b. Find a 99% confidence interval for mean age of ICU patients using the facts that, in the sample, the mean is 57.55 years and the standard error for such means is 1.42

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Answer:

99% of confidence intervals for mean age of ICU patients

(53.8920 , 61.2079)

Step-by-step explanation:

Explanation:-

Given sample mean [tex]x^{-} = 57.55[/tex]

Given standard error is determined by

                                                             S.E =[tex]\frac{S.D}{\sqrt{n} }[/tex]

Given data standard error = 1.42

99% of confidence intervals for mean is determined by

[tex](x^{-} - Z_{0.01} \frac{S.D}{\sqrt{n} } , x^{-} + Z_{0.01} \frac{S.D}{\sqrt{n} } )[/tex]

[tex]Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.05} = 2.576[/tex]

[tex](x^{-} - 2.576 S.E , x^{-} + 2.576 S.E)[/tex]

[tex](57.55 - 2.576X 1.42 , 57.55+ 2.576 X1.42)[/tex]

(53.8920 , 61.2079)

Conclusion:-

99% of confidence intervals for mean is determined by

(53.8920 , 61.2079)