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, but 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)

I was trying to get the sample standard deviation from the population sd, but I'm not sure if that's possible at all. And that's where I'm stuck. How do I need to approach this?

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    $\begingroup$ Where is the 4.7 population mean coming from? Where is the .76 population standard deviation coming from? Also, if this is homework please add the "self-study" tag. $\endgroup$ – TrynnaDoStat Feb 4 '15 at 21:51
  • $\begingroup$ Yes, this is a homework task - sorry, I didn't know about the self study tag. I took th emean from the provided confidence interval (95%) (3.2+6.2)/2 to get the population mean, and then tried to calculate the population standard deviation based on the fact that the upper lever equals µ+1.96σ. And please correct me if I already screwed up on this. How would I approach this to conclude that the sample group has a different cholesterol value than the population? $\endgroup$ – derBrain Feb 4 '15 at 22:10

The assumption of a normal distribution may be heroic here. If the assumption may be made then I assume that your population standard deviation is half the range of the population 95% confidence range divided by 1.96. In that case the standard deviation of an estimate of the mean of a sample of size $n$ is (population s. d.)/$\sqrt{n}$. You can then use this estimate to test if the sample mean is within the 95% range for a samp[le mean drayn from a normal distriution.

  • $\begingroup$ Thanks for your reply. So this is how I came up with the population sd: upper limit = 6.2, pop.mean is half the range of CI = 4.7 -> 6.2=4.7+1.96*pop.sd -> solving for pop.sd = 0.7653. But how do I continue from here? how does the pop.sd help me with the sample mean of 4.1? Sorry, I'm absolutely brain frozen, I've been staring at this for hours now.. is the sample.sd = pop.sd/sqrt(36) = 0.12755? $\endgroup$ – derBrain Feb 4 '15 at 22:28
  • $\begingroup$ based on this, the 95% confidence interval i calculated is 4.06 <= 4.1 <= 4.1359. how does this help me with the conclusion that the sample cholesterol level is different than the one of th epopulation (it seems like it's not different at all..) $\endgroup$ – derBrain Feb 4 '15 at 22:32
  • $\begingroup$ If the population mean is 4.7 then a 95% confidence limit for the mean of size n is (4.7-1.96 * (populatio s.d.)/$\sqrt(n)$ , 4.7+1.96 * (populatio s.d.)/$\sqrt(n)$ ). What is the probability that a sample of 36 with mean 4.1 was drawn from that population $\endgroup$ – John C Frain Feb 5 '15 at 23:33

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