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Q.24.A sample of 55 research cotton samples resulted in a sample average percentage elongation of 8.11 and a sample standard deviation of 1.46. Calculate a 95% large-sample CI for the true average percentage elongation μ. (Round your answers to three 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 σ unknown.
We assume the distribution of percentage elongation is uniform.
We assume the distribution of percentage elongation is normal with the value of σ known.

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Answer #1

Solution :

Given that,

Point estimate = sample mean = = 8.11

sample standard deviation = s = 1.46

sample size = n = 55

Degrees of freedom = df = n - 1 = 55 - 1 = 54

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,54 = 2.005

Margin of error = E = t/2,df * (s /n)

= 2.005 * ( 1.46 / 55)

Margin of error = E = 0.395

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

  ± E

= 8.11  ± 0.395

= ( 7.715, 8.505 )

We assume the distribution of percentage elongation is normal with the value of σ unknown.

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