Question

Data 1. Clearly do it with nice handwriting you can use excel

Day	Speed (m/s)	Power (kW)
1	5	0.25
2	6	0.38
3	5	0.25
4	5	0.25
5	3	0.08
6	4	0.14
7	3	0.08
8	6	0.38
9	5	0.25
10	6	0.38
11	6	0.38
12	4	0.14
13	3	0.08
14	3	0.08
15	5	0.25
16	4	0.14
17	3	0.08
18	8	0.64
19	6	0.38
20	4	0.14
21	3	0.08
22	6	0.38
23	5	0.25
24	4	0.14
25	4	0.14
26	2	0.04
27	5	0.25
28	6	0.38
29	3	0.08
30	3	0.08
31	10	0.97

Q.1 Why is it “clearly not” Poisson? (a) Calculate summary statistics for Data1 and use them to argue that the distribution is not a Poisson distribution. (b) Use the method of moments to estimate what the parameter of a Poisson distribution would be to give you those values. (c) Collate how many values there are in the range 0-4, 5-9, 10-14, etc. and plot the resulting histogram. (d) Use the Chi-square test to determine whether the data fit a Poisson distribution.  Find a better distribution for Data1: (f) Show your reasoning for which distribution you choose, and remember, you may need to compare several distributions. (g) Give all measures of fit you use, including at least the Chi-square measure of fit, calculated as you did in the question above. You are encouraged to use other measures also.

Answer 1

(a)

Speed (m/s)
Mean	4.677419
Standard Error	0.301919
Median	5
Mode	3
Standard Deviation	1.681014
Sample Variance	2.825806
Kurtosis	2.090492
Skewness	1.08827
Range	8
Minimum	2
Maximum	10
Sum	145
Count	31

Since for Poisson distribution, mean=variance. Here we observed that sample mean=4.677419>sample variance=2.825806 hence the distribution is not a Poisson distribution.

(b)

(c)

(d)

Since frequency of the class limit 10-14 is 1 so we marge this class with 5-9.

Class limit	O	Probability	E=31*Probability	(O-E)^2/E
0--4	15	0.49879155	15.46253797	0.013836
5 and more	16	0.50120845	15.53746203	0.013769
			Total	0.027605

where E=expected frequency, O=observed frequency