Question

A sample of 57 research cotton samples resulted in a sample average percentage elongation of 8.12 and a sample standard deviation of 1.44. Calculate a 95% large-sample CI for the true average percentage elongation μ. (Round your answers to three decimal places.)    .....................................   ,        .....................................

Answer 1

Solution :

Given that,

= 8.12 = 1.44

n = 57

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96* ( 1.44/ 57)

= 0.374

At 95% confidence interval is,

- E < < + E

8.12-0.374 < < 8.12+0.374

7.746< < 8.494

(7.746, 8.494)