Question

(a) find the margin of error for the values of c, s, and n, and (b) construct the confidence interval for m using the t-distribution. Assume the population is normally distributed.

16. c = 0.99, s = 16.5, n = 20, x = 25.2

Answer 1

Solution :

Given that,

=25.2

s = 16.5

n = 20

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

A ) At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,19 = 2.861

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

= 2.861 * (16.5 / 20)

= 10.55

Margin of error = 10.55

B )The 99% confidence interval estimate of the population mean is,

- E < < + E

25.20 - 10.55 < < 25.2 + 10.55

14.65 < < 35.75