Question

Using the following data (already sorted), use a goodness of fit test to test whether it comes from an exponential distribution. The exponential distribution has one parameter, its mean, μ (which is also its standard deviation). The exponential distribution is a continuous distribution that takes on only positive values in the interval (0,∞). Probabilities for the exponential distribution can be found based on the following probability expression:

.

Use 10 equally likely cells for your goodness of fit test.

Data Display

0.2 0.4 0.5 0.5 0.7 0.8 1.0 1.2 1.2 1.2

1.4 1.5 1.5 1.6 1.7 1.7 1.7 1.8 1.8 1.9

2.0 2.3 2.6 2.7 2.7 2.8 2.8 2.8 2.8 2.8

2.8 2.9 2.9 3.0 3.0 3.0 3.2 3.2 3.2 3.4

3.4 3.5 3.6 3.6 3.7 3.8 3.9 3.9 3.9 4.0

4.1 4.1 4.2 4.3 4.5 4.5 4.5 4.6 4.7 4.8

4.8 4.9 4.9 4.9 5.0 5.0 5.1 5.1 5.1 5.3

5.3 5.3 5.3 5.4 5.4 5.4 5.4 5.5 5.5 5.5

5.6 5.6 5.6 5.7 5.7 5.8 5.8 5.8 5.9 5.9

6.0 6.0 6.2 6.2 6.2 6.3 6.3 6.3 6.3 6.4

6.6 6.6 6.6 6.6 6.7 6.8 6.9 6.9 6.9 7.0

7.0 7.1 7.2 7.3 7.3 7.4 7.5 7.5 7.6 7.6

7.7 7.8 7.8 7.9 8.0 8.0 8.0 8.1 8.1 8.1

8.2 8.3 8.4 8.4 8.4 8.5 8.5 8.6 8.6 8.7

8.7 8.8 9.0 9.1 9.1 9.2 9.3 9.4 9.5 9.6

9.6 9.6 9.8 9.9 9.9 9.9 10.0 10.1 10.2 10.5

10.6 10.7 10.7 10.8 10.9 10.9 11.0 11.0 11.4 11.5

11.7 11.8 11.8 11.9 12.0 12.0 12.1 12.1 12.3 12.3

12.3 12.3 12.6 12.9 13.1 13.3 13.3 13.4 13.5 13.6

13.9 14.0 14.2 14.2 14.3 14.3 14.4 15.0 15.0 15.2

15.6 15.6 15.7 15.7 15.7 15.9 16.0 16.3 16.4 16.5

16.5 16.6 16.6 16.7 17.2 17.3 17.3 17.4 17.7 17.9

18.6 18.8 19.9 19.9 19.9 20.0 20.1 20.3 20.4 21.0

21.3 21.5 22.2 23.3 23.5 23.9 24.3 24.8 25.5 25.5

25.6 25.8 27.5 28.2 30.9 35.7 36.3 37.2 40.9 52.8

Descriptive Statistics:

Variable    N   Mean

Exp?      250 9.974

Answer 1

We test the hypothesis,

Null Hypothesis : The given data comes from an exponential distribution, versus

Alternative Hypothesis : The given data does not come from an exponential distribution.

We need to form a frequency table to solve the problem. Here, I have divided the observations into ten class intervals and write the corresponding frequencies.

Class Interval	Frequency
0 - 5.5	77
5.5 - 11	89
11 - 16.5	43
16.5 - 22	23
22 - 27.5	10
27.5 - 33	3
33 - 38.5	3
38.5 - 44	1
44 - 49.5	0
49.5 - 55	1

Here, we are required to estimate the parameters of the distribution if it is not provided. We are already given the parameter, i.e., mean is equal to 9.974.

If X follows exponential distribution, then the pdf of X is given by,

Then,

Hence,

Expected frequency,

The integral can be easily calculated using calculator.

"a" and "b" are the lower and upper limits of the class interval respectively.

Class Interval	Observed Frequency ,	P(a < X < b)	Expected Frequency,
0 - 5.5	77	0.424	106	7.934
5.5 - 11	89	0.244	61	12.852
11 - 16.5	43	0.141	36	1.3611
16.5 - 22	23	0.081	20	0.45
22 - 27.5	10	0.047	12	0.333
27.5 - 33	3	0.027	7	3.2667
33 - 38.5	3	0.016	4
38.5 - 44	1	0.009	2
44 - 49.5	0	0.005	1
49.5 - 55	1	0.003	1

If the frequency values are less than five, chi-square test does not work well. Hence, we have to pool the values less than 5 to get a larger value.

For the pooled values,

The test statistic, under the null hypothesis is

From the above table, U = 26.1968

Critical value, ( from the chi - square table)

where is the level of significance = 0.05

Degrees of freedom = k - r - 1, where k is the number of classes after pooling(i.e.,k=6)and r is the number of parameters estimate(i.e., r=0).

We reject the null hypothesis if  

Here, .

Hence, we reject the null hypothesis and conclude that exponential is not a good fit for the above data.