Question

The formula used to compute a confidence interval for the mean of a normal population when n is small is the following.

x^^\_ +/-\(t text( critical value)\)s/sqrt(n)

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

Answer 1

a)
`\pm 1.745`

b)
`\pm 1.795`

c)
`\pm 2.807`

d)
`\pm 2.796`

Explanation:

We are given the following information in the question:

Confidence interval:

`\bar{x} \pm t_(critical)\displaystyle(s)/(√(n))`

We have to find the appropriate t critical values for each of the following confidence levels and sample sizes:

a) 90% confidence, n = 17

Degree of freedom = n - 1 = 16

`t_(critical)\text{ at degree of freedom 16 and}~\alpha_(0.10) = \pm 1.745`

b) 90% confidence, n = 12

Degree of freedom = n - 1 = 11

`t_(critical)\text{ at degree of freedom 11 and}~\alpha_(0.10) = \pm 1.795`

c) 99% confidence, n = 24

Degree of freedom = n - 1 = 23

`t_(critical)\text{ at degree of freedom 23 and}~\alpha_(0.01) = \pm 2.807`

d) 99% confidence, n = 25

Degree of freedom = n - 1 = 24

`t_(critical)\text{ at degree of freedom 24 and}~\alpha_(0.01) = \pm 2.796`