Do we need to divide by sqrt(n) in this Hypothesis Test

[monsieuruzairo3]

Dear David Harper, CFA, FRM, CIPM

I came across the following questions and don't think the answer is correctly calculated

A headset making company claims that the resistance of their new headset is 70 ohms. A dealer wants to test this hypothesis before placing the order. He knows that the higher the resistance, the better it is for him. He takes 36 sample headsets and finds out that sample mean is 73 ohms and it is known that population standard deviation is 1.4%. What are the correct hypothesis test and test statistic he should use?
Choose one answer.
a. Right-tailed z-test and z-statistic = 2.14 Correct
b. Left-tailed t-test and t-stat = 2.14 Incorrect
c. Right-tailed z-test and z-stat=12.9 Incorrect
d. Two-tailed t-test and t-stat = 2.14 Incorrect
The correct answer is A
Sol. The dealer is interested only in finding out whether the resistance is greater or equal to the claimed 70 Ohms. So the lower tail is insignificant and it’s a right tailed test. Sample size is greater than 30 and population standard deviation is known. So Z-test.

Z-stat = ( 73-70 )/1.4 = 2.14

In my opinion, two things are incorrect with the problem
-> Standard deviation is not in Ohms, it's in percentage
-> Even if we take Std. Dev to be 1.4(as they have done in the question), we still need to divide it by Sqrt(36)= 6 (standard error of the sample) as here we are talking about a sample. The answer therefore should be C

What do you think?

Best
Uzi

 

[cdbsmith]

Hi Uzi,

Couple things I would like to point out that might help. First, in the very first sentence of the question it states that the "headset making company claims that the resistance of their new headset is 70 ohms". My interpretation of this statement is that the 70 ohms resistance claim is based on the entire population as the claim is made by the maker of the headsets. Second, the standard deviation given in the question is the population standard deviation. This, in my opinion, further confirms that the 70 ohms mean resistance claim and the 1.4% population standard deviation are indeed population parameters.

The dealer decides to take sample from the population to test the headset maker's claim. How he got access to 36 headsets to test even before placing an order is another matter. That said, based on the the dealer's test, the mean resistance is 73 ohms. But, the sample sample standard deviation was not given. Therefore, you do not have enough information to even calculate the standard error of the estimate since the sample standard deviation (not the population standard deviation) and square root of the sample size are required to calculate it.

So, since the sample size is 36 (>30) and the population mean and standard deviation are both known, it is entirely appropriate to use the the z-statistic and z-value from the normal distribution to test the hypothesis.

Does that makes sense? If not, let me know what still concerns you?

David - of course, you can chime and let me know if my logic is flawed.

Thanks,

Charles

 

[David Harper CFA FRM]

cdbsmith I agree with you that the question gives us the population mean and standard deviation. However, if we have a sample size which is large (here, n = 36), we have enough to calculate the standard error of the sample mean. Actually, the standard error (of the sample mean) is, by definition, the population standard deviation divided by SQRT(n); it's only because we typically do not know it, that we substitute the sample standard deviation. So, the knowledge of population standard deviation plus the large sample (note the question does not actually specify that the population distribution is normal, such that we are relying on the CLT to approximate normality) is enough here to warrant the Z.

monsieuruzairo3 With respect to the calculation of the test statistic, the question is unusual (to my knowledge, the FRM has never mixed the units). The Z-stat here is (73 observed sample sample - 70 hypothesized mean)/SE, where SE is sample S.D/SQRT(36). Given the question, I would use SE = (1.4%*70)/SQRT(36) = 0.1633, with Z-stat = 18.37

 

[monsieuruzairo3]

David Harper, CFA, FRM, CIPM Why have you taken .014*70(Hypothesized mean) and not 0.014*0.73(sample mean). I understand that 1.4% is population std. dev

 

[David Harper CFA FRM]

monsieuruzairo3 The units of the mean are ohms (70 ohms) so the variance units are ohms-squared (omhs^2) and the standard deviation units are ohms. We can instead defer to the 1.4% and signify the numerator in % terms, but with the same result: [73/70 - 1]/[1.4%/SQRT(36)] = [4% above population mean]/SE = 18.37. Thanks,

 

[cdbsmith]

David,

Thanks for clarifying that standard error calculation for me. It was always my understanding to use the sample standard deviation to calculate the sample error. It is good to know that the population standard is what should be used if known.

One question, what should we do with sample error (if asked) if the question has both the sample standard deviation and population standard deviation?

Thanks,

Charles

 

[David Harper CFA FRM]

cdbsmith We would use the population standard deviation (the sample standard deviation is "merely" an estimator of the population s.d., it's our realistic "fallback") per the formula at http://en.wikipedia.org/wiki/Sampling_distribution#Standard_error (in addition to the text, the greek sigma signifies population; when we don't have it, the sample s.d. is signified by a Roman S, so this is a rare instance of the Greeks beating the Romans!)

 

[cdbsmith]

Great! Thanks!