Chapter 10, Problem 14RE

To determine

Perform a test of the null hypothesis at the level of significance of $\alpha =0.05$.

Answer to Problem 14RE

There is enough evidence to conclude that the numbers are equally likely to come up.

Explanation of Solution

Calculation:

The observed frequencies for 59 Powerball for a sequence of 755 draws are given.

Consider $p_1$, $p_2$,…, $p_{59}$ denotes the proportions or probabilities of occurrence of 59 different balls.

Step 1:

The hypotheses are:

Null Hypothesis:

$H_0:p_1=p_2=\mathrm{...}=p_{59}=\frac{1}{59}$.

That is, each numbers is equally likely to come up.

Alternative Hypothesis:

$H_1:\text{Not all }{p}_i\text{'}\text{s}\text{are same as }{H}_0\text{for }i=1,2,\mathrm{...},59.$

That is, each numbers is not equally likely to come up.

Step 2:

Expected frequencies:

The expected frequencies are defined as $E_1=n{p}_1$, $E_2=n{p}_2$ and so on, for the probabilities of $p_1$, $p_2$,… are specified by null hypothesis $H_0$ with the total number of trial n.

The total number of trial is obtained as $n=755$.

Thus, the expected frequencies are obtained as,

Number	Observed Frequencies	Expected frequencies
1	15	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
2	7	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
3	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
4	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
5	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
6	14	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
7	17	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
8	17	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
9	10	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
10	14	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
11	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
12	9	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
13	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
14	17	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
15	10	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
16	14	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
17	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
18	10	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
19	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
20	14	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
21	9	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
22	14	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
23	21	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
24	8	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
25	6	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
26	19	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
27	9	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
28	14	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
29	15	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
30	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
31	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
32	12	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
33	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
34	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
35	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
36	17	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
37	6	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
38	8	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
39	15	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
40	10	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
41	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
42	9	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
43	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
44	15	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
45	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
46	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
47	9	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
48	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
49	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
50	11	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
51	9	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
52	12	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
53	12	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
54	12	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
55	17	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
56	18	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
57	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
58	13	$\left(755\right)\left(\frac{1}{59}\right)=12.79$
59	16	$\left(755\right)\left(\frac{1}{59}\right)=12.79$

Here, all the expected frequencies are more than 5. Hence, the goodness-of-fit test can be applicable.

Step 3:

Level of significance:

The level of significance is given as 0.05.

Step 4:

Chi-Square statistic:

The chi-square statistic is obtained as ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ where $O_1,O_2,\mathrm{...},O_k$ be the observed frequencies and $E_1,E_2,\mathrm{...},E_k$ be the expected frequencies for k number of categories.

Now,

Number	Observed frequencies (O)	Expected frequencies (E)	${\left(O-E\right)}^2$	$\frac{{\left(O-E\right)}^2}{E}$
1	15	12.79	4.8841	0.381869
2	7	12.79	33.5241	2.621118
3	16	12.79	10.3041	0.805637
4	11	12.79	3.2041	0.250516
5	16	12.79	10.3041	0.805637
6	14	12.79	1.4641	0.114472
7	17	12.79	17.7241	1.385778
8	17	12.79	17.7241	1.385778
9	10	12.79	7.7841	0.608608
10	14	12.79	1.4641	0.114472
11	16	12.79	10.3041	0.805637
12	9	12.79	14.3641	1.123073
13	16	12.79	10.3041	0.805637
14	17	12.79	17.7241	1.385778
15	10	12.79	7.7841	0.608608
16	14	12.79	1.4641	0.114472
17	13	12.79	0.0441	0.003448
18	10	12.79	7.7841	0.608608
19	13	12.79	0.0441	0.003448
20	14	12.79	1.4641	0.114472
21	9	12.79	14.3641	1.123073
22	14	12.79	1.4641	0.114472
23	21	12.79	67.4041	5.270063
24	8	12.79	22.9441	1.793909
25	6	12.79	46.1041	3.604699
26	19	12.79	38.5641	3.015176
27	9	12.79	14.3641	1.123073
28	14	12.79	1.4641	0.114472
29	15	12.79	4.8841	0.381869
30	11	12.79	3.2041	0.250516
31	11	12.79	3.2041	0.250516
32	12	12.79	0.6241	0.048796
33	11	12.79	3.2041	0.250516
34	11	12.79	3.2041	0.250516
35	11	12.79	3.2041	0.250516
36	17	12.79	17.7241	1.385778
37	6	12.79	46.1041	3.604699
38	8	12.79	22.9441	1.793909
39	15	12.79	4.8841	0.381869
40	10	12.79	7.7841	0.608608
41	16	12.79	10.3041	0.805637
42	9	12.79	14.3641	1.123073
43	13	12.79	0.0441	0.003448
44	15	12.79	4.8841	0.381869
45	11	12.79	3.2041	0.250516
46	16	12.79	10.3041	0.805637
47	9	12.79	14.3641	1.123073
48	13	12.79	0.0441	0.003448
49	13	12.79	0.0441	0.003448
50	11	12.79	3.2041	0.250516
51	9	12.79	14.3641	1.123073
52	12	12.79	0.6241	0.048796
53	12	12.79	0.6241	0.048796
54	12	12.79	0.6241	0.048796
55	17	12.79	17.7241	1.385778
56	18	12.79	27.1441	2.122291
57	13	12.79	0.0441	0.003448
58	13	12.79	0.0441	0.003448
59	16	12.79	10.3041	0.805637
Total	755	$754.61\approx 755$	639.5619	50.00484

Thus, the value of ${\chi }^2$ is approximately 50.

Degrees of freedom:

It is known that under the null hypothesis $H_0$ the test statistic ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$ follows chi-square distribution with $k-1$ degrees of freedom for k number of categories, provided that all the expected frequencies are greater than or equal to 5.

In the given question there are 59 categories (balls are numbered from 1 to 59). Thus, $k=59$.

Hence, the degrees of freedom is $59-1=58$.

Thus, the degrees of freedom is 58.

Step 5:

Critical value:

In a test of hypotheses the critical value is the point by which one can reject or accept the null hypothesis.

Software procedure:

Step-by-step software procedure to obtain critical value using MINITAB software is as follows:

• Select Graph > Probability distribution plot > view probability.
• Select Chi -Square under distribution.
• In Degrees of freedom, enter 58.
• Choose Probability Value and Right Tail for the region of the curve to shade.
• Enter the Probability value as 0.05 under shaded area.
• Select OK.
• Output using MINITAB software is given below:

Hence, the critical value at $\alpha =0.05$ is 76.78.

  Rejection rule:

If the ${\chi }^2$ value is greater than or equal to critical value, that is ${\chi }^2\ge {\chi }_{\alpha ;k-1}^2$, then reject the null hypothesis $H_0$. Otherwise, do not reject $H_0$.

Step 6:

Conclusion:

Here, the ${\chi }^2$ value is less than the critical value.

That is, ${\chi }^2\left(=50\right)<{\chi }_{0.05,58}^2\left(=76.78\right)$

Thus, the decision is “fail to reject the null hypothesis”.

Thus, there is enough evidence to conclude that the numbers are equally likely to come up.