# 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

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 $95\mathrm{%}$ large-sample CI for the true average percentage elongation $\mu$ (Round your answer to three decimal places.)

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stuth1
Step 1
Given that
Sample size, $n=51$
Sample mean $\stackrel{―}{x}=8.19$
Sample standard deviation, $s=1.45$
The objective is to construct $95\mathrm{%}$ large-sample confidence intervals for the true average elongation $\mu$.
The confidence level is,
$C=95\mathrm{%}=\frac{95}{100}=0.95$
The level of significance is,
$\alpha =1-C=1-0.95=0.05$
From standard normal table, at 0.05 level of significance the two-tailed critical value is,
${z}_{\alpha \text{/}2}={z}_{0.05\text{/}2}=1.96$
Step 2
The confidence interval is computed below.
$\stackrel{―}{x}±{z}_{\alpha \text{/}2}\left(\frac{s}{\sqrt{n}}\right)$
$8.19±{z}_{0.05\text{/}2}\left(\frac{1.45}{\sqrt{51}}\right)$
$8.19±1.96\left(0.2030\right)$
$8.19±0.39788$
$\left(8.19-0.39788,8.19+0.39788\right)$
$\left(7.79212,8.58788\right)$
$\left(7.792,8.588\right)$ (rounded to three decimals)
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