Question

In a 17 month period the powerball was drawn from a collection of 35 balls numbered 1-35. Total of 150 drawings were made. For the purpose of this exercise we grouped numbers into five categories. test the hypothesis that each of the categories is equally likely. Use the 0.025 level of signifigance and the critical value method with the table.

Category:|1-7| |8-14| |15-21| |22-28| |29-35|

Observed: |21 | 35 | | 33 | | 27 | | 34 |

Expected frequencies for each category????

1)What are expected frequencies?

2)Compute the value of x^2. Round the answer to three decimal places

3)How many degrees of freedom are there?

4)State the null and alternate hypotheses

H0= p1=p2=p3=p4=p5= _____

Some or all of the actual probabilities (differ or ____) from the specifies by H0

5)Find the critical value. Round the answer to three decimal places.

6)determine whether to reject H0

7)State a conclusion

There (is or is not) enough evidence to conclude that the distribution differs from what was expected.

Answer 1

We created 5 groups of equal size. Let be the probability that a random drawing falls in categories 1,2,3,4,5 respectively.

We want to test the hypothesis that each of the categories is equally likely. The means

1) Since there are 150 drawings, the expected frequencies in each of the categories is 150*(1/5) = 30

Ans: The expected frequencies are

Category	Observed (fo)	Expected (fe)
1-7	21	30
8-14	35	30
15-21	33	30
22-28	27	30
29-35	34	30

2) The chi-square statistics is

ans: The value of is 4.667

3) There are 5 classes. Hence the degrees of freedom is 5-1=4

ans: There are 4 degrees of freedom

4) The hypotheses are

5) We want to test the hypotheses at a significance level , which is the area under the right tail. We want the value

From the chi-square table for df=4 and area under the right tail=0.025, we get a value of 11.143

ans: The critical value is 11.143

6) We will reject the null hypothesis if the test statistics is greater than the critical value.

Here, the test statistics is 4.667 and it is not greater than the critical value 11.143. Hence we do not reject the null hypothesis.

ans: We fail to reject

7) There is not enough evidence to conclude that the distribution differs from what was expected.