Question

Suppose you have selected a random sample of ?=13 measurements from a normal distribution. Compare the standard normal z values with the corresponding t values if you were forming the following confidence intervals.

a)    80% confidence interval
?=
?=

(b)    90% confidence interval
?=
?=

(c)    99% confidence interval
?=
?=

Answer 1

Solution :

Given that,

a ) At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.01

t _/2,df = t_0.01,12 = 2.681

b ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,12 =1.782

c ) At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,12 = 3.054