Question

Excel was asked to generate 50 Poisson random numbers with mean λ = 5.

x	Frequency
0	1
1	1
2	4
3	7
4	7
5	8
6	10
7	3
8	3
9	4
10	1
11	1

(a) Calculate the sample mean. (Round your sample mean value to 2 decimal places.)

Sample mean            

(b) Carry out the chi-square test at α = .05, combining end categories as needed to ensure that all expected frequencies are at least five. (Round your test statistic value to 3 decimal places and the p-value to 4 decimal places.)

Test statistic
d.f.
p-value

Answer 1

Date:--18/05/2019 Answer: -- a) Calculate the sample mean. To find the sample mean, prepare the following table. X Frequency, fifx CD CD 4 8 3 7 21 28 4 7

40 8 6 10 65) 7 3 21 8 3 24 9 4 10 1 10 Sum 50 260 From the above table, n-f 50, fx, -260 Then

2f 260 50 = 5.2 So, the sample mean is λ =-= 5.2、 it is very close to the desired value 5

b) Calculate the expected frequencies assuming a Poisson distribution with λ-5 1. First calculate Poisson probabilities using the formula, 2. Multiply P(x)by n to get the expected frequencies, with n 50 0.00674 0.337 CD 0.03369 1.6845 2 0.08422 4.211 0.14037 7.0185 0.17547 8.7735 4

0.175478.7735 0.175478.7735 0.14622 7.311 6 0.10444 5.222 0.06528 3.264 0.036271.8135 0.01813 0.9065 10 11 or more 0.0137 0.685

Null hypothesis, Ho:Generated numbers follows a Poisson distribution with mean2-5 Alternative hypothesis, H, : Generated numbers do not follow a Poisson distribution with mean 2-5 Obtain the chi-square statistic, and combining end categories as needed to ensure that all expected frequencies are at least 5 Cochran's Rule that expected frequencies (e) should be at least 5. To ensure this, it is necessary to combine the top classes to end up with only 7 classes. -2 6.2325 3 7.0185 4 8.7735

5 8.7735 6 7.311 7 5.222 -8 6.669

Compute the value of chi-square test statistic. 6 6.2325 0.0087 2 7 7.0185 0.0000 7 8.7735 0.3585 4 8 8.7735 0.0682 10 7.311 0.9890 6 7 3 5.222 0.9455 8 or more 9 6.669 0.8147 Then, the value of chi-square test statistic is

0.00870.000050.35850+0.068190.989020.945480.81475 3.18 Number of classes, c 7 Number of parameters estimated, m-0 Then, the degrees of freedom is df-c-m-1 -б At α = 0.05 with df-6 , the critical value from the chi-square table is, -12.592 05,6 The value of test statistic (3.18) is less than the critical value (12.592), so fail to reject the null hypothesis of Poisson distribution with λ-5 Excel's function CHIDIST (3.18,6) gives the p-value is 0.7859, which indicates an excellent fit. Generated numbers conform to a Poisson distribution with mean 15

Even if take the original mean, the decision will be same, because the sample mean is 5.2 , it is very close to the desired value 5. λ