Question

17a. Suppose we do not know σ and the data are the same as in Exercise 14a (The date are: Xi: 10, 30, 20, 25); 17b. Determine the 95% confidence interval. 17c. Is the confidence interval in Exercise 17 narrower or wider than the one in Exercise 14a? Explain why.

Answer 1

Since the population std deviation is not known, we will use t distribution to find the 95%CI for mean.

Given sample: 10,30,20,25

Sample Mean = 21.25

Sample Std Dev = 8.54

Sample Size = 4

t_σ/2 at 95% and n-1 df = 3.182

Hence, 95% CI for Mean = Sample Mean +/- t_σ/2 * Sample Std/n^1/2 = 21.25 +/- 3.182 * (8.54/2)

= {7.66,34.84}

The Confidence Interval is wider than the standard normal as the degrees of freedom is very less. As the degrees of freedom increases, the CI approached the standard normal.