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I have 10 years monthly returns. I calculated annualized return multiplying the mean return over the period for 12. Then I calculated the excess returns as difference between the annualized mean return and the benchmark annualized mean return (no risk correction). Annualized standard deviations are obtained multiplying standard deviations for sqrt(12). Data points = n are the same. Now I want to test the hypothesis the difference of the two means (excess return) is greater than 0. Should I use annualized data or not? If I consider annualized data, also variances should be annualized, but the resultant t stat is different depending on the way I calculated it (annualized or not).

Mean is 0.0033 and sample std deviation is 0.0225, benchmark mean is 0.0065 and benchmark std deviation is 0.0197 (not annualized). I have 120 data points (10 years). I want to test (0.0033-0.0225) or (0.0033*12-0.0225*12) significance. Can you give me the formula I should use in this case and also the result?

Thank you in advance for your answer.

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You might use a paired t test on the 120 pairs of monthly returns for your series minus the benchmark monthly returns.

For example, if your 120 monthly returns are in the vector x and the corresponding returns are in the vector b, then a paired t.test gives the following results.

set.seed(828)  # for reproducibility
u = rnorm(120, 10, 2);  e =rnorm(120, .05, .2)
x = u+e;  b = u
t.test(x, b, paired=T)

        Paired t-test

data:  x and b
t = 2.5802, df = 119, p-value = 0.01109
alternative hypothesis: 
  true difference in means is not equal to 0
95 percent confidence interval:
 0.01144941 0.08700116
sample estimates:
mean of the differences 
             0.04922529 

Notice that I simulated the data so that your month-to-month series is a little higher than the benchmark on average (no compounding).

However, if you multiply both x and b by 12 to annualize them (obtaining series X and B, respectively), then the P-value of the test is the same. Means, standard deviations, standard errors, and ends of the CI all just get multiplied by 12. So it makes no difference which method you use.

        Paired t-test

data:  X and B
t = 2.5802, df = 119, p-value = 0.01109
alternative hypothesis: 
  true difference in means is not equal to 0
95 percent confidence interval:
 0.1373929 1.0440140
sample estimates:
mean of the differences 
              0.5907034 
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  • $\begingroup$ So I should assume the two sample standard deviations are equal? $\endgroup$ – CB18 Aug 29 at 7:50

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