0 0 5 = -5 – 5 5 5 = 0 – 5 5 5 = 0 – 5 5 5 = 0 – 5 5 5 = 0 – 5 5 5 = 0 – 5 5 5 = 0 – 5 5 5 = 0 – 5 5 5 = 0 – 10 10 5 = 5 – We would like to get an idea of the deviation from the mean, but if we find the “average” average of the values in the second column, the negative and positive values cancel each other out (this will always happen), so to prevent this we square the deviations. Data Value Deviation: (Data Value Deviation Squared –Mean) 0 0 5 = -5 (-5)2 = 25 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 5 5 5 = 0 02 = 0 – 10 10 5 = 5 52 = 25 – Next, we add the squared deviations and we get . Ordinarily, we would then divide by the number of scores, n, (in this case, 10) to find the mean of the deviations, but the division by n is only done if the data set represents a population. When the data set represents a sample (as it almost always does), we instead divide by n (in this case,). –1 We assume Section D represents a sample, so we will divide by 9. Note that our units are now points-squared since we squared all of the deviations. It is much more meaningful to use the units we started with, so to convert back to points we take the square root. The sample standard deviation is
0 0 – 5 = -5 5 5 – 5 = 0 5 5 – 5 = 0 5 5 – 5 = 0 5 5 – 5 = 0 5 5 – 5 = 0 5 5 – 5 = 0 5 5 – 5 = 0 5 5 – 5 = 0 10 10 – 5 = 5 We would like to get an idea of the “average” deviation from the mean, but if we find the average of the values in the second column, the negative and positive values cancel each other out (this will always happen), so to prevent this we square the deviations. Data Value Deviation: (Data Value –Mean) Deviation Squared 0 0 – 5 = -5 (-5)2 = 25 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 5 5 – 5 = 0 02 = 0 10 10 – 5 = 5 52 = 25 Next, we add the squared deviations and we get . Ordinarily, we would then divide by the number of scores, n, (in this case, 10) to find the mean of the deviations, but the division by n is only done if the data set represents a population. When the data set represents a sample (as it almost always does), we instead divide by n –1 (in this case,). We assume Section D represents a sample, so we will divide by 9. Note that our units are now points-squared since we squared all of the deviations. It is much more meaningful to use the units we started with, so to convert back to points we take the square root. The sample standard deviation is