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Vivek Malhotra · Answer 1 · 2023-05-06T08:42:27+0000

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Step-by-step explanation

Given;

$\begin{gathered} n=82 \\ \bar{x}=228 \\ \sigma=19.0 \\ \end{gathered}$

At 95% confidence level;

$\begin{gathered} \propto=1-95\% \\ =1-0.95 \\ =0.05 \\ (\propto)/(2)=0.025 \\ Z_{\frac{\operatorname{\propto}}{2}}=Z_(0.025)=1.96 \\ \end{gathered}$

At 95% confidence interval for true mean;

$\begin{gathered} \bar{x}\pm Z_{\frac{\operatorname{\propto}}{2}}(\sigma)/(√(n)) \\ =228\operatorname{\pm}1.96*(19)/(√(82)) \\ =228+1.96*(19)/(√(82))<228-1.96*(19)/(√(82)) \\ =228-4.1124<228+4.1124 \\ =223.88<\mu<232.11 \end{gathered}$

Therefore, 95% confidence interval for the true mean cholesterol content

(223.88,232.11)

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