In: Statistics and Probability
A sample of 58 research cotton samples resulted in a sample average percentage elongation of 8.11 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average percentage elongation μ. (Round your answers to three decimal places.)
Solution :
Given that,
Point estimate = sample mean = = 8.11
Population standard deviation = = 1.45
Sample size n =58
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z / 2 = Z0.025 = 1.96 ( Using z table )
Margin of error = E = Z/2 * (
/n)
= 1.96 * ( 1.45/ 58 )
= 0.373
At 95% confidence interval
is,
- E < < + E
8.11 - 0.373 <
< 8.11 + 0.373
7.737 <
< 8.483