Chapter 16, Problem 6SR

To determine

State whether there is a difference in the number of personal checking account transactions among the four branches.

Answer to Problem 6SR

There is a difference in the number of personal checking account transactions among the four branches.

Explanation of Solution

The test hypotheses are given as follows:

Null hypothesis:

$H_0:$ There is no difference in the number of personal checking account transactions among the four branches.

Alternate hypothesis:

$H_1:$ There is a difference in the number of personal checking account transactions among the four branches.

Here, the test statistic H follows the chi-square distribution with $k-1$ degrees of freedom where k is the number of populations and H is the test statistic.

In order to formulate the decision rule, the chi-square is used.

In this context, the number of populations is 4 and the significance level is 0.01.

Degrees of freedom for populations:

$\begin{array}{c}k-1=4-1\\ =3\end{array}$

From Appendix B, Table B.7 Critical Values of Chi-Square:

Procedure:

1. 1. In the table, first locate 3 in the first z column.
2. 2. Locate the value of 0.01 in the second row.
3. 3. Locate the value by the intersection of the row and column values, which gives the critical value.

The critical value for 3 df for 0.01 significance level is 11.345.

Decision rule:

• If $H>11.345$, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

In this context, the test statistic for the Kruskal–Wallis test is denoted as H.

Test statistic, H:

 $H=\frac{12}{n\left(n+1\right)}\left[\frac{{\left({\displaystyle \sum R_1}\right)}^2}{n_1}+\frac{{\left({\displaystyle \sum R_2}\right)}^2}{n_2}+\mathrm{...}+\frac{{\left({\displaystyle \sum R_k}\right)}^2}{n_k}\right]-3\left(n+1\right)$.

Where k is the number of populations,

$R_1,R_2,\mathrm{...}{R}_k$ are ranks of samples 1, 2,…k.

$n_1,n_2,\mathrm{...}{n}_k$ are sizes of samples 1, 2,…k.

n is the combined number of observations for all samples.

In this context, it is necessary to rank all the observations.

The following table represents the ranks of all the samples:

Branch E	Rank 1	Branch W	Rank 2	Branch G	Rank 3	Branch S	Rank 4
208	17	91	5	302	19	99	7
307	20	62	1	103	9.5	116	11
199	16	86	3	319	21	189	15
142	13	91	5	340	22	103	9.5
91	5	80	2	180	14	100	8
296	18					131	12

Here, the sum of Rank 1 is given below:

$\begin{array}{c}{\displaystyle \sum R_1}=17+20+16+13+5+18\\ =89\end{array}$

The sum of Rank 2 is given below:

$\begin{array}{c}{\displaystyle \sum R_2}=5+1+3+5+2\\ =16\end{array}$

The sum of Rank 3 is given below:

$\begin{array}{c}{\displaystyle \sum R_3}=19+9.5+21+22+14\\ =85.5\end{array}$

The sum of Rank 4 is given below:

$\begin{array}{c}{\displaystyle \sum R_4}=7+11+15+9.5+8+10\\ =62.5\end{array}$

The combined number of observations for all samples is given below:

$\begin{array}{c}n=n_1+n_2+n_3+n_4\\ =6+5+5+6\\ =22\end{array}$

The test statistic will be obtained as given below:

Substitute the corresponding values in order to get the test statistic.

$\begin{array}{c}H=\frac{12}{22\left(22+1\right)}\left[\frac{{\left(89\right)}^2}{6}+\frac{{\left(16\right)}^2}{5}+\frac{{\left(85.5\right)}^2}{5}+\frac{{\left(62.5\right)}^2}{6}\right]-3\left(22+1\right)\\ =0.023715\left[1,320.167+51.2+1,462.05+651.0417\right]-3\left(23\right)\\ =82.6354-69\\ =13.6354\end{array}$

Thus, the test statistic is 13.6354.

Conclusion:

Here, the test statistic is greater than the critical value.

Therefore, by the decision rule, reject the null hypothesis.

Therefore, there is a difference in the number of personal checking account transactions among the four branches.