Quantitative Methods - Sampling & Estimation Question

C

chippp

New member
Joined
Jun 18, 2026
Messages
0
Reaction score
0
Question
A sample of size 600 is drawn from a population. The sample mean equals 329.
The total width of the 99% confidence interval for the population mean is 893. The estimated population variance
equals ________.

A) 7.3
B) 6.3
C) 4.9
D) 6.7

I couldn't even come near these values....My calculation is as such;

2 TAILED 99% CI = 2*2.575*Std Dev = 893
therefore Std Dev of population = 173.4
I'm lost from here on since no matter how I manipulate (divide by n, power 2, etc etc) the figure it's never going to get near the answers....Is the question wrong? Can anyone help?
 
Something is wrong with the question here and your solution.
It's not quite as you say -
2 TAILED 99% CI = 2*2.575*Std Dev = 893
It's
2 TAILED 99% CI = 2*2.575*Std Dev/Sqrt(n) = 893 where n = sample size

But that means the answer is some gigantor number not one of these answers. I suspect that there is a decimal point omitted in that 893, but I am too lazy to figure out where.
 
I saw that question too...it is a fubar question inaccurate information even if there was a decimal point out of place you should still get something resembling one of those numbers.

446.5 = 2.575 (SD/24.49)

173.39 = SD/24.49

SD = 4,246...so yeh something is screwy with the numbers.
 
The solution was given as below, but didn't explain how it arrived at the numerical value of 6.3;

If z is the z-value corresponding to the specified confidence level, the sample mean is M and the standard deviation is D in a sample size N, the confidence interval is specified as
[M - z * D / root (N), M + z * D / root(N)].
Thus, the width of the interval equals 2 *z*D / root(N).

Answer is B.


Thanks for the reply, JoeyDVivre



Edited 2 time(s). Last edit at Wednesday, April 26, 2006 at 12:26PM by chippp.
 
893 / 2 = 446.5

noseykibitzer is just using one tail instead of 2...same difference...
 
893 is population mean, why it need be devided by 2?
 
893 is the width of the 99% interval. Not the population mean...
 
u r right, sorry, i'm careless.
it seems not a normally distribution.
 
You must log in or register to reply here.
Back
Top