In: Statistics and Probability
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 tα/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. | |
(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 = t0.05,22 =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 = t0.025,22 =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 = t0.05,20 =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 = t0.025,20 =2.086