# Confidence Intervals using Normal Distribution with variance known

From a sample of $32$ observations from a population with $σ^2$ $= 24.602$, we obtained $x = 79.47$. Construct a $99$% confidence interval for $\mu$.

$(a) 79.47±1.4424$
(b) 79.47±1.7186
$(c) 79.47±2.0430$
$(d) 79.47±2.2622$

Why is the bolded (b) the answer? I keep getting (d).

Using R

 n = 32
sigma = 4.96
SE = sigma/sqrt(n)
ME = qnorm(.995)*SE
x = 79.47
x+c(-ME,ME)


=($77.21, 81.73$) Answer D

Is it possible the book made an error??

With some further explanation, I am certain the book made an error

Here is what the book used. They used a 95% confidence interval

 n = 32
sigma = 4.96
SE = sigma/sqrt(n)
ME = qnorm(.975)*SE
x = 79.47
x+c(-ME,ME)


= ($77.75, 81.19$)

You were right, the answer is in fact D. Hopefully the code isn'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

• Thanks Brandon, I feel less stupid now haha! – EconDude Dec 5 '16 at 6:44
• @EconDude You're welcome. Sorry I had commented a few times and deleted those because I thought there was an issue, but everything is all good. – Brandon Dec 5 '16 at 7:07

Respectively, answers (A) through (D) correspond to symmetric two-sided $90\%$, $95\%$, $98\%$, and $99\%$ confidence intervals. The critical values for each are:

(A) $1.64485$, (B) $1.95996$, (C) $2.32635$, (D) $2.57583$,

with standard errors being $\sqrt{24.602/32} \approx 0.87682$. Thus the respective margins of error are ($\pm$)

(A) $1.44224$, (B) $1.71853$, (C) $2.03979$, (D) $2.25854$.