Question

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

Answer 1

Answer:

99% of confidence intervals for mean age of ICU patients

(53.8920 , 61.2079)

Step-by-step explanation:

Explanation:-

Given sample mean $x^{-} = 57.55$

Given standard error is determined by

                                                             S.E =$\frac{S.D}{\sqrt{n} }$

Given data standard error = 1.42

99% of confidence intervals for mean is determined by

$(x^{-} - Z_{0.01} \frac{S.D}{\sqrt{n} } , x^{-} + Z_{0.01} \frac{S.D}{\sqrt{n} } )$

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.05} = 2.576$

$(x^{-} - 2.576 S.E , x^{-} + 2.576 S.E)$

$(57.55 - 2.576X 1.42 , 57.55+ 2.576 X1.42)$

(53.8920 , 61.2079)

Conclusion:-

99% of confidence intervals for mean is determined by

(53.8920 , 61.2079)