Question

The value of t_α/2 when determine the 99% confidence interval for µ when the sample standard deviation of the population is unknown and a sample size of 35 was taken from a normally distribution random variable

Answer 1

Since it is given that there is a need for a 99% confidence interval. So, the level of significance will be 1-0.99=0.01. The value of the population standard deviation is not given. So, in order to find a 99% CI for µ, we will use the t-test.

Again the sample size is 35. So, required degree of freedom for critical value of t = n-1 = 35-1 = 34

So, from the t-table, t_α/2 at 0.01 level of significance and at 34 degrees of freedom = 2.73

Note In some standard books, you may not get this above value of t from t-table because most of the books give t value up to 30 degrees of freedom. So, I have used R to obtain this value. You may see my R-code as below:

> abs(qt(0.01/2,34))
[1] 2.728394