A sample of size $n = 22$ is drawn from a normal population. Find the critical value $t_{\alpha/2}$ needed to construct a $98\%$ confidence interval.
I have tried everything I know how to figure out this t value for $98\%$ confidence interval and I cannot figure it out given so little information.
So from my notes I the value of $t= df = n-1$ = a value, which you then look up using the t distribution calculator. the distribution calculator only has values for the 90% 95% and 99% confidence levels.
$t=df=22-1= 21$ then input into t 99% calc = 2.831 and from here I dont know what value to subtract or where to go from here. I input .98 into the normal distribution calculator to see if that would give me a useful value but I think that further confused me.
this is from a graded worksheet so the correct answer is 2.518 but I have no idea how they arrived to that solution.
Yes you are right. You have $$df = n - 1 = 21$$ degrees of freedom for the $t-$distribution. Here we have a 2-sided test, so we should be looking at a right (or left) tailed area of $\frac{\alpha}{2} = \frac{0.02}{2}=0.01$ with $21$ degrees of freedom. Using the table below, we can extract the critical value for the desired confidence interval. (At the rows, you should be looking at $df = 21$ and at the columns you should be reading $0.01 $).