In: Statistics and Probability
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.
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.