Question

A sample of 57 research cotton samples resulted in a sample average percentage elongation of 8.16 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average percentage elongation u. (Round your answers to three decimal places What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is normal with the value of o known. We make no assumptions about the distribution of percentage elongation. We assume the distribution of percentage elongation is uniform. We assume the distribution of percentage elongation is normal with the value of o unknown. You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 8.16

sample standard deviation = s = 1.45

sample size = n = 57

Degrees of freedom = df = n - 1 = 57 - 1 = 56

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

t$\alpha$/2,df = t_0.025,56 = 2.003

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.003 * ( 1.45/ $\sqrt$ 57)

Margin of error = E = 0.385

The 95% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 8.16 ± 0.385

=( 7.775, 8.545 )

We assume the distribution of percentage elongation is normal with the value of $\sigma$ unknown.