method 4 Chi square in statestical R prog

D E M O N S T R AT I O N
PROBLEM 9.4
A small business has 37 employees. Because of the uncertain demand for its prod-
uct, the company usually pays overtime on any given week. The company assumed
that about 50 total hours of overtime per week is required and that the variance on
this figure is about 25. Company officials want to know whether the variance of over-
time hours has changed. Given here is a sample of 16 weeks of overtime data (in
hours per week). Assume hours of overtime are normally distributed. Use these data
to test the null hypothesis that the variance of overtime data is 25. Let a = .10.
57
46
48
63
56
53
51
53
52
44
55
51
44
44
48
50
Solution
HYPOTHESIZE:
STEP 1. This test is a two-tailed test. The null and alternative hypotheses are
H 0 : s 2 = 25
H a : s 2 Z 25
TEST:
2. The test statistic is
STEP
x 2 =
STEP
STEP
(n - 1)s 2
s 2
3. Because this test is two tailed, a = .10 must be split: a>2 = .05
4. The degrees of freedom are 16 - 1 = 15. The two critical chi-square values are
x 2(1 - .05),15 = x 2.95,15 = 7.26093
x 2.05.15 = 24.9958
The decision rule is to reject the null hypothesis if the observed value of the test
statistic is less than 7.26093 or greater than 24.9958.
STEP 5. The data are as listed previously.
STEP 6. The sample variance is
s 2 = 28.0625
The observed chi-square value is calculated as
x 2 =
(n - 1)s 2
(15)(28.0625)
=
= 16.84
25
s 2
ACTION:
STEP 7. This observed chi-square value is in the nonrejection region because
x 2.95,15 = 7.26094 6 x 2 observed = 16.84 6 x 2.05,15 = 24.9958. The company fails to reject the
null hypothesis. The population variance of overtime hours per week is 25