Question

for parts a and b, use technology to estimate the following
a) the critical value of t for a 99% confidence interval with df = 6
b) the critical value of t for a 95% confidence interval with df = 106
a) what is the critical value of t for a 99% confidence interval with df = 6?
3.71 (round to two decimal places as needed)
b) what is the critical value of t for a 95% confidence interval with df = 106?
(round to two decimal places as needed)

Explanation:

Step1: Recall the formula for confidence interval

The confidence interval formula for \(t -\)distribution is \(\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(t_{\alpha/2}\) is the critical value. For a \(95\%\) confidence interval, \(\alpha = 1 - 0.95=0.05\), and \(\alpha/2=0.025\). The degrees of freedom \(df = 106\).

Step2: Use technology (e.g., t - table or statistical software)

Using a t - table or software (such as R: qt(1 - 0.025, 106) or Excel: T.INV.2T(0.05, 106)).
When we calculate, we find that \(t_{0.025,106}\approx1.98\)

Answer:

\(1.98\)