Question

The Springfield Emergency Medical Service keeps records of emergency telephone calls. A study of 150 five-minute time intervals resulted in the distribution of number of calls as follows. For example, during 18 of the 5-minute intervals, no calls occurred. Use the chi-square goodness-of-fit test and α = .01 to determine whether this distribution is Poisson.
Number of Calls

(per 5-minute interval)
Frequency
0
18
1
28
2
47
3
21
4
16
5
11
6 or more
9

Answer 1

The null and the alternative hypothesis is given by,

Ho: The distribution of number of calls follows Poisson.

Ha: The distribution of number of calls does not follows Poisson.

First, calculate the mean :

Number of calls (x)	Frequency (f)	f*x
0	18	0
1	28	28
2	47	94
3	21	63
4	16	64
5	11	55
6 or more	9	54
Total	150	358

First we calculate hypothesized probabilities related to each class by using the poisson distribution mean equal to 2.4.

Calculate the expected frequency of each class:

  

Here N = 150

So,

Now, the observed (Oi) and expected frequencies (Ei) shown in the following table:

Number of calls (x)	Oi	Ei
0	18	13.605
1	28	32.655
2	47	39.195
3	21	31.35
4	16	18.81
5	11	9.03
6 or more	9	5.355
Total	150	150

Test statistic can be calculated as:

Thus the test statistic is 10.385.

The degrees of freedom is = k – 1 = 7 – 1 = 6

Using chi-square table, and the degrees of freedom 6, the critical value obtained at the significance level 0.01 is 16.812.

Since the calculated value (10.385) is less than critical value (16.812), the decision is fail to reject null hypothesis at the significance level 0.01.

Therefore, it can be concluded that distribution of number of calls follows Poisson.