Restoring Confidence in Usability Results by Jeff Sauro | October 18, 2004 :: 2 Related Articles:: 48 Related Questions
Summary
Adding confidence intervals to completion rates
in usability tests will temper both excessive skepticism and overstated
usability findings. Confidence intervals make testing more efficient
by quickly revealing unusable tasks with very small samples. Examples
are detailed and downloadable calculators are available.
Are you Lacking Confidence?
You just finished a usability test. You had 5 participants
attempt a task in a new version of your software. All 5 out of 5 participants
completed the task. You rush excitedly to tell your manager the results
so you can communicate this success to the development team. Your
manager asks, “OK, this is great with 5 users, but what are
the chances that 50 or 1000 will have a 100% completion rate?”
You stop, think for a minute and say “pretty good.”
Fortunately there is a way to be a bit more precise than “pretty
good” and you don’t need to be a statistician (or bribe
the one in your company with jelly-donuts to help you). The way to
precision is through confidence intervals. Not only are they precise,
but they make you sound smart when you talk about them. This article
has two parts. In part one confidence intervals for task completion
will be discussed and in part two (coming soon) confidence intervals
for task times will be illustrated.
Need a CI refresher? Try the adaptive
confidence interval lessons.
Part 1: Confidence Intervals
for Task Completion
Even if the only metric you're gathering during a usability test is
a binary assessment of task completion (complete, didn’t complete)
you can and should still provide a confidence interval.
Here’s why. If five out of five users complete a task, you can
be 95% confident that the completion rate could be as high as 100%
but it also could be as low as 48%. In other words, if you tested
another five users, their completion rate will fall somewhere between
48% and 100%. Or if you're testing 100 users, as many as 52 of the
100 could fail the task. Calculation Note(added 8/13/2008)
Most analysts would agree that a 48% completion rate
is unacceptable. What is often overlooked is the as unacceptable degree
of uncertainty. The question every analyst must also ask is, “How
much risk are we willing to accept?” That risk is described
in two places:
-
The confidence level
-
The width of the confidence interval
Both are easy to calculate and present along with the completion
rate. Here’s how.
Step 1: Determine your
Confidence Level
Quite simply, the confidence level represents the likelihood that
another sample will provide the same results. It is the percent likelihood
statement that accompanies the width of the confidence interval. It
is set often to the 95% level by convention but can be adjusted. You
would want to lower it to 90% or 85% or raise it to 99% depending
on the impact of being wrong. A confidence level of 99% means that
1 out of 100 times your sample completion rate will NOT fall within
your confidence intervals. We will set it to 95% for these examples.
Step 2: Calculate the Confidence
Interval
Example 1: 50% Completion Rate (Symmetric
Confidence Intervals)
To create the confidence intervals for task completion we use the
binomial distribution, since the event is binary/binomial. To start
with a simple example lets assume you tested 40 users attempting to
complete one task. Half of them completed the task and half failed,
making the completion rate 50%.
First we calculate the standard error of the mean. Since we’re
sampling to determine the real mean, the standard error is as the
name suggests, the estimate of the error between the true mean of
the population and our sample mean. Think of the standard error as
the standard deviation of the mean—or the area where the sample
mean can “float” as we take multiple samples. The bigger
the standard deviation, the bigger the error. The standard error is
calculated by dividing the standard deviation by the square root of
the sample size. To obtain the standard deviation we multiply the
proportion of people who completed the task (p) times the proportion
of people who failed the task (1-p or q).
(.5)×(.5) = .25
This value is the variance of the sample. The variance is the standard
deviation squared, so to obtain the standard deviation we take the
square root of .25 which is .50. Then divide that figure by the square
root of the sample—in this case 40. The square root of 40 is
6.325.
| |
StDev. |
|
SQRT Sample Size |
|
| Standard Error = |
.50 |
÷ |
6.325 |
= .0791 |
Now that you have the standard error, you
need to multiply this value times a critical value from the t-distribution.
This value is derived from the confidence level set above (.05) and
the sample size minus one, 39 (called degrees of freedom).
To get this critical value you can either look the value up in a
t-distribution
table or use the excel function [=TINV(.05, 39)]. This will return
(2.023) which is the critical value for a 95% confidence level with
(40-1) degrees of freedom.
