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FRM 2008 Practice PII question 18 - Confidence Interval

fullofquestions

New Member
QUESTION
Suppose the standard deviation of a normal population is known to be 10 and the mean is hypothesized
to be 8. Suppose a sample size of 100 is considered. What is the range of sample means that allows the
hypothesis to be accepted at a level of significance of 0.05?
a. Between -11.60 and 27.60
b. Between 6.04 and 9.96
c. Between 6.355 and 9.645
d. Between -8.45 and 24.45
Answer: b
a. Incorrect. This answer is the range of values that contain 95% of observations drawn from a normal
distribution of mean 8 and standard deviation 10. However, to accept the hypothesis that the mean is
8 at a 0.05 significance level, the test statistic Z = (X - 8) / (10 / ÷ (n)) must fall between -1.96 and
1.96, where X is the sample mean.
b. Correct. To accept the hypothesis at a 0.05 significance level, the test statistic Z must fall between
-1.96 and 1.96 -1.96 ? Z = (X - 8) / (10 / ÷ (100)) ? 1.96, which implies that the sample mean X
must be between 6.04 and 9.96.
c. Incorrect. This answer is the range of values the sample mean must fall within to accept the
hypothesis at a 0.10 significance level.
d. Incorrect. This answer is the range of values that contain 90% of observations drawn from a normal
distribution of mean 8 and standard deviation 10. However, to accept the hypothesis that the mean
is 8 at a 0.05 significance level, the test statistic Z = (X - 8) /(10/ ÷(n)) must fall between -1.96 and
1.96, where X is the sample mean.

Answer b is clear once you divide standard deviation by sqrt(n). In terms of wording, is the simple fact that since the number of samples is provided then you should use the sample std deviation (10/sqrt(100)) as opposed to using the standard deviation (10)?
Furthermore, what in the world are they trying to say in answer A and B? So far, I have seen many typos and rather mediocre explanations in these practice exams so I'm reluctant to think the answers always correct. Anyway, if A and B have any useful information please confirm. Otherwise, this is just a simple 'trick' question to get you to use sample std. dev. How useful...
 

hsuwang

Member
Hello,

I think we are not dealing with sample stdev here, but rather, we are dealing with "standard error", which is the stdev of the "sample mean", and that is why we divide the stdev by the square-root of n.

Thanks.
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
Hi FoQ,

Jack is right ... (a) does not look helpful to me ...
The thing you want to notice is that the statistic is the *sample mean* rather than just a random observation drawn from the population. The popluation is N(mean = 8, variance = 10^2) ... so you could if the question were to ask about the interval for a random drawn from this population, your answer would be 8 +/- 1.96 * 10

...but the sample mean isn't a draw from the population, it's the mean of a sample. As a sample statistic, it has it's own distribution (per the CLT). I think the hard thing is seeing one sample mean as a draw from the distribution of sample means (i.e., take a sample from the population, compute sample mean1. Okay, take another sample from the popluation, compute sample mean 2 ... now we have 2 sample means and are starting to collect a set of sample means)...so, when Gujarati uses the phrase "sampling distribution of sample means" you can start to see how this is a distribution constructed of sample means, not the distribution of the underlying popluation...in any case, IMO, the keyword to give you the signal is "sample mean"

David
 
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