A sample of 51 research cotton samples resulted in a sample average percentage elongation of 8.19 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average

Chesley

Chesley

Answered question

2021-03-06

A sample of 51 research cotton samples resulted in a sample average percentage elongation of 8.19 and a sample standard deviation of 1.45. Calculate a 95% large-sample CI for the true average percentage elongation μ (Round your answer to three decimal places.)

Answer & Explanation

stuth1

stuth1

Skilled2021-03-07Added 97 answers

Step 1 
We have:
Sample size, n=51 
Sample mean x=8.19 
Sample standard deviation, s=1.45 
The objective is to construct 95% large-sample confidence intervals for the true average elongation μ
The degree of confidence is
C=95%=95100=0.95 
The level of significance is, 
α=1C=10.95=0.05 
The two-tailed critical value for the standard normal table at 0.05 level of significance is
zα/2=z0.05/2=1.96 
Step 2 
The computed confidence interval is shown below 
x±zα/2(sn) 
8.19±z0.05/2(1.4551) 
8.19±1.96(0.2030) 
8.19±0.39788 
(8.190.39788,8.19+0.39788) 
(7.79212,8.58788) 
(7.792,8.588) (rounded to three decimals)

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