LAB SESSION 12
ANALYZING THE POPULATION VARIANCE
INTRODUCTION: In
this lab we will present the hypothesis test for the standard deviation for a
normal population. When sample data are
skewed, just one outlier can greatly affect the standard deviation. It is very important, especially when using
small samples, that the sampled population be normal; otherwise the procedures
are not reliable. However, unlike the
analysis for the mean you will not have convenient computer commands to help
you.
To
use Illustration 9-16 as an example of using Excel to aid in completion of the
hypothesis test, let's assume the 12 samples tested yielded the following data:
165 172 180
189 181 173
167 192 212
169 198 171
Enter
the data into Column A.
Determine
the descriptive statistics by the following:
Choose: Tools > Data Analysis > Descriptive Statistics
This
gives you the following:
|
165 |
|
Column1 |
|
|
172 |
|
|
|
|
180 |
|
Mean |
180.75 |
|
189 |
|
Standard
Error |
4.152627865 |
|
181 |
|
Median |
176.5 |
|
173 |
|
Mode |
#N/A |
|
167 |
|
Standard
Deviation |
14.38512489 |
|
192 |
|
Sample
Variance |
206.9318182 |
|
212 |
|
Kurtosis |
0.37412902 |
|
169 |
|
Skewness |
1.005775368 |
|
198 |
|
Range |
47 |
|
171 |
|
Minimum |
165 |
|
|
|
Maximum |
212 |
|
|
|
Sum |
2169 |
|
|
|
Count |
12 |
|
|
|
Confidence
Level(95.0%) |
9.139876928 |
From
the table we see that n = 12, s = 14 and we calculate C2* = 21.56
To
calculate the p-value, activate Cell B1.
Choose: Insert > fx > Statistical > CHIDIST >
OK
Enter: C2*:
21.56
Df: 11 >
OK
This
gives you the value 0.0280.
Recall
that the manufacturer claims “shelf life” is normally distributed.
Why
is this important?
What
decision should be made?
Does
your conclusion match that for Illustration 9-16?
ASSIGNMENT: Do
Exercises 9.116 and 9.119 in your text.
Use the following data for 9.116
31.6 31.9 32.6 31.9
31.5 32.5 32.0
32.2 31.9 32.0
32.2 31.8 31.8 32.3
31.1 31.8 31.5
31.7 31.8 31.8