Question

In Exercises 7.5 to 7.8 , the categories of a categorical variable are given along with the observed counts from a sample. The expected counts from a null hypothesis are given in parentheses. Compute the \(\chi^{2}\) -test statistic, and use the \(\chi^{2}\) -distribution to find the p-value of the test. $$ \begin{array}{l} \hline \begin{array}{l} \text { Category } \\ \text { Observed } \\ \text { (Expected) } \end{array} & \begin{array}{c} \mathrm{A} \\ 38(30) \end{array} & \begin{array}{c} \mathrm{B} \\ 55(60) \end{array} & \begin{array}{c} \mathrm{C} \\ 79(90) \end{array} & \begin{array}{c} \mathrm{D} \\ 128(120) \end{array} \\ \hline \end{array} $$

Short answer

The \(\chi^{2}\)-test statistic for the given data is 4.45. Based on this and the degrees of freedom (3), we can say that p-value is greater than 0.05. Therefore, we cannot reject the null hypothesis.

Step-by-step solution

Identify the observed and expected counts

The first step involves identifying the observed and expected counts for each category. For example, the observed count for category A is 38 and the expected count is 30. The same is done for categories B, C, and D.

Compute the \(\chi^{2}\) -test statistic

The second step is to compute the \(\chi^{2}\)-test statistic. This is calculated using the formula \(\chi^{2} = \sum((O-E)^2 / E)\), where O is the observed frequency and E is the expected frequency. For category A, the \(\chi^{2}\) contribution would be \((38-30)^2 / 30 = 2.13\). Repeat this process for categories B, C and D, and sum all these four values to obtain the total \(\chi^{2}\)-value which is \(2.13 + 0.42 + 1.37 + 0.53 = 4.45 \).

Find the p-value of the test

Finally, refer to a \(\chi^{2}\) distribution table or use a statistical computation tool to find the p-value that corresponds to the total \(\chi^{2}\)-value under degree of freedom = 3 (The degree of freedom is calculated as 4 - 1, because there are 4 categories). You will find that this \(\chi^{2}\)-value is greater than the critical value for significance level 0.05, implying that the null hypothesis can be accepted. However, the exact p-value requires more detailed tables or computation software.