Question

A sample of size 20 is drawn from a normal distribution with unknown variance and unknown mean µ. The sample mean is ¯ x = 1.25 and the sample standard deviation is s = 0.25. The 99% lower confidence bound on µ is

A. 1.108 ≤ µ

B. 1.120 ≤ µ

C. µ ≤ 1.392

D. µ ≤ 1.380

Answer 1

)solution

Given that,

= 1.25

s =0.25

n = 20

Degrees of freedom = df = n - 1 =20 - 1 = 19

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

t   df = t0.01,19= 2.539    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.539* (0.25 / 20) = 0.142

The 99%LOWER confidence interval mean is,

- E <

1.25 - 0.14 <

1.108<