Q.1:
The Bradfield Container Company makes “cardboard” boxes for commercial
use (i.e. pizza boxes). One of the big issues for the company is the
set-up time required to change over from one order to the next. At one
particular machine, the set-up time is thought to be uniformly distributed
between 10 and 21 minutes. To test whether this is true or not, a random
sample of 180 set-ups on this machine was selected with set-up time rounded
to the nearest two-minute intervals. The following results occurred:
Set-up Time Frequency
10-11 minutes
13
12-13 minutes
23
14-15 minutes
40
16-17 minutes
44
18-19 minutes
40
20-21 minutes
20
a. What is the appropriate null and alternative hypothesis to be tested?
ANSWER:
The null and alternative hypotheses are:
: set-up times on this machine are uniformly distributed between 10 and 21
minutes
: set-up times are not uniformly distributed between 10 and 21 minutes.
b. Based on the null and alternative hypotheses stated in part a,
determine the expected frequencies for each set-up time category.
ANSWER:
Because the hypothesized distribution is a uniform distribution, the
expected number of values at each set-up time level is found by dividing
number of levels by the sample size:
Expected = 180/6 = 30
Set-up Time Frequency Expected
10-11 minutes
13
30
12-13 minutes
23
30
14-15 minutes
40
30
16-17 minutes
44
30
18-19 minutes
40
30
20-21 minutes
20
30
c. Assuming that we wish to conduct the hypothesis test at the .05 level,
what is the critical value that should be used?
ANSWER:
The critical value will be a chi-square value with degrees of freedom equal
to the number of levels minus one. The degrees of freedom is 6-1 = 5.
From the chi-square table with alpha = .05 and 5 degrees of freedom, the
critical value is = 11.0705.
Thus, if the test statistic > 11.0705, reject the null hypothesis.
d. Compute the test statistic and carry out the hypothesis test.
ANSWER:
The test statistic is found using:
.
The computations are shown as follows:
Set-up Time Frequency Expected
(O – E)2
(O – E)2/E
10-11 minutes
13
30 (13-30)2 = 289 289/30 = 9.63
12-13 minutes
23
30 (23-30)2 = 49 49/30 = 1.63
14-15 minutes
40
30 (40-30)2 = 100 100/30 = 3.33
16-17 minutes
44
30 (44-30)2 = 196 196/30 = 6.53
18-19 minutes
40
30 (40-30)2 = 100 100/30 = 3.33
20-21 minutes
20
30 (20-30)2 = 100 100/30 = 3.33
Total =27.78
Since = 27.78 > 11.07, we reject the null hypothesis and conclude that the
distribution of set-up times is not uniformly distributed.
Q.2:
Recently a survey was conducted involving customers of a fitness
center in Dallas, Texas. Participants were asked to indicate how often
they use the club by checking one of the following categories: 0 – 1 time
per week; 2-3 times per week; 4-5 times per week; more than 5 times. The
following data show how males and females responded to this question.
One of the purposes of the survey was to determine whether there is a
relationship between the gender of the customer and the number of visits
made each week.
a. State the appropriate null and alternative hypothesis.
ANSWER:
: number of visits is independent of gender
: number of visits is related to gender
b. What test procedure is appropriate to use to conduct this test?
ANSWER:
Because the data are observed frequencies in various discrete categories,
the appropriate test to use is the chi-square contingency analysis. This
involves determining the expected frequencies assuming the null hypothesis
is true and then comparing these expected frequencies, cell by cell, to the
observed frequencies. If these closely match, then the null hypothesis
should not be rejected. However, if there is a big difference between the
expected and observed cell frequencies, we should reject the null
hypothesis.
c. Conduct the hypothesis test using an alpha = .05 level.
ANSWER:
The test statistic for performing a chi-square contingency analysis is
computed as follows:
with d.f. = (r – 1)(c – 1).
The first step needed is to compute the expected cell frequencies. This is
done under the assumption that the null hypothesis is true and that the
proportion of customers in each use level is the same regardless of
gender. The expected frequencies can be computed using:
Expected Frequency = .
For example for the cell corresponding to males who use the center 0-1
times per week, we get:
Expected Frequency = = 54.3158.
The following shows the expected cell frequencies for each cell:
Next for each cell we compute: . For example in the cell for males and
use between 0 and 1, we get:
.
Below we show the computation for each cell:
0-1
2-3
4-5
over 5
Males
3.264433
0.822572
2.59585
0.531475
Females
1.853077
0.466939
1.473552
0.301696
The chi-square test statistic is computed by summing these values giving
=11.309.
The critical value for the contingency analysis test with (2-1) x (4-1) = 3
degrees of freedom and alpha equal .05 is found in the chi-square table to
be 7.8147.
The decision rule is:
If > 7.8147, reject the null hypothesis
Otherwise, do not reject.
Since =11.309 > 7.8147, reject the null hypothesis and conclude that use
rate is related to gender of the customer