Percentage Improvement

[markj3]

Hi there,

I am reading a thesis and I am struggling to compute the results section.

Before a correction factor was applied to a measurement, the average measurement error was -9.8%. After applying a correction factor, the average measurement error is now -0.9%. The desired error is 0%. The author has then said that the percentage improvement between -9.8% and -0.9% is 91.1%.

I am lost at how this improvement has been calculated, can anyone help make sense of this, or is this a mistake?

 

[Deleted member 4993]

Hi there,

I am reading a thesis and I am struggling to compute the results section.

Before a correction factor was applied to a measurement, the average measurement error was -9.8%. After applying a correction factor, the average measurement error is now -0.9%. The desired error is 0%. The author has then said that the percentage improvement between -9.8% and -0.9% is 91.1%.

I am lost at how this improvement has been calculated, can anyone help make sense of this, or is this a mistake?

Please tell us how you would compute % change. Let's do a simple problem:

A pen was originally priced $50. The price was reduced to $37.

How much was the "% reduction" of the price of the pen?

Please share your work in detail.

 

[markj3]

Please tell us how you would compute % change. Let's do a simple problem:

A pen was originally priced $50. The price was reduced to $37.

How much was the "% reduction" of the price of the pen?

Please share your work in detail.

For this problem I would use (50-37/50)x100, which would give a 26% decrease in cost.
Using similar in the numbers above I calculate a 90.8% increase, as opposed to 91.1%

 

[Deleted member 4993]

For this problem I would use (50-37/50)x100, which would give a 26% decrease in cost.
Using similar in the numbers above I calculate a 90.8% increase, as opposed to 91.1%

Correct.

I suspect the difference in the "reported" number and the "calculated" number could have resulted from rounding the actual numbers.