Question

find the critical value $t_{c}$ for the confidence level $c = 0.98$ and sample size $n = 22$.
click the icon to view the t - distribution table.
$t_{c}=\square$ (round to the nearest thousandth as needed)

Explanation:

Step1: Calculate the degrees of freedom

The degrees of freedom \(df=n - 1\). Given \(n = 22\), then \(df=22-1 = 21\).

Step2: Calculate the significance level \(\alpha\)

The confidence level \(c = 0.98\). Using the formula \(\alpha=1 - c\), we get \(\alpha=1 - 0.98=0.02\).

Step3: Find \(\frac{\alpha}{2}\)

\(\frac{\alpha}{2}=\frac{0.02}{2}=0.01\).

Step4: Look up the \(t\) - value in the \(t\) - distribution table

We look for the row with \(df = 21\) and the column corresponding to the area of \(\frac{\alpha}{2}=0.01\) (one - tail area) in the \(t\) - distribution table.

Answer:

\(t_{c}=2.518\)