 # Here are summary stastistics for randomly selected weights of newborn girls: n=224,overline{x}=28.3 text{hg},s=7.1 text{hg}. Construct a confidence interval estima Khaleesi Herbert 2020-11-07 Answered

Here are summary stastistics for randomly selected weights of newborn girls: $$\displaystyle{n}={224},\overline{{{x}}}={28.3}\text{hg},{s}={7.1}\text{hg}$$. Construct a confidence interval estimate of mean. Use a 98% confidence level. Are these results very different from the confidence interval $$\displaystyle{26.5}\text{hg}{<}\mu{<}{30.7}\text{hg}$$ with only 14 sample values, $$\displaystyle\overline{{{x}}}={28.6}$$ hg, and $$\displaystyle{s}={2.9}$$ hg? What is the confidence interval for the population mean $$\displaystyle\mu$$?

$$\ {27,2 \ hg<}\mu{<29,4 \ hg}?$$ (Round to one decimal place as needed).

Are the results between the two confidence intervals very different?

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Step 1 From the provided information, Sample size $$\displaystyle{\left({n}\right)}={244}$$ Sample mean $$\displaystyle{\left(\overline{{{x}}}\right)}={28.3}$$ hg Sample standard deviation $$\displaystyle{\left({s}\right)}={7.1}$$ hg Step 2 Since, the population standard deviation is unknown, therefore, the t distribution will be used. Confidence level = 98% Level of significance $$\displaystyle{\left(\alpha\right)}={1}-{0.98}={0.02}$$ The degree of freedom $$\displaystyle={n}–{1}={244}–{1}={243}$$ The critical value of t at 243 degree of freedom with 0.01 level of significance from the t value table is 2.34. Step 3 The required 98% confidence interval can be obtained as: $$\displaystyle{C}{I}=\overline{{{x}}}\pm{t}_{{\frac{\alpha}{{2}},{n}-{1}}}{\frac{{{s}}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={28.3}\pm{\left({2.34}\right)}{\frac{{{7.1}}}{{\sqrt{{{224}}}}}}$$
$$\displaystyle={28.3}\pm{1.1}$$
$$\displaystyle={\left({27.2},{29.4}\right)}$$ Thus, the confidence interval is $$\displaystyle{27.2}{<}\mu{<}{29.4}$$. No, the results between the two confidence intervals are not very different. The confidence interval limits are almost similar.