CFA=NOLIFE Wrote:
-------------------------------------------------------
> st.dev/sq rt of n is the error and that's what
> you used estimating from a sample.
>
> if you have entire population then you use st dev
> since you can't have sample error (there is no
> sample).
Sorry, can anyone confirm this? I am looking at Schweser answer solutions that dont seem to always follow this rule. Can anyone please help. I've spent a long time trying to figure this out.
It has nothing to do with sample/population size, right?
And then I see this language below, which i can't figure out.
Thanks in advance!
Schweser Question ID 22958:
A sample of 100 recently hired employees shows an average starting salary of $50,000 and a standard deviation of $3,000.
Assuming the population has a normal distribution, construct a 90% confidence interval for the starting salary of a recently hired
employee.
Answer:
90% confidence interval is X � 1.65s = 50,000 � 1.65(3,000) = $45,050 to $54,950.
Note that this is a confidence interval for a single observation, which we estimate as a number of standard deviations from the mean.
If we were constructing a confidence interval for the population mean, we would need to use the standard error (3,000 / sq rt of 100).
Schweser Question ID 1770:
Question: A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 20 cars at rush
hour, he finds that the mean number of occupants per car is 2.5, with a standard deviation of 0.4. If the population is normally
distributed, what is the confidence interval at the 5% significance level for the number of occupants per car?
Solution:
We can use standard distribution tables because the sample is so large.
From a table of area under a normally distributed curve, the z value corresponding to a
95%, one-tail test is: 1.65. (We use a one-tailed test because we are not concerned with passengers arriving too early, only arriving too late.)
Here, we do not divide by the standard error, because we are interested in a point estimate of making our flight.
The answer is one hour, twenty minutes + 1.65(30 minutes) = 2 hours, 10 minutes.