Answer to Question #223314 in Statistics and Probability for Tom

Question

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

Answer 1

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

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

number of runs of $H, R_1=6$

number of runs of $T, R_2=5$

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

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

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

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

$$s^2=\dfrac{2n_1n_2(2n_1n_2-n_1-n_2)}{(n_1+n_2)^2(n_1+n_2-1)}$$

$$=\dfrac{2(8)(8)(2(8)(8)-8-8)}{(8+8)^2(8+8-1)}=\dfrac{56}{15}$$

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

For $\alpha=0.05,$ $z_c=1.96.$

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