How Do You Calculate a Z-Score/ Sigma Level?
by Jeff Sauro | June 14, 2004 :: RSS
:: 2 Related Articles
The benefit of using a z-score in usability metrics was explained in "What's a Z-Score and why use it in Usability Testing?" this article discusses different ways of calculating a z-score.
The short answer is: It depends on your data and what you're looking for. If you've encountered the z-score in a statistics book you usually get some formula like:
Calculating a Z-Score Example
For example, lets say you took the GRE a few weeks ago and got scores of 630 Verbal and 700 Quantitative. How good are these scores? Which is better, the Verbal or Quantitative score? Using a z-score can tell you how far you are from the mean and thus how well you performed. If you know the mean and standard deviations for a set of GRE test takers you can compare your scores. ETS publishes the means and standard deviations of a set of test takers on the GRE website.| Verbal | Quantitative | |
| Mean | 469 | 591 |
| StDev | 119 | 148 |
By plugging in your scores you get the following:
Verbal z = (630 - 469) ÷ 119 = 1.35σ
Quantitative z = (700 - 591) ÷ 148 = .736σ
To convert these sigma values into a percentage you can look them up in a standard z-table, use the Excel formula =NORMSDIST(1.35) or use the Z-Score to Percentile Calculator (choose 1-sided) and get the percentages : 91% Verbal and 77% Quantitative. You can see where your score falls within the sample of other test takers and also see that the verbal score was better than the quantitative score. Assuming the sample data was normally distributed, here's how the scores would look graphically:Figure 1: Verbal Score
Figure 2: Quantitative Score
Z-Scores and Process Sigma
An interactive Graph of the Standard Normal Curve similar to Figures 1 & 2 is available for you to visualize how the z-scores and the area under the normal curve correspond. The graphs also allow you to see the difference between one and two-sided (also called two-tailed) areas. In Six Sigma the process sigma metric is derived using the same method as a z-score. However, in Six Sigma you are measuring the distance a sample mean is above a specification limit--there can be an upper and lower spec limit that a sample must fall between as well. As in the z-score, you still use the same normal-deviates from the z-table to approximate the area under the curve. The process sigma metric is essentially a Z equivalent. When testing software with users, task times are usually a good metric that will reveal the individual differences in performance. For task times there typically is only an upper spec limit. That is, it usually doesn't matter how fast a user completes a task, but it does matter if a user takes too long. For example, say you and your product team determined that a task should be completed in 120 seconds. 120 seconds becomes your Upper Spec Limit (USL). You sampled 10 users and got these task times:| Sample |
USL: 120 |
To calculate the process sigma you subtract the mean (104) of the sample from the target (120) and divide by the sample standard deviation (12). For Sample 1 the process sigma is -1.32σ. The visual representation of the data can be seen below:
In the case of task times, a negative process sigma is ideal--as you want more people completing the task below the task time, not above it. You can simply drop the negative when communicating the results in the event it causes confusion. If you were to make radical improvements to the UI and then sampled another set of ten users, here are more results:
| Sample 2 |
| 60 75 99 88 65 72 75 72 87 65 |
| USL: 120 Mean: 75.8 StDev: 12.14 |
In the redesign, the average of the new sample is well below the spec limit and the process sigma is now very high. The corresponding defect area is now only .01% and the quality area is 99.98%
Of course having users perform that much below the spec limit is not very common due to the inherent variability in user performance.
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What's a Z-Score and Why Use it in Usability Testing?
Calculating a Sigma Level from Task Success
| January 24, 2010 | ali wrote: |
| that is very summerized unable to understand nad i raed difrent formula |
| January 14, 2010 | Dr. Al-Ani wrote: |
| I am confused in the z score of body weight to age when every one z zcore equal to 1 standard deviation.how this can be decided? |
| January 9, 2010 | Vijay Kumar wrote: |
| Please provide Z score calculation with standard daviation and Roll up machanism |
| December 6, 2009 | Page wrote: |
| hello if you get a score wich is 22 what is the posabal level you would get? |
| October 28, 2009 | maggie wrote: |
| z-score of z = +3.00 indicates a location that is the following. 1slightly above the mean near the center of the distribution the location depends on the mean and standard deviation for the distribution far above the mean in the extreme right-hand tail of the distribution |
| October 14, 2009 | Jessie Ratliff wrote: |
| Do you think Dr. Zak’s choice of scaling is appropriate |
| October 14, 2009 | Jessie Ratliff wrote: |
| What scale of measurement is Dr. Zak using |
| September 7, 2009 | Oliver wrote: |
| How do you find the Z sigma value without knowing the population size? I have a mean of 5 and a standard dev. of 1.5 and I need to find out 95% of the breaking strengths. I know I should use the formula for z=(n-mean)/stand. dev but I don't know the population size. it is not given. I tried using the 95% but I'm not getting what my instructor says I should be getting. |
| August 28, 2009 | SUMERA wrote: |
| HI I AM SUMERA,SCIENTIFIC OFFICER IN A RESEARCH ORGANIZATION.I FIND THIS WEB PAGE VERY INFORMATIVE AND LEARNING AS I HAVE PROBLEM IN CALCULATING Z-SCORE FOR MY RESULTS OF SPECTROMETER. |
| July 12, 2009 | Theresa wrote: |
| 200 students took a test. The scores were normally distributed. Your score was in the 60th percentile. How many people scored at or below your score? |
| June 27, 2009 | jimboy wrote: |
| gud day. i would like to ask how a standard score (general scholastic aptitude) is transformed back to its raw score. thanks for the help |
| June 25, 2009 | Tasha wrote: |
| What is the z score for each of the five numbers 1,2,3,4,5 if their standard deviation is 1.41? |
| June 4, 2009 | jormin wrote: |
| for a given process, short term stanard deviation is 1.0, and mean is 4.5. what is short-term process Z score if upper specification limit for the process is 9.0? - thanks |
