Question

A computer random number generator was used to generate 550 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
58	55	45	50	53	50	57	57	46	79

Test the claim that all the outcomes are equally likely using the significance level ?=0.05?=0.05.
The expected frequency of each outcome is E=E=  
The test statistic is ?2=?2=  
The p-value is  
Is there sufficient evidence to warrant the rejection of the claim that all the outcomes are equally likely?
A. Yes
B. No

Answer 1

As there are a total of 550 random digits generated, therefore the expected frequency for each of the digit should be 550/10 = 55 for each of them to be equally likely.

Now the chi square test statistic here is computed as:

Therefore 15.0545 is the test statistic value here.

Now degrees of freedom = Number of digits - 1 = 9

Therefore from chi square distribution tables, the p-value here is computed as:

Therefore 0.0894 is the p-value here.

As the p-value here is 0.0894 > 0.05 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here.

Therefore we dont have sufficient evidence here to warrant the rejection of the claim that all the outcomes are equally likely