I have a question for an exam:

You read a journal article on TOTAL-cholesterol, HDL-cholesterol and LDL-cholesterol from a study of n=100 patients where TOTAL-cholesterol = HDL + LDL cholesterol. Assume that for HDL, LDL and TOTAL cholesterol, the distributions are iid multivariate normal.
The article reports for:

HDL: sample-mean = 145 and p-value for the population mean to be 150 is 0.10
LDL: sample-mean = 48 and p-value for the population mean to be 50 is 0.05
TOTAL: sample-mean = 193 and p-value for the population mean to be 200 is 0.05 All P-values are Two-Sided

You must derive the standard deviations and covariances for HDL-cholesterol and LDL-cholesterol in a single observation from this information

A. Work backwards from the p-values for the population means be equal to the specified values based on the observed sample means to obtain the standard deviations for both HDL-cholesterol and LDL-cholesterol in a single observation?

B. What is the covariance of HDL and LDL cholesterol in a single observation?

How does one figure out the standard deviation with only the information given is the mean and p-value?

  • $\begingroup$ You'd also need sample size. $\endgroup$
    – Glen_b
    Nov 30, 2018 at 4:00
  • $\begingroup$ @Glen_b He said the sample size is 100 in Question. $\endgroup$
    – user158565
    Nov 30, 2018 at 22:45
  • $\begingroup$ @user158565 Thanks, though I was responding to the explicit question in the title which seems to propose ignoring that information. $\endgroup$
    – Glen_b
    Dec 1, 2018 at 1:27

1 Answer 1



Suppose p-values were got from t-test.

A: Step 1: from sample size and p-value, you can get t value

step 2: from Sample mean, population means, and t value, you can get standard error

Step 3: From standard error and sample size, you can get estimated variance, then standard deviation.

B: Total = HDL +LDL, so Var(Total) = Var(HDL) + Var(LDL) +2Cov(HDL,LDL)

  • $\begingroup$ From step 2, do u mean t value=sample mean / sd/sqrt(n) ? if so, why does this equality hold? $\endgroup$ Nov 30, 2018 at 4:44
  • 1
    $\begingroup$ $|\bar Y - \mu|/se = t$ $\endgroup$
    – user158565
    Nov 30, 2018 at 4:46

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