A sample of 56 research cotton samples resulted in a sample average percentage e

Question

A sample of 56 research cotton samples resulted in a sample average percentage elongation of 8.16 and a sample standard deviation of 1.15. Calculate a 95% large-sample CI for the true average percentage elongation . (Give answers accurate to 2 decimal places.) What assumptions are you making about the distribution of percentage elongation? We make no assumptions about the distribution of percentage elongation We assume the distribution of percentage elongation is normal with the value of o unknown. We assume the distribution of percentage elongation is uniform. We assume the distribution of percentage elongation is normal with the value ofo known.

Explanation / Answer

x' = 8.16 s = 1.15 z = 1.96 n = 56

LCL = x' - z s / n = 8.16 - 1.15 / 56 = 8.0063

UCL = x' + z s / n = 8.16 + 1.15 / 56 = 8.3137

The 95% confidence interval is (8.0063, 8.3137).

We assumt that the distribution of percentage elongation is normal with the value of unknown.

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