And finally, the confidence interval is the standard error times
the critical value.
| |
Stnd
Err |
|
t-value |
|
| Confidence Interval = |
.0791 |
× |
2.023 |
= 16% |
So we can say with 95% confidence that the sample proportion of 50%
could be as low as 34% (p – CI) or as high as 66% (p +CI). If
you were to continue sampling users to complete a task, 95 times out
of 100, the proportion will fall somewhere between 34% and 66%.
That’s the basic concept behind calculating the confidence
interval. The one unfortunate wrinkle in this neat calculation is
that the confidence interval is NOT usually symmetric. That is, it’s
not always a neat + or – one number (16% in the example). As
the proportion successful or unsuccessful gets closer to 100% complete
or 0% complete, the binomial confidence interval departs from symmetry.
Try it for yourself. What if 38 out of 40 users completed a task (p
=.95). Using the formula above would provide a confidence interval
of 7%.
Standard Error = SQRT[(.95)×(.05)]÷
SQRT(40) = .0344
t-statistic = (.05, 39) =
Confidence Interval = .0344× 2.023 =.0697
You can see immediately something’s awry as it doesn’t
make sense to say that that we are 95% confident the true mean lies
between 88% and 102% since there is no way to have more users complete
a task than attempt it. The larger issue is that in binomial distributions
the confidence interval is only symmetric when p is .5 or
very close to .5. More often, the proportion is far from
.5 and TWO confidence intervals need to be derived—one for above
the proportion and one below the proportion. If this sounds confusing,
don’t worry, it is and so is the calculation to derive them.
Example 2: 95% Completion Rate (Asymmetric
Confidence Intervals)
There are at least two ways I’ve found to derive these asymmetric
binomial confidence intervals. The first method is using a technique
called the Paulson-Takeuchi approximation and the
second uses the incomplete beta function and the F distribution. Don’t
worry if you’ve never heard of either one of them, unless you’re
a total stats geek, you shouldn’t have. The important point
is that either one of these methods provides accurate asymmetric confidence
intervals. Even better, there are calculators on the web that will
do the work for you and I’ve also built an Excel spreadsheet
you can download where all you have to do is plug in the values but
can still see the workings of the formula. I used the incomplete beta
function approach since there is a paper
publicly available that shows the formulas.
Try this confidence interval calculator, or
if you have Minitab you can get the confidence intervals from Stats
> Basic Statistics > 1 Proportion.
Download Excel Binomial
Confidence Interval Calculator
Figure1: Binomial Confidence Interval Calculator
Using either the Web, Minitab or the Excel calculator, take that
same 95% completion rate and obtain the confidence intervals for the
38 out of 40 task (use .05 for the alpha error). You should get the
lower interval as 83% and upper interval should be 99.4%. Compare
that with the 88% and 102% intervals calculated earlier.
For those of you who don’t want to download my spreadsheet
or enter the values, you can also see a
large table with many common values and their upper and lower
95% confidence intervals.
Using Asymmetric Confidence Intervals:
A real example
All these calculations are nifty, but you may have a hard time being
convinced that a 100% completion rate with five users could turn into
a 48% completion with 1000 users (something the 95% confidence interval
indicates is probable.) Here are the completion rates from of a summative
usability evaluation on a task from a financial application.
After testing five users we had a 100% completion rate. We know the
lower bounds of the interval could be 48% and continue testing to
see what happens.
| |
5 Users |
| Lower CI |
48% |
| p |
100% |
| Upper CI |
100% |
| CI Width |
52% points |
After 10 users the completion rate decreases to 80% and the lower
CI is 44%. Notice how the width of the confidence interval actually
got slightly wider.
| |
10 Users |
| Lower CI |
44% |
| p |
80% |
| Upper CI |
97% |
| CI Width |
53% points |
We continued testing and below are the results of testing 45 users.
Notice how the confidence interval begins to narrow and the mean begins
to stabilize as we test more users.
