, III, & IV

            Different – V & III  /  V & II

 

Duncan:  Same – I & V  /  II, III, & IV  /  I, III, & IV

               Different – V & II  /  I & II  /  V & III  / V & IV

6.          Based on your findings write a short (one paragraph) summary of your results, and make recommendations about which compound(s) you would use if your goal was to create ball-bearings with the longest time to fatigue failure.

 

Compounds I and V clearly had the highest times to fatigue failure.  Also, there appears to be no difference between the compounds made by US companies and foreign companies.  There is a difference between the lead alloy and iron alloy; and, the iron alloy compounds had higher average fatigue failure times than the lead alloy compounds.  If I were to recommend compounds to use to create ball-bearings with the longest time to fatigue failure, I would recommend compounds I and V.  Compound V could be the best, while compound I is also acceptable.  The decision between the two could be based on other matters such as price and desired material.