Testing Hypothesis Regarding a Population Mean
Authors
A randomly selected sample of XYZ Investement Advisors' portfolios show the following:
Sample size 25
Average Sharpe Ratio 2.6
Standard Deviation 1.6
The firm claims its methodology produces portfolios with an average Sharpe ratio of 3.0. Verify.
Null hypothesis: H0: µ0 = 3.0
Alternative Hypothesis H1: µ0 is not equal to 3.0
This is a two tailed test.
Degrees of freedom = n –1 = sample size –1 = 25-1 = 24
Probability = alpha/2 = .05/2 = .025
From students t-Distribution table, the critical t is found to be 2.064. Hence acceptance range of t-statistic is –2.064 to 2.064.
T-calculated for the sample
Standard error = Sx/Square root of n = 1.6/5 = 0.32
T calculated = (X -µ0)/Sx = (2.6 – 3.0)/0.32 = -1.25
Since the –1.25 falls within the acceptance range, we cannot reject the null hypothesis.
Acceptance range around the hypothesized value =
1.0 + (2.064*0.32) = 2.34 to 3.66
The sample average falls within this range, so we cannot reject the null hypothesis.
One-Tailed Hypothesis Tests
We want to do the above test with the idea of the firm that their Sharpe ratio are greater than or equal to 3.0. Hence it becomes a one-tail test. Also the population variance is known as 2.44.
Null hypo = H0 = Mu > 3.0
Alt. Hypo = H1 = Mu < 3.0
Now p = alpha/1 = 0.050
From standard normal tables critical value is –1.645. The acceptance range for the test statistic is –1.645 to +infinity.
The test statistic is population standard deviation/Square root of the sample size. = S.R(2.44)/S.R.(25) = 1.56/5 = 0.31
Test Statistic calc = Z calc = (X - )/ Sx = (2.6 – 3.0)/0.31
= -1.29
The –1.29 is within the range, thus we cannot reject the null hypothesis.
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Included in Research methodology Knol Book Chapter
http://knol.google.com/k/narayana-rao-k-v-s-s/-/2utb2lsm2k7a/3579
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