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# 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

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Confidence intervals
asked 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 $$\mu$$ (Round your answer to three decimal places.)

## Answers (1)

2021-03-07
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)

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