Figure 2: Proportions from completion rates
and 95% Binomial Confidence Intervals
Most usability testing, even summative benchmark testing, doesn’t
usually have a sample as large as 45 users. Even with a larger than
normal sample, the width of the confidence interval is still rather
large. What becomes immediately evident is that it is much easier
to proclaim a task completion rate unacceptable than it is to declare
it acceptable. That is, it’s hard to show usability,
it’s much easier to show un-usability. This is a point
Jim Lewis made in his 1996 article on Binomial Confidence intervals
and it's worth repeating—
[Binomial Confidence Intervals] cannot
be used with a small sample to prove that a success rate is acceptably
high. With small samples, even if the observed defect percentage is
0 or close to 0 percent, the interval will be wide, so it will include
defect percentages that are unacceptable. Therefore, it is relatively
easy to prove (requires a small sample) that a product is unacceptable,
but it is difficult to prove (requires a large sample) that a product
is acceptable (Lewis p.735).
What Confidence can do for
you
Here's what you should take away from this article. First, binomial
confidence intervals are a resource saving tool during formative evaluations.
When refining a new feature that needs a high completion rate, say
90% for first time users, you’ll know when to reject a design
earlier. If only two out of five users complete the task, there’s
less than a 5% chance that the completion rate will ever be above
85%.
Second, it should temper both overconfidence and excessive skepticism
in usability findings. It will show you and the readers of usability
reports the true sense of confidence in a completion rate. Instead
of leaving the door wide open for attacks on sample size issues, the
door now is partially closed (it’s only open as wide as your
confidence intervals). Instead of arguing about undefined possibilities,
you can discuss probabilities. If the lower limits of the confidence
interval are too low, then you can either sample more users or know
the unacceptable completion rate is indicative of a problem that needs
to be addressed.
Calculation Note: When the observed completion rate is 100% or 0% there cannot be a two sided confidence interval (since you cannot have more than 100% or less than 0%). In these cases it is necessary to use a z-critical value for a one-sided confidence interval. For example, a 95% two sided confidence interval uses the z-score of approximately 1.96, a one sided interval uses a z-score of approximately 1.64. It is often the case that statistics text or examples you encounter do not make this adjustment for simplicity in instruction. The 5/5 users completing a task example in this article would then have an interval of between 55% and 100%. The Confidence Interval Calculator will make this adjustment automatically.
References
-
Fujino, Y. (1980).
Approximate binomial confidence limits. Biometrika, vol. 67, 677-681.
-
Harte, David, “Non
Asymptotic Binomial Confidence Intervals” from the Statistics
Research Associates (1992). Publication downloaded June 2004
-
Lewis, James “Binomial
Confidence Intervals for Small Sample Usability Studies”
in Proceedings of the 1st International Conference on Applied
Ergonomics. Istanbul, Turkey, May (1996)
-
Landauer, Thomas K, “Research
Methods in Human-Computer Interaction” in The Handbook
of Human Computer Interaction M. Helander (1998)
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| Hi Jeff thanks for all your help so far, your answers have been great and very timely. It is great to be able have help from someone with great professionalism as yourself.
I have another question which I am unsure as how to tackle it. I hope you can help me once again – thanks. So here goes.
For traffic signs, measurement of retro reflective performance with a reflectometer will be needed as well as visual assessment during Street Surveillance inspections.
The Service provider (that’s me!) shall collate and report on the results of actual reflectivity readings for all signs visually identified as required a measured reflectivity assessment to ensure that the Contract Manager and the Service provider have confidence in the visual assessment of retro-reflective performance.
The Service provider shall ensure that its procedures to determine the retro reflective performance of Signs provide valid assessments.
(So what we have done is taken random readings of various signs for the quarter, see below.)
The Australian Standards for these signs states that for Class 1W signs the minimum level of luminous intensity for white, yellow & red signs are 380, 265 & 75 respectively which are to replace the old standard Class 1 signs; white, yellow and red of min. levels of 250, 75 & 50 respectively.(This is a works in progress. It was not noted when taking the readings whether the signs were the new or old standard? I assume the the old standard signs would be the ones with readings of lower values? But they could also be old faded signs with low readings. Do we need to separate the two classes of signs or should we just keep it simple with a general observation?)
Further more there are also Class 2 signs which the standards state that the minimum luminous intensity for white, yellow & red signs are 70 50 & 14 respectively.
Finally there is one other question which I don’t believe we are in the position to answer without further historical data but I would appreciate your comments.
The question asks to assess and report on the degradation of reflectivity performance as a method of predicting the future reflectivity performance of signs.
