Test the following sequence for randomness. $$\text { B A A A A B A B B B B A A A }$$ You may use the following abbreviated table of critical values for runs (at = .05). Reject the null hypothesis if the number of runs is less than or equal to the smaller value or greater than or equal to the
A) Do not reject because the test value 6 is not between the critical values 3 and 11
B) Do not reject because the test value 5 is between the critical values 3 and 12
C) Do not reject because the test value 6 is between the critical values 3 and 12
D) Reject because the test value 5 is not between the critical values 3 and 11

Question

Test the following sequence for randomness. $$\text { B A A A A B A B B B B A A A }$$ You may use the following abbreviated table of critical values for runs (at = .05). Reject the null hypothesis if the number of runs is less than or equal to the smaller value or greater than or equal to the 
A) Do not reject because the test value 6 is not between the critical values 3 and 11 
B) Do not reject because the test value 5 is between the critical values 3 and 12 
C) Do not reject because the test value 6 is between the critical values 3 and 12 
D) Reject because the test value 5 is not between the critical values 3 and 11

Quizplus · Accepted Answer

To test for randomness using runs test, we need to count the number of runs in the sequence. A run is a sequence of consecutive numbers or values that are increasing or decreasing, for example, "B A A A A B" has 2 runs, one of A's and one of B's, while "A B A A B B A" has 4 runs: AB, A, B, and A.

To apply the runs test, we need to compute the expected number of runs, which is given by the formula:

$$ E(R) = \frac{2n_1 n_2}{n_1 + n_2} + 1 $$

where $n_1$ and $n_2$ are the number of 1's and 2's in the sequence, respectively, and $n = n_1 + n_2$ is the total sample size. In our case, the sequence has 7 A's and 6 B's, so $n_1 = 7$ and $n_2 = 6$. Therefore,

$$ E(R) = \frac{2 \cdot 7 \cdot 6}{7 + 6} + 1 = 4.545 $$

The expected number of runs is a theoretical value that we use as a reference to compare with the observed number of runs, which we count in the sequence. To count the runs, we start with the first value and compare it with the next value, and so on until the end of the sequence. Each time the value changes from 1 to 2 or from 2 to 1, we count a run. The final count is the observed number of runs.

In our case, we have:

$$ 	ext{ B A A A A B A B B B B A A A } $$
$$ 	ext{ 1 2 2 2 2 1 2 1 1 1 1 2 2 2 } $$

So, the observed number of runs is 6. To determine whether this is a random sequence or not, we need to compare it with the critical values for runs, which depend on the sample size and the significance level. In this case, we have a sample size of 13 and a significance level of .05, so we look up the critical values in the table given:

\begin{array}{|c|c|c|}
\hline
n & 	ext{Lower CV} & 	ext{Upper CV} \
\hline
\hline
10 & 2 & 8 \
\hline
11 & 2 & 9 \
\hline
12 & 3 & 10 \
\hline
13 & 3 & 11 \
\hline
14 & 4 & 12 \
\hline
\end{array}

We see that the lower critical value for 13 observations is 3, and the upper critical value is 11. Since our observed number of runs is 6, which is between these two values, we cannot reject the null hypothesis of randomness. Therefore, the answer is C: Do not reject because the test value 6 is between the critical values 3 and 12.

Test the Following Sequence for Randomness B A A A A B A B B B B A A A \text { B A A A A B A B B B B A A A } B A A A A B A B B B B A A A

Test the Following Sequence for Randomness $\text { B A A A A B A B B B B A A A }$