Confidence Intervals using Normal Distribution with variance knownFrom a sample of 32 observations from a population with σ2=24.602, we obtained x=79.47. Construct a 99% confidence interval for μ.(a)79.47±1.4424(b)79.47±1.7186(c)79.47±2.0430(d)79.47±2.2622

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kaosimqu5t · Accepted Answer

Respectively, answers (A) through (D) correspond to symmetric two-sided 90%, 95%, 98%, and 99% confidence intervals. The critical values for each are:a) 1.64485b) 1.95996c) 2.32635d) 2.57583with standard errors being 24.60232≈0.87682. Thus the respective margins of error are (±).(A) 1.44224, (B) 1.71853, (C) 2.03979, (D) 2.25854.

pautmndu · Accepted Answer

n=32σ=4.96SE=σnME=q‖.995‖⋅SEx=79.47x+c(−ME,ME)=(77.21,81.73) Answer DWith some further explanation, I am certain the book made an errorHere is what the book used. They used a 95% confidence intervaln=32σ=4.96SE=σnME=q‖.995‖⋅SEx=79.47x+c(−ME,ME)=(77.75,81.19)You were right, the answer is in fact D. Hopefully the code isn&#039;t overwhelming, I could have done the same process by hand, but the book used z=1.96 for a 95% confidence interval, rather than using 2.58 for a 99% confidence interval

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