Do you have a standard deviation from your sample? With that, you should be able to perform a 2-sample t-test.
Hi everybody,
I have a question here that I'm completely brain frozen with.. (I'm not the youngest anymore, and I can't seem to wrap my brain around this)
I have a sample of people (n=36) with a mean cholesterol level of 4.1.
the reference table states that the normally distributed population has values between 3.2 and 6.2 (95% confidence interval)
Can I conclude that the sample group has a different cholesterol value than “the normal population”. I have to solve this using a significance level of 0.05, showing all steps required.
I honestly have no idea where to start, bu this is how I tried:
xbar = 4.1
n = 36
pop.mean = 4.7 (right?)
pop. standard deviation = 0.7653 (not sure about this one either)
and that's where I'm stuck. How do I need to approach this?
any help would be greatly appreciated!!
Mark
Do you have a standard deviation from your sample? With that, you should be able to perform a 2-sample t-test.
Miner,
thanks for your reply. The problem is, that I don't have the standard deviation for the sample. I tried to derive it from the population, but I'm not sure if that is a valid way to do:
sample mean xbar = 4.1
sample size n = 36
population mean µ = (3.2+6.2)/2 = 4.7
population s.d. σ = 0.7653
based on 6.2 = µ+1.96 σ → 6.2 = 4.7+1.96σ → 1.5 = 1.96σ → σ = 0.7653
sample s.d. s = 0.7653/(SQRT(36)) = 0.12755
Cheers!
Mark
I misspoke earlier. Since you are comparing a sample to a population mean, you should use a 1-sample t-test. Your only option is to make the (very big) assumption that the sample standard deviation is equivalent to the population standard deviation and use that in your test. However, recognize that you are taking a risk. How big that risk is ...?
Is there a way to calculate the sample standard deviation from the population s.d.?
I have the population mean and the population s.d. calculated (see previous post, not 100% if they're correct though), and used the pop.s.d./sqrt(36) to get the sample standard deviation, but I'm not sure if this is a valid way to do that.
Again, thanks a ton for your input!!
No. Each sample that you take from a population would have a slightly different standard deviation. Also, if the sample did not represent that population, it might have a drastically different standard deviation.
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