Defects –
Is the day of the week important?
Question #6 TGIF
Intent of the Question
The primary goals of this question were to assess a
student's ability to
(1) determine whether the conditions for applying a
particular inference procedure are satisfied;
(2) calculate the p-value for a binomial test and
draw an appropriate conclusion; and
(3) calculate a test statistic based on signed rank
data; and
(4) use simulation results to draw an appropriate
conclusion.
Scoring
This question is scored in four sections.
Section 1 consists of parts (a) and (b),
section 2 consists of part (c),
section 3 consists of part (d), and
section 4 consists of part (e).
Each of the four sections is scored as essentially
correct (E), partially correct (P), or incorrect (I).
Question #6
In a large manufacturing company
every item produced is inspected
for defects and will go through a
repair process if there are serious
defects. Management wanted to
investigate whether items
produced on Mondays are more
likely to require repairing than
items produced on the midweek
day Wednesday. A random
sample of 9 weeks from the past
5 years was taken and the
number of items which required
repairing for the 9 weeks are
shown in the table below.
Week
Monday
Wednesday
Difference
More Repairing
on Monday
Signed Rank
of
|Difference|
A
17
18
−1
NO
−1
B
14
17
−3
NO
−2
C
19
12
7
YES
6
D
21
17
4
YES
3
E
18
13
5
YES
F
19
10
9
YES
G
24
4
20
YES
H
25
17
8
YES
I
22
16
6
YES
Part (a) – Section 1
A boxplot of the differences in number of
items which required repairing on
Monday and Wednesday in this sample
of 9 weeks is shown below.
Explain why management determined
that the matched pair t-test of
H0: μ difference = 0
Ha: μ difference > 0
(where μ difference is the mean of the
differences in the number of produced
items which required repairing on
Monday and on Wednesday for all weeks
in the past 5 years) was not appropriate
after seeing the boxplot of the
differences.
Part(b) –Section 1
A different possible set of hypotheses for this
investigation could be
H0: p = 0.5
Ha: p > 0.5
where p is the proportion of weeks where Monday
had more produced items which required repairing
than Wednesday.
Explain why the one-sample proportion z-test
would not be appropriate for these data.
Section 1 Solutions
Solution Part (a)
Solution Part (b)
One of the conditions for the
matched pair t-test is a check for
normality which for a small data
set (n < 30 or n < 25) is usually a
graphical check for the absence
of extreme skewness and outliers.
The boxplot identifies the
difference value of 20 as an
outlier, and hence, it may not be
reasonable to believe that the
differences are from a population
which has an approximately
normal distribution.
The normal approximation to
the binomial distribution in the
one sample proportion z-test is
generally used when np ≥ 10
and n(1 – p) ≥ 10 (or ≥ 5)
where p is the hypothesized
value in the null hypothesis.
For these data np = 9(.5) = 4.5
is too small.
Scoring Section 1
Essentially correct (E) if the response includes the following two
components:
1) Concludes that it is not reasonable to assume the differences are
from a population that is normally distributed because of the outlier.
2) Calculates the value of np in part (b) AND concludes that np is too
small to use the normal approximation to the binomial distribution.
Partially correct (P) if the response includes only one of the two
components listed above.
Incorrect (I) if the response does not meet the criteria for E or P.
Part (c) Section 2
A sign test of the
hypotheses
H0: p = 0.5
Ha: p > 0.5
can be used when the onesample proportion z-test is
not appropriate. The test
statistic for the sign test is X
= the number of weeks of
the 9 sampled weeks where
more items required
repairing on Monday than
Wednesday.
Assuming that the null
hypothesis is true (that
Mondays and Wednesdays
are equally likely to have the
most produced items which
require repairing), calculate
the p-value P(X ≥ 7) and use
this p-value to provide the
conclusion of the sign test
for a significance level of 𝛼 =
0.05.
Part (c) – Section 2 Solution
Assuming that the null hypothesis is true, X will
follow a binomial distribution with n = 9 and p =
0.5, with P(X ≥ 7) = 1 – P(X ≤ 6) = 0.08984375
(from the calculator). Since the p-value of
approximately 0.09 is larger than 𝛼 = 0.05, the
null hypothesis is not rejected. Thus, from the
sign test there is not sufficient evidence to
conclude that the proportion of weeks where
more items which require repairing is greater on
Monday than on Wednesday.
Section 2 is scored as follows:
Essentially correct (E) if the response includes the following two
components:
1)The p-value is correctly calculated from the binomial distribution.
