Question

5. Suppose we were formulating a 90% confidence interval for mean for a particular random variable that we know is normally distributed. We are going to use a random sample of n=15 to formulate our confidence interval. What value for t should we have?

Answer 1

solution:

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,14 =  1.761    ( using student t table)