Answer to Question #223314 in Statistics and Probability for Tom

Question #223314
A sequence of heads(H) and tails (T) in tossing of a coin 16 times are given below HTTTHTHTHHTHTTHH count the number of runs and also test whether the heads and the tails occur in a random order alpha =0.05
1
Expert's answer
2021-08-05T13:48:48-0400

For the given sequence, the sample size n=16,n=16, number of heads n1=8,n_1=8, number of tails n2=8.n_2=8.


(H)(TTT)(H)(T)(H)(T)(HH)(T)(H)(TT)(HH)(H)(TTT)(H)(T)(H)(T)(HH)(T)(H)(TT)(HH)

number of runs of H,R1=6H, R_1=6

number of runs of T,R2=5T, R_2=5

So that number of runs is R=R1+R2=6+5=11.R=R_1+R_2=6+5=11.

H0:H_0: the sequence was produced in a random manner.

H1:H_1: the sequence was not produced in a random manner.



μ=2n1n2n1+n2+1=2(8)(8)8+8+1=9\mu=\dfrac{2n_1n_2}{n_1+n_2}+1=\dfrac{2(8)(8)}{8+8}+1=9

s2=2n1n2(2n1n2n1n2)(n1+n2)2(n1+n21)s^2=\dfrac{2n_1n_2(2n_1n_2-n_1-n_2)}{(n_1+n_2)^2(n_1+n_2-1)}

=2(8)(8)(2(8)(8)88)(8+8)2(8+81)=5615=\dfrac{2(8)(8)(2(8)(8)-8-8)}{(8+8)^2(8+8-1)}=\dfrac{56}{15}

z=Rμs=11956151.0351z=\dfrac{R-\mu}{s}=\dfrac{11-9}{\sqrt{\dfrac{56}{15}}}\approx1.0351

For α=0.05,\alpha=0.05, zc=1.96.z_c=1.96.

Since z=1.0351<1.96=zc,z=1.0351<1.96=z_c, we accept H0.H_0. It means that the heads and tails occur in random order or it can be said that the coin is unbiased at the α=0.0\alpha=0.0 significance level.



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