Question

We use the t distribution to construct a confidence interval for the population mean when the underlying population standard deviation is not known. Under the assumption that the population is normally distributed, find t_α/2,df for the following scenarios. (You may find it useful to reference the t table. Round your answers to 3 decimal places.)

		tα/2,df
a.	A 90% confidence level and a sample of 23 observations.
b.	A 95% confidence level and a sample of 23 observations.
c.	A 90% confidence level and a sample of 21 observations.
d.	A 95% confidence level and a sample of 21 observations.

Answer 1

(a) n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,22 =1.717

(b) n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,22 =2.074

( c) n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,20 =1.725

(d) n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,20 =2.086