Question

The formula used to compute a confidence interval for the mean of a normal population when n is small is the following. What is the appropriate t critical value for each of the following confidence levels and sample sizes? (Round the answers to two decimal places.) (a) 90% confidence, n = 17 (b) 90% confidence, n = 12 (c) 99% confidence, n = 24 (d) 90% confidence, n = 25 (e) 80% confidence, n = 13 (f) 95% confidence, n = 9Please explain where the final numbers/answers are coming from.

Answer 1

Answer:

We calculated the t-critical values from t-table.

Step-by-step explanation:

We are given the following information in the question:

$\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}$

We have to find appropriate t-critical values.

a) 90% confidence, n = 17

Degree of freedom = n - 1 = 16

$t_{critical} \text{ at 0.10 level of significance, 16 degree of freedom } = \pm 1.745$

b) 90% confidence, n = 12

Degree of freedom = n - 1 = 11

$t_{critical} \text{ at 0.10 level of significance, 11 degree of freedom } = \pm 1.795$

c) 99% confidence, n = 24

Degree of freedom = n - 1 = 23

$t_{critical} \text{ at 0.01 level of significance, 23 degree of freedom } = \pm 2.807$

d) 90% confidence, n = 25

Degree of freedom = n - 1 = 24

$t_{critical} \text{ at 0.10 level of significance, 24 degree of freedom } = \pm 1.710$

e) 80% confidence, n = 13

Degree of freedom = n - 1 = 12

$t_{critical} \text{ at 0.20 level of significance, 12 degree of freedom } = \pm 1.356$

f) 95% confidence, n = 9

Degree of freedom = n - 1 = 8

$t_{critical} \text{ at 0.05 level of significance, 8 degree of freedom } = \pm 2.306$