The number of times a machine broke down each week was observed over a period of
ID: 2908775 • Letter: T
Question
The number of times a machine broke down each week was observed over a period of 100 weeks and recorded as shown in the table below. Complete parts a and b below.
Number of
Breakdowns
0
1
2
3
4
5 or More
Number of
weeks
9
22
32
26
6
5
a. Using alphaequals?0.10, perform a? chi-square test to determine if the population distribution of breakdowns follows the Poisson probability distribution.
What is the null? hypothesis, Upper H 0??
A.
The distribution of breakdowns differs from the expected distribution.
B.
The distribution of breakdowns follows the Poisson probability distribution.
C.
The distribution of breakdowns follows the normal probability distribution.
D.
The distribution of breakdowns follows a uniform probability distribution.
What is the alternate? hypothesis, Upper H 1??
A.
The distribution of breakdowns follows a uniform probability distribution.
B.
The distribution of breakdowns follows the Poisson probability distribution.
C.
The distribution of breakdowns differs from the claimed or expected distribution.
D.
The distribution of breakdowns follows the normal probability distribution.
Determine the estimated mean of the frequency distribution. The average number of breakdowns per week over this period is
?(Type an integer or a? decimal.)
Calculate the test statistic.
(Round to two decimal places as? needed.)
Determine the critical? value, (Round to three decimal places as? needed.)
Choose the correct rejection region below.
A.
chi squared less than or equals chi Subscript alpha Superscript 2
B.
chi squared greater than or equals chi Subscript alpha Superscript 2
C.
chi squared less than chi Subscript alpha Superscript 2
D.
chi squared greater than chi Subscript alpha Superscript 2
Draw a conclusion.
?
the null hypothesis. Based on the? results, it is reasonable to assume the distribution of breakdowns
?
does not follow
follows
the Poisson probability distribution for making business decisions.
b. Determine the? p-value and interpret its meaning.
?p-valueequals
nothing ?(Round to three decimal places as? needed.)
Interpret the? p-value.
The? p-value is the probability of observing a test statistic
?
less than
equal to
greater than
the test? statistic, assuming
?
the distribution of the variable is the same as the given distribution.
at least one expected frequency differs from 5.
the distribution of the variable differs from the given distribution.
the expected frequencies are all equal to 5.
the distribution of the variable is the normal distribution.
the distribution of the variable differs from the normal distribution.
At alphaequals?0.10, what is the correct? conclusion?
?
Reject
Do not reject
the null hypothesis. There is
?
insufficient
sufficient
evidence to conclude that the population distribution of breakdowns is not Poisson.
Explanation / Answer
Solution:
Here, we have to use Chi square test for goodness of fit. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The distribution of breakdown follows the Poisson probability distribution.
Alternative hypothesis is given as below:
Alternative hypothesis: H1: The distribution of breakdowns differs from the claimed or expected distribution.
Test statistic formula is given as below:
Chi square = ?[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
Calculation tables for this test are given as below:
x
f
xf
Poisson Prob.
Exp.
0
9
0
0.118837
11.88373
1
22
22
0.253123
25.31234
2
32
64
0.269576
26.95765
3
26
78
0.191399
19.13993
4
6
24
0.10192
10.19201
5 or more
5
25
0.065143
6.514341
Total
100
213
1
100
Estimated Mean = ?xf/?f = 213/100 = 2.13
Estimated mean = 2.13
(Poisson probabilities are calculated using Poisson table/excel.)
O
E
(O - E)
(O - E)^2/E
9
11.88373
-2.88373
0.699771765
22
25.31234
-3.31234
0.433448519
32
26.95765
5.04235
0.943156897
26
19.13993
6.86007
2.458763454
6
10.19201
-4.19201
1.724188638
5
6.514341
-1.514341
0.352027728
Total
6.611357001
Chi square = ?[(O – E)^2/E] = 6.611357001
Test statistic = 6.61
N = 6
DF = N – 1 = 6 – 1 = 5
? = 0.10
Critical value = 9.236(by using chi square table/excel)
Rejection region:
Chi squared greater than Chi square alpha
(Correct Answer: D)
Conclusion:
Here, test statistic chi square value is less than critical value, so we do not reject the null hypothesis.
We do not reject the null hypothesis. Based on the results, it is reasonable to assume the distribution of breakdowns follows the Poisson probability distribution for making business decisions.
P-value = 0.251 (by using chi square table/excel)
P-value > ? = 0.10
The p-value is the probability of observing a test statistic is greater than the test statistic, assuming the distribution of the variable is the same as the given distribution.
So, we do not reject the null hypothesis
At ? = 0.10, we do not reject the null hypothesis; there is insufficient evidence to conclude that the population distribution of breakdowns is not Poisson.
x
f
xf
Poisson Prob.
Exp.
0
9
0
0.118837
11.88373
1
22
22
0.253123
25.31234
2
32
64
0.269576
26.95765
3
26
78
0.191399
19.13993
4
6
24
0.10192
10.19201
5 or more
5
25
0.065143
6.514341
Total
100
213
1
100
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