A sample of 59 research cotton samples resulted in a sample average percentage e

Question

A sample of 59 research cotton samples resulted in a sample average percentage elongation of 8.14 and a sample standard deviation of 1.48. Calculate a 95% large-sample CI for the true average percentage elongation mu. What assumptions are you making about the distribution of percentage elongation? We assume the distribution of percentage elongation is normal with the value of sigma unknown. We assume the distribution of percentage elongation is normal with the value of sigma known. We make no assumptions about the distribution of percentage elongation. We assume the distribution of percentage elongation is uniform.

Explanation / Answer

mean = 8.14 , s = 1.48 , n = 59
z value at 95% CI = 1.96

CI = mean +/- z * ( s / sqrt(n))
= 8.14 =/- 1.96 * ( 1.48/sqrt(59))
= (7.762 , 8.517)

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