Question

Professor Cornish studied rainfall cycles and sunspot cycles. (Reference: Australian Journal of Physics, Vol. 7, pp. 334-346.) Part of the data include amount of rain (in mm) for 6-day intervals. The following data give rain amounts for consecutive 6-day intervals at Adelaide, South Australia.

7 27 7 3 65 2 1 1 22 8 19 4 51 162 61 71 27 4 3 2 9 141 109 1 91 44 1 7 1 20 0 58 116 53 4 156 41 22 13 42 1 22 19 70 28 63 110 36 34 71 51 36 7 4 16 3 12 7 0 4 3 7 8 14

(i) Find the median. (Use 1 decimal place.)

(ii) Convert this sequence of numbers to a sequence of symbols A and B, where A indicates a value above the median and B a value below the median. Test the sequence for randomness about the median at the 5% level of significance. (a) State the null and alternate hypotheses. Ho: The numbers are randomly mixed about the mean. H1: The numbers are not randomly mixed about the mean. Ho: The numbers are not randomly mixed about the median. H1: The numbers are randomly mixed about the median. Ho: The numbers are randomly mixed about the median. H1: The numbers are not randomly mixed about the median. Ho: The numbers are not randomly mixed about the mean. H1: The numbers are randomly mixed about the mean. (b) Find the number of runs R, n1, and n2. Let n1 = number of values above the median and n2 = number of values below the median. R n1 n2 (c) In the case, n1 > 20, we cannot use Table 10 of Appendix II to find the critical values. Whenever either n1 or n2 exceeds 20, the number of runs R has a distribution that is approximately normal, as follows. $ \mu_R = \dfrac{2n_1n_2}{n_1+n_2}+1 $ and $ \sigma_R = \sqrt{\dfrac{(2n_1n_2)(2n_1n_2 - n_1 - n_2)}{(n_1+n_2)^2(n_1+n_2-1)}} $ We convert the number of runs R to a z value, and then use the normal distribution to find the critical values. Convert the sample test statistic R to z using the following formula. (Use 2 decimal places.) z = R – μR σR (d) The critical values of a normal distribution for a two-tailed test with level of significance α = 0.05 are -1.96 and 1.96 (see Table 5(c) of Appendix II). Reject H0 if the sample test statistic z ≤ -1.96 or if the sample test statistic z ≥ 1.96. Otherwise, do not reject H0.

Answer 1

Following is the ordered data set:

S.No.	Orderd data set
1	0
2	0
3	1
4	1
5	1
6	1
7	1
8	1
9	2
10	2
11	3
12	3
13	3
14	3
15	4
16	4
17	4
18	4
19	4
20	7
21	7
22	7
23	7
24	7
25	7
26	8
27	8
28	9
29	12
30	13
31	14
32	16
33	19
34	19
35	20
36	22
37	22
38	22
39	27
40	27
41	28
42	34
43	36
44	36
45	41
46	42
47	44
48	51
49	51
50	53
51	58
52	61
53	63
54	65
55	70
56	71
57	71
58	91
59	109
60	110
61	116
62	141
63	156
64	162

(a)

Since there are 64 data values so median will be average of 32nd and 33rd data values. That is

median = (16+ 19) /2 = 17.5

(ii)

Following table shows the symbol assigned to each data value:

Data	Symbol
7	B
27	A
7	B
3	B
65	A
2	B
1	B
1	B
22	A
8	B
19	A
4	B
51	A
162	A
61	A
71	A
27	A
4	B
3	B
2	B
9	B
141	A
109	A
1	B
91	A
44	A
1	B
7	B
1	B
20	A
0	B
58	A
116	A
53	A
4	B
156	A
41	A
22	A
13	B
42	A
1	B
22	A
19	A
70	A
28	A
63	A
110	A
36	A
34	A
71	A
51	A
36	A
7	B
4	B
16	B
3	B
12	B
7	B
0	B
4	B
3	B
7	B
8	B
14	B

(a)

Hypotheses are:

Ho: The numbers are randomly mixed about the median.

H1: The numbers are not randomly mixed about the median.

(b)

n1= 32, n2= 32

Following table shows the runs:

Runs
1	B
2	A
3	BB
4	A
5	BBB
6	A
7	B
8	A
9	B
10	AAAAA
11	BBBB
12	AA
13	B
14	AA
15	BBB
16	A
17	B
18	AAA
19	B
20	AAA
21	B
22	AAA
23	B
24	AAAAAAAAAAA
25	BBBBBBBBBBBB

So test statistics is

R = 25

The distribution of runs will be approximately normal with following mean and SD:

Hy = 2.32.32 32+32

=397

Therefore test statistics is

z = (25 -33) /3.97 = -2.02

Since z < -1.96 so we reject the null hypothesis. That is we cannot conclude that the numbers are randomly mixed about the median.