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In: Statistics and Probability

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.)

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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    * ( /n)
= 1.96 * ( 1.45/ 58 )

= 0.373
At 95% confidence interval
is,

- E <   < + E

8.11 - 0.373 <   < 8.11 + 0.373

7.737 <   < 8.483


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