# 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: $n=224,\stackrel{―}{x}=28.3\text{hg},s=7.1\text{hg}$. Construct a confidence interval estimate of mean. Use a 98% confidence level. Are these results very different from the confidence interval $26.5\text{hg}<\mu <30.7\text{hg}$ with only 14 sample values, $\stackrel{―}{x}=28.6$ hg, and $s=2.9$ hg? What is the confidence interval for the population mean $\mu$?

(Round to one decimal place as needed).

Are the results between the two confidence intervals very different?

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Bentley Leach

Step 1 From the provided information, Sample size $\left(n\right)=244$ Sample mean $\left(\stackrel{―}{x}\right)=28.3$ hg Sample standard deviation $\left(s\right)=7.1$ hg Step 2 Since, the population standard deviation is unknown, therefore, the t distribution will be used. Confidence level = 98% Level of significance $\left(\alpha \right)=1-0.98=0.02$ The degree of freedom $=n–1=244–1=243$ 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: $CI=\stackrel{―}{x}±{t}_{\frac{\alpha }{2},n-1}\frac{s}{\sqrt{n}}$
$=28.3±\left(2.34\right)\frac{7.1}{\sqrt{224}}$
$=28.3±1.1$
$=\left(27.2,29.4\right)$ Thus, the confidence interval is $27.2<\mu <29.4$. No, the results between the two confidence intervals are not very different. The confidence interval limits are almost similar.