Promising research shows that a new treatment will delay the progression of Amyotrophic lateral sclerosis.
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With the new treatment, here are the number of months by which progression of ALS is delayed: 3, 5, 6, 6, 8, 8, 9, 9, 9, 10, 11, 45
With the old treatment, the mean number of months before a significant disability is present is 9.6, with a standard deviation of 3.2.
The researchers are overjoyed that the mean number of months the patients in their study had delayed progression was 10.75, with a standard deviation of 11.02.
Mean and standard deviation
From the data set, the mean and standard deviation showing the amount of time that the Amyotrophic lateral sclerosis (ALS) disease could be delayed under the new treatment is computed as follows: Mean (µ) = Sum of the total data set ÷ the number of observations made (n) (McClave, Benson & Sincich, 2008)). In this case, µ= 3+5+6+6+8+8+9+9+9+10+11+45 = 129 while n=12. Therefore, µ = 129 ÷ 12 = 10.75.
Under the ALS new treatment, the standard deviation (∂) for the ALS progression delayed data set is computed as in the table below:
|N = 12||Total= 1336.2875|
But the standard deviation for the data set is given by
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The effects of extreme or outlier values
From the above mean and standard deviation calculations, it is apparent that the mean for the old ALS treatment is less than that of the new ALS treatment. That is, the old ALS treatment means of 9.6 < the new ALS treatment given by 10.75. This does not signify any perfection of the new treatment regarding delaying the progression of ALS because as the data indicates, the larger extreme value might have distorted the value of the mean.
Larger extreme values in the data tend to increase the meanwhile smaller values decrease the mean (Johnson & Kuby, 2011). Thus, the old ALS treatment means value could have been affected by very smaller numbers in the data set.
The omission of the outlier value in mean and standard value calculation
On the other hand, mean is used to compute the standard deviation for any given data set (Rumsey, 2005). However, given that the extreme data values significantly affect the mean, the standard deviation will similarly be affected by extremely larger values. For instance, the standard deviation for the old ALS treatment is smaller than that of the new ALS treatment and this difference could have occurred as a result of the outlier value.
Thus, before funding any future research, variation in the data must be gauged. An assessment of the new ALS delayed progression treatment data set shows that there was an outlier number of 45 months. The value was significantly unlike any other new ALS treatment values. In fact, if this value was excluded from the new treatment data set, the mean could have been: µ= 3+5+6+6+8+8+9+9+9+10+11= 84 while n=11. Therefore, µ = 84 ÷ 11 = 7.64. The standard deviation for the new ALS treatment without the outlier value (45 months) could have been obtained as in the table below:
|N = 11||Total= 56.5456|
Nevertheless, the standard deviation for the data set is given by
This implies that, without that one outlier number, the mean monthly value that the ALS progression could have been delayed would have essentially been less than the old ALS progression treatment meanwhile the standard deviation is somewhat less.
From these mean and standard deviation computations, it emanates that the funding of future research on the treatment cannot be justified on the basis of a single observation. If the Board recommends the funding of the new ALS treatment research, such recommendations would be made on a single outlier value, which in this case is 45 months. This should not be the case.
Johnson, R. R. & Kuby, P. (2011). Elementary statistics. Florence, KY: Cengage Learning.
McClave, J. T., Benson, P. G. & Sincich, T. (2008). Statistics for business and economics. New Delhi, India: Pearson Custom Publishing.
Rumsey, D. (2005). Statistics workbook for dummies. New York, NY: For Dummies.