Question

8-80 Consider the suspension rod diameter measurements described in Exercise 8-40 (use the modified data of 8-40 as given in Chapter 8 homework problems), compute a 99% prediction interval on the diameter of the next rod tested. Compare the length of the prediction interval with the length of the 99% CI on the population mean.

A machine produces metal rods used in an automobile suspension system. A random sample of 15

rods is selected, and the diameter is measured. The resulting data (in millimeters) are as follows:

8.20 8.25 8.18 8.25 8.22 8.20 8.28 8.28 8.18 8.24 8.25 8.25 8.17 8.26 8.22

Answer 1

First we need to find the mean and SD of data:

Descriptive statistics
	X
count	15
mean	8.2287
sample standard deviation	0.0360
sample variance	0.0013
minimum	8.17
maximum	8.28
range	0.11

So we have

-= 8.2287, n-15, s = 0.0013

Since there 15 data values in the sample so degree of freedom is df=15-1=14 and critical value of t for 99% confidence interval, using excel function "=TINV(0.01,14)" is 2.977.

The 99% prediction interval is

$\bar{x}\pm t_{\alpha/2,df}s\cdot \sqrt{1+\frac{1}{n}}=8.2287\pm 2.977\cdot 0.0013\cdot \sqrt{1+\frac{1}{15}} =8.2287\pm 0.0040=(8.2247, 8.2327)$

-------------------------

The 99% CI is

Here we have T-8.2287, s0.0013, n-15 Since populaiton SD is unknown so t critical value will be used. Level of significance: α-0.01 Critical value: 2.977 Degree of freedom df -n-1-14 The requried confidence interval is: tt8.2287+ (0.0013*2.977/sqrt (15)) -8.2287 + 0.001 -(8.228,8.23) Excel function for critical value: TINV(0.01,14) Hence, the required confidence interval is (8.228,8.23)

Prediction interval is wider.