Question

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 \(\mu\) (Round your answer to three decimal places.)

Answer 1

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