Question

Sign Runs Test/G

Consider the following sequence of 1 and 6:

6 111 66 111 66 11 6 11 666 11 6 1111 666 11 66

With a 0.01 significance level, we wish to test the claim that the above sequence was produced in a random manner. Answer each of the following questions

(a) The null hypothesis H0 is given by

A. n1=n2
B. ρ=0
C. β=0
D. The data are in an order that is not random
E. The data are in a random order
F. Median=0
G. G=0
H. r=0
I. None of the above.

(b) The null hypothesis H1 is given by

A. ρ≠0
B. G≠0
C. n1≠n2
D. The data are in an order that is not random
E. The data are in a random order
F. r≠0
G. Median ≠0
H. β≠0
I. None of the above.

(c) The number of runs G is

A. 27
B. 30
C. 15
D. 17
E. 21
F. 19
G. None of the above.

(d) What kind of test should you conduct

A. Both sign and goodness of fit tests
B. Sign test
C. Independence Test
D. Goodness of fit test
E. Runs test for randomness and the test statisc is G
F. Runs test for randomness and the test statistic is a z-score
G. None of the above.

(e) What is/are the critical(s) value(s)

A. The smallest one is -1.96 and the largest one is 1.96
B. The smallest one is 11 and the largest one is 24
C. The only critical value is 11
D. The negative one is -1.645 and the positive one is 1.645
E. The smallest one is -2.575 and the largest one is 2.575
F. The only critical value is 24
G. The only critical value is 23
H. The only critical value is -2.575
I. None of the above.

(f) The test statistic is

A. z=1.65 with n1=15 and n2=18.
B. G=15 with n1=15 and n2=18.
C. z=1.49 with μG=19.37 and σG=4.80.
D. z=1.56 with n1=18 and n2=15.
E. z=−.84 with μG=17.36 and σG=2.8.
F. z=1.56 with n1=15 and n2=18.
G. z=−.49 with μG=19.37 and σG=2.8.
H. G=15 with n1=18 and n2=15.
I. z=−.49 with G=17.36 and σG=4.80.
J. None of the above.

(g) The conclusion:

A. We reject H0 and then there is enough evidence to support the calim that the above sequence was in a random order
B. We fail to reject H0 and then there is enough evidence to support the calim that the above sequence was in a random order
C. We reject H0 and then there isn't enough evidence to support the calim that the above sequence was in a random order
D. We fail to reject H0 and then there isn't enough evidence to reject the calim that the above sequence was in a random order.
E. We fail to reject H0 and then there isn't enough evidence to support the calim that the above sequence was in a random order
F. We fail to reject H0 and then there is enough evidence to reject the calim that the above sequence was in a random order
G. We reject H0 and then there is enough evidence to reject the calim that the above sequence was in a random order
H. We reject H0 and then there isn't enough evidence to reject the calim that the above sequence was in a random order
I. None of the above.

Answer 1

(a)

E. The data are in random order.

(b)

D. The data are in an order that is not random.

(c)

There are 15 sequences of 6s and 1s so the number of runs:

C: G=15

(d)

There are 15 6s and 18 1s. So,

Since n1 and n2 are greater than 12 so correct option is

F: Runs test for randomness and the test statistic is a z-score

(e)

E: The smallest one is -2.575 and the largest one is 2.575

(f)

The distribution of G will be approximately normally distributed with mean and SD as follow:

The z-score is

Correct option:

E: z=−.84 with μG=17.36 and σG=2.8.

(g)

Since z lies between the critical values so we fail to reject the null hypothesis.

D: We fail to reject H0 and then there isn't enough evidence to reject the calim that the above sequence was in a random order.