A sample of 58 research cotton samples resulted in a sample average percentage elongation of 8.11...

A sample of 58 research cotton samples resulted in a sample average percentage elongation of 8.11 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average percentage elongation μ. (Round your answers to three decimal places.)

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 8.11

Population standard deviation = = 1.45

Sample size n =58

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z / 2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)
= 1.96 * ( 1.45/ $\sqrt$ 58 $\sqrt$ )

= 0.373
At 95% confidence interval
is,

- E < < + E

8.11 - 0.373 < < 8.11 + 0.373

7.737 < < 8.483

orchestra answered 2 years ago

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