Confidence Interval Calculator for a Completion Rateby Jeff Sauro | October 1, 2005 :: 2 Related Articles:: 74 Related Questions
Use this calculator to calculate a confidence interval and best point
estimate for an observed completion rate. This calculator provides the
Adjusted Wald,
Exact,
Score
and
Wald intervals.
Explanation
The Adjusted Wald method should be used almost
all the time. For exceptions, see below.
For a detailed discussion of binomial confidence intervals with
small samples, see the
HFES paper
and for a discussion on the best point estimate see the
JUS
paper.
Adjusted Wald Method
The adjusted Wald interval (also called the modified Wald interval),
provides the best coverage for the specified interval when samples
are less than about 150. In other words, if you want a 95% confidence
interval then this formula will produce an interval that will contain
the observed proportion on AVERAGE about 95 percent of the time.
It uses the Wald Formula but is "adjusted" in that it
adds half of the squared
Z-critical value to
the numerator and the entire squared critical value to the denominator
before computing the interval i.e (x+z
2/2)/(n+z
2).
For example, a 95% confidence level uses the
Z-critical
value of 1.96 or approximately 2. If you observe 9 out of 10
users completing a task, this formula computes the proportion as(
9 + (1.96
2/2) )/ (10 + (1.96
2)) = approx.
11/14 and builds the interval using the Wald formula.
Note: Prior
to March 1st 2006, this calculator computed this interval by adding
one z-value to the numerator and a squared z-value to the denominator.
Exact Method
The Exact method was designed to guarantee at least 95% coverage,
whereas the approximate methods (adjusted Wald and Score) provide
an average coverage of 95% only in the long run. Use the Exact method
when you need to be sure you are calculating a 95% or greater interval
- erring on the conservative side. For example, at the population
completion rate of 97.8% both the Score and adjusted Wald methods
had actual coverage that fell to 89%. When the risk of this level
of actual coverage is inappropriate for an application, then the
Exact method provides the necessary precision.
Score Method
The Score method provided coverage better than the Exact and
Wald methods but falls short of the adjusted Wald method. Additionally,
its drawback is its computational difficulty and its poor coverage
for some values when the population completion rate is around 98%
or 2%, regardless of sample size (Agresti and Coull, 1998). The
only advantage in using the Score method is that it provides more
precise endpoints when the ends of the intervals are close to 0
or 1. For some values (e.g. 9/10) the adjusted Wald's crude intervals
go beyond 0 and 1 and a substitution of >.999 is used. For the
score method, the upper interval is .9975.
Wald Method
The Wald method should be avoided if calculating confidence
intervals for completion rates with sample sizes less than 100.
Its coverage is too far from the nominal level to provide a reliable
estimate of the population completion rate. As the sample size increases
above 100, all four methods converge to similar intervals. Use the
Wald as a point of reference or for larger sample sizes.
* The "Margin of Error"
values are half the width of the Confidence Intervals. For the adjusted
wald and wald formulas, you can use the proportion +/- the confidence
interval. For the exact method, the intervals are not symmetrical
as the proportion complete gets further from 50% (e.g. 90% or 15%).
Therefore the margin of error should be only used at as an approximation
for the exact method and the actual values above and below the proportion
should be reported.
When All Users Pass or Fail
With small sample sizes, it is a common occurrence that all users
in the sample will complete a task (100% completion rate) or all
will fail the task (0% completion rate). For these scenarios, it
is often unpalatable to report 100% or 0%. After all, how likely
is it that the true population parameter is as extreme as 100% or
0%? The Best Estimate box provides the best point estimate under
these conditions and uses the
LaPlace method
for calculation. While this value may seem too far from the observed
100%, its attractiveness is that it is a function of the sample
size-- the greater the sample size, the closer this value will be
to 100%.
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.
Likely Population Completion Rate
The two options in this drop-down:
Between .5 and 1
If you conduct usability tests in which your task completion rates
are roughly restricted to the range of .5 to 1.0, then select "Between
.5 and 1" in the drop-down. See the
Best
Estimates section below for how the point estimate is calculated
with this option.
Unknown
If your task completion rates typically take a wide range of values,
uniformly distributed between 0 and 1, then select "Unknown"
from the drop down. If you don't know either way then leave it at
"Unknown." This selection will use the LaPlace method
for the best estimate of the completion rate.
Point Estimates
Whereas a confidence interval describes a likely range or
interval
of values, a point estimate describes a single value-
a point
as an estimate of an unknown parameter in the population. The
chance that the sample point estimate is the same as the unknown
population completion rate is extremely unlikely. For that reason,
you should always compute a confidence interval when reporting a
completion rate. It is much more informative than a point estimate
since it provides a reasonably likely boundary for the population
completion rate.
Although it receives little attention in introductory
statistics classes and has had little influence on measurement practices
in the field of usability engineering, there is a rich history of
alternative methods developed to achieve a more accurate point estimate
of p than simply dividing the number of successes by the number
of attempts (for example, see Chew, 1971; Laplace, 1812; Manning
& Schutze, 1999). This need is most evident when there is an
extreme outcome, specifically, when x=0 (0%) or x=n (100%) - especially,
but not exclusively, when sample sizes are small.
