The number of times a machine broke down each week was observed over a period of
ID: 3311736 • 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
11
24
30
24
7
4
Determine the estimated mean of the frequency distribution.
The average number of breakdowns per week over this period is
nothing.
(Type an integer or a decimal.)
Calculate the test statistic.
chi squared2equals=nothing
(Round to two decimal places as needed.)
Determine the critical value,
chi Subscript alpha Superscript 22.
chi Subscript alpha Superscript 22equals=nothing
(Round to three decimal places as needed.)
Choose the correct rejection region below.
A.
chi squared less than or equals chi Subscript alpha Superscript 222
B.
chi squared greater than or equals chi Subscript alpha Superscript 222
C.
chi squared greater than chi Subscript alpha Superscript 22>2
D.
chi squared less than chi Subscript alpha Superscript 2
Draw a conclusion.
Reject
Fail to reject
the null hypothesis. Based on the results, it is reasonable to assume the distribution of breakdowns
follows
does not follow
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
the test statistic, assuming
the distribution of the variable differs from the normal distribution.
the expected frequencies are all equal to 5.
the distribution of the variable is the normal distribution.
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.
At
alphaequals=0.10,
what is the correct conclusion?
the null hypothesis. There is
insufficient
sufficient
evidence to conclude that the population distribution of breakdowns is not Poisson.
Number of
Breakdowns
0
1
2
3
4
5 or More
Number of
weeks
11
24
30
24
7
4
Explanation / Answer
Using minitab:
Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: Number of week
Test Contribution
Category Observed Proportion Expected to Chi-Sq
1 11 0.166667 16.6667 1.9267
2 24 0.166667 16.6667 3.2267
3 30 0.166667 16.6667 10.6667
4 24 0.166667 16.6667 3.2267
5 7 0.166667 16.6667 5.6067
6 4 0.166667 16.6667 9.6267
N DF Chi-Sq P-Value
100 5 34.28 0.000
test stat X2 =34.28
and critical value for 5 degree of freedom at0.1 level=9.2364
as test stat falls in critical region we reject that above follow poisson