For a confidence level of 99% with a sample size of 22, find the critical t value for a confidence interval for one mean. Round to the nearest thousandth.

Answer 1

Hello, thank you for taking the time to post your question!

You can use either technology like Microsoft Excel or a t-table to find the value on this one! But the underlying idea is that we’re looking at the 99% confidence level and then df = n – 1 = 22 – 1 = 21, so the corresponding t-value here should be 2.831.

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

alpha = 1-.99=.01

df = n-1 = 22-1 =21

for 2 tails

critical t value = 2.831

found by using a t table

reading off the row for df=21

and the column for .01

a very, very rough check would be

use the empirical rule and a modified form

with 68-95-99 instead of 68-95-99.7

then

99% fall within 3 standard deviations from the

mean and 3.000 would be the z value for .01=alpha

but 21 =df is too low to approximate t with z (need at least 30)

still 3 is close to 2.8, at least to let you know you're

in the ball park, the right general range for t critical value