Question

Exercises 4 and 5. Use the runs test for randomness and complete a hypothesis test with all 5 steps listed

in the module summary for this unit to answer each question.

Exercise 5. Xs and Ys. A researcher listed a sequence of Xs and Ys as shown in the table below. The list is read from left to right and row by row, top to bottom. Another researcher claims that the sequence of Xs and Ys is random. At α = 0.05, can you reject the other researcher’s claim?

X

Y

X

X

X

Y

Y

X

X

X

X

X

Y

Y

Y

X

X

Y

Y

X

X

X

X

X

X

Y

X

X

Y

Y

X

Y

X

X

X

X

Y

X

Y

X

Answer 1

Claim: The sequence of Xs and Ys is random.

The null and alternative hypothesis are:

H₀: The sequence of Xs and Ys is random.

H₁: The sequence of Xs and Ys is not random.

Here there are 19 total runs that is R = 19

n₁ = number of Xs = 26 and n₂ = number of Ys = 14

The formula of test statistics is,

$Z = \frac{R - E(R)}{\sqrt{V(R)}}$

Where E(R) = expected value of run

$E(R) = \frac{2n_{1}n_{2}}{n_{1}+n_{2}} + 1 = \frac{2*26*14}{26+14} +1 =19.2$

V(R) = variance of the run

$V(R) = \frac{2n_{1}n_{2}(2n_{1}n_{2}-n_{1}-n_{2})}{(n_{1}+n_{2})^{2}(n_{1}+n_{2}-1)} = \frac{2*26*14(2*26*14-26-14)}{(26+14)^{2}(26+14-1)}$

V(R) = 8.026667

Test statistics Z follows normal distribution.

$Z = \frac{R - E(R)}{\sqrt{V(R)}} = \frac{19-19.2}{\sqrt{8.026667}} = -0.07$

The critical value using normal distribution is,

Alpha = 0.05

alpha/2 = 0.025

1-(alpha/2) = 0.975

Using z table the critical value for the area 0.975 is 1.96

Decision rule: If |Z test statistics| > Z critical value then reject the null hypothesis otherwise fail to reject the null hypothesis.

Here |-0.07| = 0.07 is not greater than 1.96, so fail to reject the null hypothesis.

Fail to reject the null hypothesis means in favor of the null hypothesis.

Conclusion: There is sufficient evidence to support the researcher's claim that the sequence of Xs and Ys is random.

That is we can not reject the researchers claim.