Chapter D, Problem 1E

To determine

Test whether the data provide that the numbers of bills are not equally valued among three different servers.

Answer to Problem 1E

There is sufficient evidence to conclude that the numbers of bills are not equally valued among three different servers.

Explanation of Solution

Denote $p_1,p_2,\text{ and }{p}_3$ be the proportions for A, B, and C, respectively.

The null and alternative hypotheses can be written as follows:

Null hypothesis:

$H_0:p_1=p_2=p_3=\frac{1}{3}.$

Alternative hypothesis:

$H_a:\text{At least one }{p}_i\left(i=1,2,3\right)\text{ is not specified as in }{H}_0.$

The total observed counts for A, B, and C are 60, 65, and 32, respectively.

By adding all the observed counts, the total numbers of observations are 157.

If all the proportions are equally likely, the expected count for each answer is given below:

$\begin{array}{c}\text{E}\left(p_i\right)=n\cdot p_i\\ =157\left(\frac{1}{3}\right)\\ \approx 52.3\end{array}$

The ${\chi }^2-\text{test statistic}$ is calculated as follows:

$\begin{array}{c}{\chi }^2={\displaystyle \sum \frac{{\left(\text{Observed}-\text{Expected}\right)}^2}{\text{Expected}}}\\ =\left[\frac{{\left(60-52.3\right)}^2}{52.3}+\frac{{\left(65-52.3\right)}^2}{52.3}+\frac{{\left(32-52.3\right)}^2}{52.3}\right]\\ =1.134+3.084+7.879\\ \approx 12.09\end{array}$

There are 3 attributes in the test. Therefore, the degrees of freedom for the Chi-square distribution for the table are given below:

$\begin{array}{c}\text{df}=k-1\\ =3-1\\ =2\end{array}$

Step-by-step procedure to obtain the p-value using StatKey software:

• Click on ${\chi }^2$, enter degrees of freedom as 2.
• Click Ok.
• Select right tail at the top-left of the screen.
• Double click on the cutoff value on the X-axis.
• Enter cutoff value as 12.09.
• Click Ok.

Output obtained using StatKey software is given below:

From the above result, the p-value for the test is 0.002.

Decision rule:

If the p-value is less than or equal to the level of significance, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Conclusion:

Here, the p-value is 0.002.

Therefore, the p-value is less than any common levels of significance such as 0.10, 0.05, and 0.01.

Hence, reject the null hypothesis.

Therefore, there is sufficient evidence to conclude that the numbers of bills are not equally valued among three different servers.