SOLUTION: Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 306 days. Number of Credit Frequency Applications (Num

Question 977793: Banner Mattress and Furniture Company wishes to study the number of credit applications
received per day for the last 306 days.

Number of Credit Frequency
Applications (Number of Days)
0: 51
1: 87
2: 74
3: 49
4: 31
5 or more: 14

To interpret, there were 51 days on which no credit applications were received, 87 days on which only one application was received, and so on. Would it be reasonable to conclude that the population distribution is Poisson with a mean of 4.0? Use the 0.02 significance level. Hint: To find the expected frequencies use the Poisson distribution with a mean of 4.0. Find the probability of exactly one success given a Poisson distribution with a mean of 4.0. Multiply this probability by 306 to find the expected frequency for the number of days in which there was exactly one application. Determine the expected frequency for the other days in a similar manner. (Round x2 to 3 decimal places. Round fe to 4 decimal places. Round (f0 - fe)2/fe) to 4 decimal places.)
H0: Distribution with Poisson with � = 4. H1: Distribution is not Poisson with � = 4

a. Decision rule: If _____?___ > reject H0.
b.
Applications f0(for each) fe(for each) (f0-fe)2/fe (for each)
1
2
3
4
5 or more
C.
(reject/do not reject) H0. There (is/isnot) sufficient evidence to reject the null that the distribution is Poisson with � =4.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Poisson Probability Distribution Table

k	P(X = k)
0	0.0183156389
1	0.0732625556
2	0.1465251111
3	0.1953668148
4	0.1953668148
5 or more	0.3711630648

-------------------------------------------------------
a)
Decision Rule: if the

(chi-squared) test statistic is larger than the chi-squared critical value, then reject H0.

alpha = 0.02

critical value = 13.3882

I used this calculator to get the critical value
Note: df = k-1 = 6-1 = 5. The area to the left of the critical value is 0.98
-------------------------------------------------------
b)

with 0 applications, we expect approximately...306*0.018315639 = 5.604585534 = 5.6046
with 1 applications, we expect approximately...306*0.073262556 = 22.418342136 = 22.4183
with 2 applications, we expect approximately...306*0.146525111 = 44.836683966 = 44.8367
with 3 applications, we expect approximately...306*0.195366815 = 59.78224539 = 59.7822
with 4 applications, we expect approximately...306*0.195366815 = 59.78224539 = 59.7822
with 5 applications, we expect approximately...306*0.3711630648 = 113.5758978288 = 113.5759

Those results above form the

column (expected frequencies column)

Now let's compute the

column

0 applications:
[(observed - expected)^2]/(expected) = ((51-5.6046)^2)/(5.6046) = 367.687674617279 = 367.6877

1 applications:
[(observed - expected)^2]/(expected) = ((87-22.4183)^2)/(22.4183) = 186.044257365188 = 186.0443

2 applications:
[(observed - expected)^2]/(expected) = ((74-44.8367)^2)/(44.8367) = 18.968792683003 = 18.9688

3 applications:
[(observed - expected)^2]/(expected) = ((49-59.7822)^2)/(59.7822) = 1.94465638333819 = 1.9447

4 applications:
[(observed - expected)^2]/(expected) = ((31-59.7822)^2)/(59.7822) = 13.8572189855843 = 13.8572

5 applications:
[(observed - expected)^2]/(expected) = ((14-113.5759)^2)/(113.5759) = 87.3016182201506 = 87.3016

Final Table

-------------------------------------------------------
c)

Refer to part b). Add up all of the numbers in the last column.

The

test statistic is 675.804

Since the chi-squared test statistic (675.804) is larger than the critical value (13.3882), this means we reject the null hypothesis.

There is sufficient evidence to reject the null.

So we conclude that this observed distribution does NOT fit with the expected distribution. Therefore, the values are NOT distributed from a Poisson distribution.
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