Question

I he accompanying data are the times to failure (in milions per cycle) of high-speed turbine engine bearings made out of two ditferent compounds. These were taken from "Analysis of Single Classincation Experiments Based on Censored Samples trom the Two parameter Weibull Distribution" bry J.I. McCool (The Journal of Statistical Planning and Inference, 1979 Compound 1 Compound 2 3.19 5.53 4.26 4.47 5.60 9.30 4.53 9.92 4.67 12.51 4.69 12.95 15.21 6.79 16.04 9.37 16.84 12.75 a) Find the 0.34 quantile of the Compound 1 failure times. b Give the coordinates of the two lower-left points that would appear on a normal plot of the compound 1 data (c) Make back-to-back stem-and-leaf plots for comparing the life length properties of bearings made from Compounds 1 and 2. (d Make (to scalel side-by-side boxplots for comparing the life lengths for the two compounds Mark numbers on the plots indicating the locations of their main features e Compute the sample meais and standard deviations of the two sets of lifetimes. r)Describe what your answers to parts (c(d, and (e)above indicate about the life lengths of these turbine bearings.

Answer 1

ANSWER:

A)

Step 1: Order the data from smallest to largest. The data in the question is already in ascending order.
Step 2: Count how many observations you have in your data set.
Step 3: Insert your values into the formula:
ith observation = q (n + 1)
ith observation = .84 (10 + 1) = 9.24
Answer: The ith observation is at 9.24, so we round down to 9. The 9th number in the set is 16.04, which is the number where 84 percent of the values fall below it.

Answer b):

To obtain the two lower left points in normal plot for data.

Find the mean of data. In this case the mean is

(3.03+5.53+5.60+9.30+9.92+12.51+12.95+15.21+16.04+16.84)/10 =10.693

Now you can select any two points which are less than 10.693. We take the lowest two values i.e. 3.03 and 5.53

If we wanted to know the numbers on the right side of normal plot we had to select the numbers greater than 10.693.

Answer c):

Stem Leaf (0.11 (1) Leaf (o.1) 665424 7 5 56 3 9 39 l,00 1.1 712 59 13 14 15 2 16 08

Answer d):

Compound 1:

16 14 12 10 4

Compund 2:

Answer e):

12 4 29 10

1 Compund 1 Compund 2 (X-Mean)12 (X-Mean)2 3.19 3.03 58.72 8.18 3.2 4 5.53 26.66 4.26 25.94 4.47 4.53 9.3 1.94 2.31 9.92 0.6 4.67 4.69 12.51 3.3 1.85 5.09 12.95 5.78 0.07 10 15.21 20.4 6.79 0.55 16.04 28.59 9.37 11.02 37.79 12 16.84 12.75 44.89 13 14 209.030 76.47 15 Mean 10.693 Mean 6.050 7.647 16 Variance 20.903 Variance SD 17 SD 4.57198 2.765321 18

Answer f):

The mean average failure time of compund 1 is higher than compund 2. The data are widely spreaded for compund 1 than compund 2.