Both Class1 sign readings are:
Reflectivity Colour
91.9 W
277 W
403 W
514 W
458 W
97.7 W
520 W
416 W
481 W
508 W
448 W
516 W
282 W
264 W
240 W
276 W
262 W
258 W
296 W
211 W
183 W
46.5 R
45.1 R
80 R
381 R
235 R
35.9 R
and for Class2:
Reflectivity Colour
32.1 Y
51.4 Y
5.7 Y
41.2 W
45.3 W
79.2 W
52.1 W
54.8 W
63.5 W
22.1 W
60.4 W
77.9 W
78 W
81.6 W
78 W
68.2 W
66.4 W
21 W
72.5 W
3.1 W
50.1 W
106 W
87.4 W
85.3 W
18.1 W
78.9 W
62.5 W
64.5 W
61.2 W
76.9 W
82 W
24.5 W
65.5 W
19.6 W
48.5 W
22.4 W
48.4 W
79.9 W
40 W
84.2 W
7.4 W
45.4 W
80 W
9 W
86.2 W
70.2 W
80.3 W
87.4 W
105 W
86.3 W
11.4 R
26.5 R |
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I would like to see a 2% improvement in my conversion rate.
|
| The width of a confidence estimate for a proportion will be:
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b. Wider for a sample size of 100 than for a sample size of 50.
c. Narrower for 90% confidence than for 95% confidence
d. Narrower when the sample proportion is .50 than when the sample proportion is .20. |
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Old: 40, 28, 35, 38, 31, 42, 26, 44, 29, 43
New: 29, 26, 31, 26, 28, 31, 19, 21, 27, 30
Calculate the mean, median, and mode.
Calculate variance and standard deviation for each set.
Calculate a 95% Confidence Interval for the two sets. |
| How can I compute a one-sided 97.5% confidence interval using SPSS for this ?
IN a cohort of 121 eyes treated with drug A, 3 eyes experience a drug related side effect, i.e 3/121. Thanks |
| Jeff:
Can you possibly share the actual formula you are using in this application to produce the Adjusted Wald calculations?
Thanks |
| Linemarking for parking bays shall be determined by the retroreflectivity performance of glass beads in the linemarking. The linemarking average level of reflectivity over the City is to be not less than 100 mcd/sqm/lx and the minimum aceptable reflectivity is 80 mcd/m2/lx. Conduct a quarterly assessment of roadmarking condition, randomly sampling 5% of the City Roads each quarter so that 20% of all the roadmarkings are assessed each year. The results shall produce the results specified below. The methods used to perform the assessment shall be sufficient to produce a confidence interval of +/- 5% with a level of confidence of 95%;
(a)Average retroreflectivity for the roadmarking
(b)Percentage of roadmarking that are below the minimum acceptable standard.
The recordings of roadmarkings for the second quarter are as follows:
260.00
263.00
254.00
180.00
174.00
229.00
230.00
56.00
209.00
260.00
309.00
359.00
389.00
491.00
202.00
387.00
440.00
Thanks again |
| Popcorn kernels take between 100 and 200 seconds to pop. What sample size (number of kernels) would be needed to estimate the true mean seconds to pop with and error of 5 seconds and 95% confidence level? |
| Why don't statisticians calculate 100% confidence intervals? |
| Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 86. (a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop. (b) Check the normality assumption. (c) Try the Very Quick Rule. Does it work well here? Why, or why not? (d) Why might this sample not be typical? |
| What is the Z score of a 98% confidence interval? |
| A random sample of n=64 children of working mothers showed that they were absent from school an average of 5.3 days per term, with a standard deviation of 1.8 days. Provide a 96% confidence interval for the average number of days absent for all students. |
| Briefly describe the concept of a confidence interval and provide an example. |
| Find a confidence interval for the mean assuming that each sample is from a normal population. Mean = 127, s = 27, n = 16. Find the 90% Confidence Interval. |
| A machine produces 3 inch nails. A sample of 100 nails is selected, and it is found that 25 are shorter than 3 inches. Find the 95% confidence interval on the proportion of all such nails that are shorter than 3 inches. |
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| December 11, 2007 | Lei wrote: |
| Very useful infomation. thanks. |
|
