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

Respuesta :

Answer:

a)[tex]\pm 1.745[/tex]

b)[tex]\pm 1.795[/tex]

c)[tex]\pm 2.807[/tex]

d)[tex]\pm 2.796[/tex]

Step-by-step explanation:

We are given the following information in the question:

Confidence interval:  

[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]

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

[tex]t_{critical}\text{ at degree of freedom 16 and}~\alpha_{0.10} = \pm 1.745[/tex]

b) 90% confidence, n = 12

Degree of freedom = n - 1 = 11

[tex]t_{critical}\text{ at degree of freedom 11 and}~\alpha_{0.10} = \pm 1.795[/tex]

c) 99% confidence, n = 24

Degree of freedom = n - 1 = 23

[tex]t_{critical}\text{ at degree of freedom 23 and}~\alpha_{0.01} = \pm 2.807[/tex]

d) 99% confidence, n = 25

Degree of freedom = n - 1 = 24

[tex]t_{critical}\text{ at degree of freedom 24 and}~\alpha_{0.01} = \pm 2.796[/tex]