Question

A computer random number generator was used to generate 750 random digits (0,1,...,9). The observed frequences of the digits are given in the table below.

0	1	2	3	4	5	6	7	8	9
81	62	74	82	76	75	70	66	80	84

Test the claim that all the outcomes are equally likely using the significance level α=0.05α=0.05.
The expected frequency of each outcome is E=

The test statistic is χ2=

The p-value is

Is there sufficient evidence to warrant the rejection of the claim that all the outcomes are equally likely?
A. No
B. Yes

Answer 1

Solution:

Here, we have to use chi square test for goodness of fit.

Null hypothesis: H₀: all the outcomes are equally likely.

Alternative hypothesis: H_a: all the outcomes are not equally likely.

We are given level of significance = α = 0.05

Test statistic formula is given as below:

Chi square = ∑[(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

We are given

N = 10

Degrees of freedom = df =N – 1 = 10 – 1 = 9

α = 0.05

Critical value = 16.91898

(by using Chi square table or excel)

Calculation tables for test statistic are given as below:

X	O	E	(O - E)^2/E
0	81	75	0.48
1	62	75	2.253333333
2	74	75	0.013333333
3	82	75	0.653333333
4	76	75	0.013333333
5	75	75	0
6	70	75	0.333333333
7	66	75	1.08
8	80	75	0.333333333
9	84	75	1.08
Total	750	750	6.24

Test statistic = Chi square = ∑[(O – E)^2/E] = 6.24

P-value = 0.715678517

(By using Chi square table or excel)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that all the outcomes are equally likely.

Is there sufficient evidence to warrant the rejection of the claim that all the outcomes are equally likely?

Answer:

A. No