| May 30, 2009 | Mona wrote: |
| which raw score corresponds to a z-score of +0.5? |
| May 26, 2009 | EDITH SHERROD wrote: |
| Determine the area under the standard normal curve that lies between (a) z = -1.05 and z= 1.05. |
| April 29, 2009 | Carmen M. Maldonado wrote: |
| I have a math problem I cannot solve. I used the same method, but it does not work. There are 60 girl: mean is 71,Std. 6, 50 boys: mean 66, Std, 5. Those students with a score greater than 75 are eligible to go to a field trip. What percent of those going to the field trip. I found the z-score for each one, but I cannot the answer. According to the book the answer is 88%. Can you explain to me how to solve this problem. Thanks |
| March 29, 2009 | m d wrote: |
| The deviation is 20,000 people. What is the z-score |
| February 7, 2009 | Vincent wrote: |
| I am confused on how to calculate a z-score. I am doing a six sigma project right now and am lost on doing the z-score. Here is the information I am provided: Calculate the average student score necessary for the district to retain its federal funding. (You may assume the standard deviation will not change). This will require some thought. Think of the z formula. You will be solving for ì (average test score in the next school year.) Think about what you already know. You know the standard deviation (because we assume that the variation in test scores stays approximately the same across years.) You also know that 70% of students must meet the cut off score. Data: The following are the results from last year's MEAP scores for EHS N = 1000 ì = 69.7 ó = 11.55 A student must score at or above 70% on the MEAP to meet the president's cut off score Project Information: The improvement program “No Child Left Behind” legislation has placed emphasis on testing students, and has tied federal funding (and thus some matching state funds) to a certain percentage of students meeting or exceeding cut off scores on state level achievement tests. The challenges that face education are very interesting indeed: unlike in manufacturing (where Six Sigma methodology was originally developed) the outcome is not a product that meets stated specifications, rather it is a student who is “prepared” to take his/her place in society. Unfortunately, the definition of “prepared” is elusive. Also unlike in manufacturing, public educational institutions cannot reject the materials they receive (students) as being defective and send them back – they must take students at whatever point they received them, and advance their state of preparedness. The challenge for this project revolves around Everyday High School in Normal, Michigan. Last year only 49% of the students at the high school met the cutoff score of 70% of questions answered correctly on the Michigan Educational Assessment Program (MEAP) – Michigan’s state level test. This percentage of students meeting the cutoff score is well below the 70% of students who must meet the cutoff score for the district to continue to receive federal funding. Because losing the federal funding (approximately $2,000,000 per year) would mean cuts in programs and maybe even staff that the board of education (BOE), parents, community, students, and staff would find unacceptable, the board of education has directed you (the Six Sigma Black Belt) to conduct a project to improve the scores on next year’s MEAP tests. Any help would be greatly appreciated. |
| January 30, 2009 | Julie wrote: |
| There are two major tests of readiness for college, the ACT and the SAT. ACT scores are reported on a scale from 1 to 36. The distribution of ACT scores for more than 1 million students in a recent high school graduating class was roughly normal with mean = 20.8 and standard deviation = 4.8. SAT scores are reported on a scale from 400 to 1600. The SAT scores for 1.4 million students in the same graduating class were roughly normal with mean = 1026 and standard deviation = 209. Jose scores 545 on the SAT. Assuming that both tests measure the same thing, what score on the ACT is equivalent to Jose's SAT score? |
| December 6, 2008 | Carol Brazzier-Auguste wrote: |
| Using the raw score of 64, calculate the Z-score and the T-score. I managed to work out the Z-score,since the mean for that score is 56 and the SD is 4 using the normal curve, but when you have to calculate the T-score I am using the formula 50 + 10(Z)/SD?? If so what figure amd I using for SD is 4 where the formula will be 50 + 10 (2)/4 or 50+10(2)/10. I theory it is said that In T-scores the mean is always 50 and standard deviation is always 10 By the way I got the Z-score of 2 by subtracting the raw score of 64 from the mean (the mean for that set of scores was 56 and the standard deviation was 4 |
| September 23, 2008 | Kelly wrote: |
| How can I calculate the z score for raw scores of 22 and 28 in with numbers 16,16,22,23,24,27,28,30,31 |
| September 8, 2008 | basua wrote: |
| How do I calculate 1Sigma = 68.26, 2Sigma = 95.44, 3Sima = 99.73 and so on? |
| August 22, 2008 | ratan wrote: |
| interested in knowing all about basic statistics - i am a new learner. thanks |
| August 18, 2008 | victoria wrote: |
| give the raw scores for a person with scores 2.4,1.5,0,-4.5 not sure how to do this |
| July 31, 2008 | Karen wrote: |
| A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be confident that its estimate is within of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about . Carry your intermediate computations to at least three decimal places. Write your answer as a whole number (and make sure that it is the minimum whole number that satisfies the requirements). |
| July 28, 2008 | Mina wrote: |
| Using the cumulative area from the left, locate the closest probability in the body of a table of normal distribution values and identify the z score corresponding to 0.99. Alternately, technology may be used to determine the most accurate z score. How does the temperature for P subscript 99 come out to be approximately 2.33? |
| June 19, 2008 | cochise180 wrote: |
| how do I calculate z score for a nominal of zero with usl 0.008" and lsl -0.008". i get good Cpk and Cp but bad Z score. |
| May 9, 2008 | Tim wrote: |
| Am confused as to how to calculate a negative Z score into a percentage using a standard Z score table |
| December 19, 2007 | anonomous wrote: |
| Why can't all math be explained this simple? Thank you for this explanation! |