2) A correct conclusion is provided in context, with justification based on
linkage between the p-value and the given 𝛼 = 0.05.
Partially correct (P) if the response includes only one of the two components
listed above.
Incorrect (I) if the response does not meet the criteria for E or P.
Notes:
The conclusion must be related to the alternative hypothesis.
If the p-value is incorrect, section 2 is scored as P if the response includes
proper linkage and a conclusion in context consistent with that p-value.
If the p-value is correct and greater than 0.05, wording that states or implies
that the null hypothesis is accepted lowers the score from E to P in section 2.
Part (d) Section 3
A signed rank test uses both the ranks of the absolute value of the
differences and the signs of the differences to test the hypotheses
H0: The distributions of the numbers of items which require repairing
for Mondays and Wednesday are the same.
Ha: The distribution of the numbers of items which require repairing
for Mondays is shifted to the right of the distribution of the number of
items which require repairing on Wednesdays.
The test statistic for the signed rank test is the sum of the positive
ranks.
d) Calculate the test statistic for the signed rank test by completing the
signed rank of difference column in the table at the beginning of the
problem and then adding up the positive ranks.
Section 3
Part (d) Solution
Part (d) Scoring
Week
Monday
Wednesday
Difference
More Repairing
on Monday
A
B
C
D
E
F
G
H
I
17
14
19
21
18
19
24
25
22
18
17
12
17
13
10
4
17
16
−1
−3
7
4
5
9
20
8
6
NO
NO
YES
YES
YES
YES
YES
YES
YES
Signed Rank
of
|Difference|
−1
−2
6
3
4
8
9
7
5
Thus, the sum of the positive ranks is 6 +
3 + 4 + 8 + 9 + 7 + 5 = 42.
Essentially correct (E) if the response
includes the following two components:
1) The signed ranks of the differences are
correct.
2) The signed rank test statistic is correctly
calculated from the given signed ranks.
Partially correct (P) if the response includes
only one of the two components listed
above.
Incorrect (I) if the response does not meet
the criteria for E or P.
Part (e) Section 4
Under the assumption that the null hypothesis of the
distributions of the numbers of items requiring repairing for
Mondays and Wednesday are the same, 1000 simulations were
performed and the signed rank test statistic was calculated for
each simulation. The frequency table below provide the
frequencies for these 1000 simulated signed rank test statistics.
Sign Rank Statistic Values
0
1
2
3
4
5
⋯
40
41
42
43
44
45
Frequency
2
3
4
5
8
15
⋯
17
10
6
4
3
1
Part (e) – Section 4
Part (e)
Based on the value of the
signed rank test statistic
calculated in part (d) and the
distribution of the 1000
simulated signed rank test
statistics above, what should
be the conclusion for the
manufacturing company for
comparing the number of
items requiring repairing on
Mondays and Wednesdays?
Part (e) solution
The proportion of the 1,000 simulations in
which the signed rank test statistic was at
least as extreme as the observed test
statistic value of 42 is equal to (6 + 4 + 3 +
1)/1,000 = 0.014. Because the simulated pvalue of 0.014 is small (much less than the
conventional significance levels of 𝛼 =
0.10, 𝛼 = 0.05, or 𝛼 = 0.01), the null
hypothesis that the distributions of the
numbers of items which require repairing
for Mondays and Wednesday are the same
is rejected. Thus, based on the signed rank
test, there is strong evidence to believe
the distribution of the numbers of items
which require repairing for Mondays is
shifted to the right of the distribution of
the number of items which require
repairing on Wednesdays.
Section 4 is scored as follows:
Essentially correct (E) if the response includes the following two components:
The value of the test statistic in part (d) is used to correctly calculate the simulated pvalue from the provided simulation results.
A correct conclusion is provided in context, with justification based on the linkage
between the size of the p-value and the conclusion.
Partially correct (P) if the response includes only one of the two components listed
above.
Incorrect (I) if the response does not meet the criteria for E or P.
Notes:
The conclusion must be related to the alternative hypothesis, and must reference the
shifting in the distribution.
If the p-value is incorrect, section 2 is scored as P if the response includes proper
linkage and a conclusion in context consistent with that p-value.
If the p-value is correct and less than 0.05, wording that states or implies the
alternative hypothesis is proven lowers the score from E to P in section 4.
Scoring
Each essentially correct (E) step counts as 1 point. Each partially correct (P)
step counts as ½ point.
4
Complete Response
3
Substantial Response
2
Developing Response
1
Minimal Response
If a response is between two scores (for example, 2½ points), score down.