For the following series, the true value is 6.0. Series: 5,4,5,6,7,5,3,8,5,9,5,4,5,6. The SDI is:

For the following series, the true value is 6.0. Series: 5,4,5,6,7,5,3,8,5,9,5,4,5,6. The SDI is:

• A. -0.3
• B. +0.3
• C. +1.0
• D. -0.7

Answer: (A)

The value being tested is how far the average result of the series is from the target value, expressed in units of the spread of the peer results. First find the mean of the series: sum is 77 across 14 values, so the mean is 77/14 = 5.5. The target true value is 6.0, so the difference is 5.5 − 6.0 = −0.5. Next determine the standard deviation of the peer group (using the same data): the deviations from the mean sum of squares is 33.5, so the population standard deviation is sqrt(33.5/14) ≈ sqrt(2.3929) ≈ 1.55. The SDI is (mean − true value) divided by this SD: (5.5 − 6.0)/1.55 ≈ −0.5/1.55 ≈ −0.32, which rounds to −0.3. Therefore, the SDI is about −0.3. The negative sign shows the mean is below the target, and the magnitude indicates it’s about 0.3 standard deviations away.

The value being tested is how far the average result of the series is from the target value, expressed in units of the spread of the peer results. First find the mean of the series: sum is 77 across 14 values, so the mean is 77/14 = 5.5. The target true value is 6.0, so the difference is 5.5 − 6.0 = −0.5. Next determine the standard deviation of the peer group (using the same data): the deviations from the mean sum of squares is 33.5, so the population standard deviation is sqrt(33.5/14) ≈ sqrt(2.3929) ≈ 1.55. The SDI is (mean − true value) divided by this SD: (5.5 − 6.0)/1.55 ≈ −0.5/1.55 ≈ −0.32, which rounds to −0.3. Therefore, the SDI is about −0.3. The negative sign shows the mean is below the target, and the magnitude indicates it’s about 0.3 standard deviations away.