Question

For each successive presidential term from Franklin Pierce (the 14th president, elected in 1853) to William J. B. Clinton, first term, (42nd president), the party affiliation controlling the White House is shown below, where R designates Republican and D designates Democrat. (Reference: The New York Times Almanac.) Historical Note: We start this sequence with the 14th president because earlier presidents belonged to political parties such as the Federalist or Wigg party (not Democratic or Republican). In cases in which a president died in office or resigned, the period during which the vice president finished the term is not counted as a new term. The one exception is the case in which Lincoln (a Republican) was assassinated and the vice president Johnson (a Democrat) finished the term. D D R R D R R R R D R D R R R R D D R R D D D D D R R D D R R D R R R D Test the sequence for randomness at the 5% level of significance. Use the following outline. (b) Find the number of runs R, n1, n2, and the critical values, c1 and c2. Let n1 = number of Republicans and n2 = number of Democrats. (Enter NONE in any unused answer blanks.) R n1 n2 c1 c2 (c) In the case, where 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. NOTE, if n1 ≤ 20, enter NA in the answer blank.) z = R – μR σR z = (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. Sample z ≤ -1.96 -1.96 < sample z < 1.96 Sample z ≥ 1.96 Reject H0 Fail to reject H0 Reject H0 Using this decision process, do you reject or fail to reject H0 at the 5% level of significance? What is the P-value for this two-tailed test? (Use 4 decimal places. NOTE, if n1 ≤ 20, enter NA in the answer blank.) P-value =

Answer 1

Following table shows the runs:

Runs
1	DD
2	RR
3	D
4	RRRR
5	D
6	R
7	D
8	RRRR
9	DD
10	RR
11	DDDDD
12	RR
13	DD
14	RR
15	D
16	RRR
17	D

So we have

R = 17

and

n1 =20, n2=16

The critical values are: c1= 13 and c2= 24

Since R lies between critical values so we fail to reject the null hypothesis. That is we can conclude that sequence is random.