Question

Suppose you have selected a random sample of n =5 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)    8080% confidence interval
z=
t=

(b)    9898% confidence interval
z=
t=

(c)    9595% confidence interval
z=
t=

ConfidenceIntervals

Answer 1

A

n = Degrees of freedom = df = n - 1 = 5- 1 = 4

a ) At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

  / 2= 0.20 / 2 = 0.10

t /2,df = t 0.10,4 = 1.533 ( using student t table)

B

Degrees of freedom = df = n - 1 = 5- 1 = 4

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,4 =3.746 ( using student t table)

Degrees of freedom = df = n - 1 =5 - 1 = 4

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,4 =2.776 ( using student t table)