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In: Statistics and Probability

A sample of 51 research cotton samples resulted in a sample average percentage elongation of 8.11...

A sample of 51 research cotton samples resulted in a sample average percentage elongation of 8.11 and a sample standard deviation of 1.49. 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 assume the distribution of percentage elongation is normal with the value of σ unknown.

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 σ known.

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 8.11

sample standard deviation = s = 1.49

sample size = n = 51

Degrees of freedom = df = 51 - 1 = 50

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,50 = 2.009

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

= 2.009* ( 1.49 / $\sqrt$ 51 )

= 0.419

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

- E < < + E

8. 11 - 0.419 < < 8.11 + 0.419

7.691 < < 8.529

( 7.691 , 8.529 )

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

orchestra answered 2 years ago

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