Question

Banner Mattress and Furniture Company wishes to study the number of credit applications received per day for the last 395 days.

The sample information is reported below.

Number of Credit	Frequency
Applications	(Number of Days)
0		10
1		26
2		58
3		79
4		79
5 or more		143

To interpret, there were 10 days on which no credit applications were received, 26 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 0.1 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 395 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.

H₀: Distribution with Poisson with µ = 4.

H₁: Distribution is not Poisson with µ = 4.

State the decision rule. Use 0.1 significance level. (Round your answer to 3 decimal places.)

Compute the value of chi-square. (Round f_e to 4 decimal places. Round (f_o - f_e)²/f_e) to 4 decimal places. Round χ² to 4 decimal places.)

What is your decision regarding H₀?

Answer 1

Here the hypothesis are

H₀: Distribution with Poisson with µ = 4.

H₁: Distribution is not Poisson with µ = 4.

so here observed values are given in the table. Now we have to calculate the expected values.

Here if µ = 4

p(x) = e^-4 4^x/x!

Here total days = 395

so here expected values are

Expected numbers for credit applications 0 = POISSON (x = 0, µ = 4) * 395 = e^-4 4⁰/0! * 395 = 7.23

Expected numbers for credit applications 1 = POISSON (x = 1, µ = 4) * 395 = e^-4 4¹/1! * 395 = 28.94

Expected numbers for credit applications 2 = POISSON (x = 2, µ = 4) * 395 = e^-4 4²/2! * 395 = 57.88

Expected numbers for credit applications 3 = POISSON (x = 3, µ = 4) * 395 = e^-4 4³/3! * 395 = 77.17

Expected numbers for credit applications 4 = POISSON (x = 4, µ = 4) * 395 = e^-4 4⁴/4! * 395 = 77.17

for expected numberss for credit applications 5 or more = 395 - (7.23 + 28.94 + 57.88 + 77.17 + 77.17) = 146.61

Here

=

Number of credit applications	frequency	Expected	X^2
0	10	7.23	1.0570
1	26	28.94	0.2984
2	58	57.88	0.0003
3	79	77.17	0.0434
4	79	77.17	0.0434
5 or more	143	146.61	0.0889
Sum	395	395.00	1.5313

= 1.53

Here degree of freedom = n - 1 = 6 -1 = 5 (here as we know the paremeter before hand so we are not discounting that factor from degree of freedom calculation)

_critical = CHIINV(0.10,5) = 9.24

Here,

<  _critical  so we would fail to reject the null hypothesis and conclude that Distribution with Poisson with µ = 4