Question

(c) when $\alpha = 0.02$ and $n = 5$,
$\chi_{\text{left}}^{2}=\square$
$\chi_{\text{right}}^{2}=13.277$

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 5\), then \(df=5-1 = 4\).

Step2: Find the left - tailed \(\chi^{2}\) value

For a left - tailed test with \(\alpha=0.02\) and \(df = 4\), we use the property that \(\chi_{left}^{2}=\chi_{1-\alpha,df}^{2}\). So we look up the \(\chi^{2}\) value with \(\alpha'=1 - 0.02=0.98\) and \(df = 4\) in the \(\chi^{2}\) distribution table.
From the \(\chi^{2}\) distribution table, \(\chi_{0.98,4}^{2}=0.484\).

Answer:

\(\chi_{left}^{2}=0.484\)