Hypothesis testing

CFA Level I
#1

Hi everyone,

I am bit lost on a question from the CFAI (taken from the topic test).

"An analyst determines that approximately 99% of the observations of daily sales for a company are within the interval from

$230,000 to $480,000 and that daily sales for the company are normally distributed. If approximately 99% of all the observations fall in the interval μ±3σ, then using the approximate z-value rather than the precise table, the standard deviation of daily sales for the company is closest to:

$83,333.

$41,667.

$62,500.

0 out of 1

Incorrect.

Given that sales are normally distributed, the mean is centered in the interval. MeanUnder a normal distribution, 99% of the observations will be approximately plus or minus three standard deviations. Next, use the following formula:

Z =(X-μ)/σ

or, by rearranging:

σ=(X-μ)/Z,

where

Z = 3,X = $480,000, and μ = $355,000.

Thus, ($480,000 – $355,000)/3.0 = $41,667.

Alternatively, use

Z = –3, X = $230,000, and μ = $355,000: ($230,000 – $355,000)/(–3.0) = $41,667.

"

Okay, so I got lost when they used the formula: Z =(X-μ)/σ

because in my formular I only have the formula: Z =(X-μ)/(σ/n1/2)

So I don’t really understand from where this formula come (didn’t found it in the CFAI books). I am sure I missed something and that it’s kind of stupid but still I got no clue. Your help would be much appreciated.

Thanks

ps: the editing is a bit bad but I wanted to write n root two

#2

They used standard error of the sample mean which is (std. deviation) / (n)^(1/2)

They are on the books, check again.

#3

hi, yeah I know this formula, but in the correction they used another formula ( Z =(X-μ)/σ), which is not the one you are indicating

#4

Hmm, yes you right! Can you believe me that I have remembered this question from Q-bank past year? And… couldn’t solve it at the first try.

Note this is not a regular “calculate CI for this data”, this is an uncommon exersice where if you look well, you are not given with sample size (n), and instead you are given with the confidence inverval spread ( u ± 3 std.dev) and the the bounds are 230,000 and 480,000. Since confidence intervals are symetrical, u = 355,000, so each side, from the lower bound to the mean, and from the upper bound to the mean have a 3 x std. dev. width.

This means that 480 - 355 = 125 = 3 x std. dev. Thus std. dev. = 125 / 3 = 41.667

Also works for the left side: 355 - 230 = 125 = 3 x std. dev. Thus std. dev. = 125 / 3 = 41.667

Any question please ask!

Regards

1 Like
#5

Ok, thanks

Would it be right the make the below assumptions:

if it’s normally distributed, the mean is centered in the interval and we can use the formula Z =(X-μ)/σ

if not, we use the formula Z =(X-μ)/(σ/n1/2)?

#6

As far as I’ve seen you must use the second one (the one with the standard error) because you using a sample. If I’m not mistaken, you use the first one when you have the data of an entire population and the variance of population is known.

#7

How did they arrive at μ = $355,000? Is it the average of the intervals or something else?

#8

yes indeed

#9

they calculated mean this way:

480,000-230,000= 250,000

50% of 250,000 is below the mean and 50% of 250,000 is above the mean

therefore 230,000 + (250,000/2) = 230,000 + 125,000 = 355,000

or

480,000 - (250,000/2) = 480,000 - 125,000 = 355,000

#10

That’s right, a normal distribution is symetrical, so if you know the bounds of the confidence interval, you can know the mean simply by calculating the average of the bounds.