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 2021-03-06 Answered

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.)

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Expert Answer

stuth1
Answered 2021-03-07 Author has 97 answers
Step 1
Given that
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 confidence level is,
C=95%=95100=0.95
The level of significance is,
α=1C=10.95=0.05
From standard normal table, at 0.05 level of significance the two-tailed critical value is,
zα/2=z0.05/2=1.96
Step 2
The confidence interval is computed 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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