Four estimation methods that pertain to situations
more common in usability testing are detailed below:
MLE:(Maximum Likelihood Estimate)(x / n)
The MLE is the sample proportion or the number of users succeeding
divided by the total attempting. It is the most common point estimate
reported.
LaPlace (x+1)/(n+2)
A famous large-sample problem comes from the seminal work of
Laplace in the early 1800s. He posed the question of how certain
you can be that the sun will rise tomorrow, given that you know
that it has risen every day for the past 5000 years (1,825,000 days).
You can be pretty sure that it will rise, but you can't be absolutely
sure. The sun might explode, or a large asteroid might smash the
Earth into pieces. In response to this question, he proposed the
Laplace Law of Succession, which is to add one to the numerator
and two to the denominator ((x+1)/(n+2)). Applying this procedure,
you'd be 99.999945% sure that the sun will rise tomorrow - close
to 100%, but slightly backed away from that extreme. The magnitude
of the adjustment is greater when sample sizes are small. For example,
if you observe two out of two successes and apply the LaPlace procedure,
then your estimate of p is 75% (x+1=3, n+2=4, p=3/4) rather than
100%. If you had observed two failures, then your estimate of p
is 25% (x+1=1, n+2=4, p=1/4) rather than 0%. LaPlace in essence
is saying, the next result is a toss up, so give each alternative
an equally likely chance of occurring.
Wilson (x+z2/2)/(n+z2)
Wilson's point estimate is the midpoint of the adjusted wald interval.
It is derived by adding half a squared critical value to the numerator
and a squared critical value to the denominator. Wilson's is the
more conservative approach.
Jeffreys (x+.5)/(n+1)
Jeffreys (1961) provided a compromise between the LaPlace and MLE
methods. See reference for technical details.
Best Estimate
The best point estimate is calculated using the following logic:
If "Unknown" is selected from the
Likely
Population Completion Rate drop-down, the
LaPlace
method is used. The smaller your sample size and the farther
your initial estimate of p is from .5, the greater the benefit over
the MLE.
If "Between .5 and 1" is selected from the Likely
Population Completion Rate drop-down and the observed completion
rate is:
-
Less than or equal to .5: the Wilson
method is used.
- Between .5 and .9: the MLE is used.
- Greater than .9: the LaPlace method
is used (Note, if 1 > x > .9 the Jefferys
method is also a viable alternative).
Need more information? Be sure to check out the online
confidence interval tutorial.
References
- Agresti, A., and Coull, B. (1998). Approximate is better than 'exact'
for interval estimation of binomial proportions. The American Statistician,
52, 119-126.
- Chew, V. (1971). Point estimation of the parameter of the binomial
distribution. The American Statistician, 25, 47-50.
- Jeffreys, H (1961) Theory of Probability (3rd Ed),
Clarendon Press, Oxford pp. 179-192.
- Laplace, P. S. (1812). Theorie analytique des probabilitites.
Paris, France: Courcier.
- Lewis, J.R. & Sauro, J. (2006) "When
100% Really Isn't 100%: Improving the Accuracy of Small-Sample Estimates
of Completion Rates" in Journal of Usability Studies
Issue 3, Vol. 1, May 2006, pp. 136-150
- Manning, C. D., & Schutze, H. (1999). Foundations
of statistical natural language processing. Cambridge, MA: MIT Press.
- Sauro, J & Lewis, J R (2005) " Estimating
Completion Rates from Small Samples using Binomial Confidence Intervals:
Comparisons and Recommendations" in Proceedings of the
Human Factors and Ergonomics Society Annual Meeting (HFES 2005)
Orlando, FL
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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.)
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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
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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
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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
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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
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count 53 47
mean 36,492.92 24,451.51
sample variance 340,313,003.72 154,893,232.30
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__________
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b. $200? __________
c. $100? __________
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a. Narrower for a 99% confidence interval than for a 95% confidence interval.
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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Could you help me determine the formula for sample size at a 95% confidence level?
My historical Conversion Rate (success): 11%
Visitors: 280 per day
I would like to see a 2% improvement in my conversion rate.
|
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The principal randomly selected six students to take an aptitude test. Their scores were:
77.9 89.1 80.7 78.6 74.4 82.0
Determine a 90 percent confidence interval for the mean score for all students.
76.36 < x < 84.54
84.64 < x < 76.26
76.26 < x < 84.64
84.54 < x < 76.36 |
| 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 |
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a. State the null and alternate hypotheses.
b. State the decision rule.
c. Compute the value of the test statistic.
d. Compute the p-value |
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A. What is the probability that 3 out of the 5 order a blueberry muffin?
B. Explain why this scenario would meet the criteria of a binomial distribution.
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A) What percent of the students scored below 30?
B) What percent scored between 30 and 55? |
| Survey to determine what proportion of new car buyers continue to have their car serviced at the dealership after warranty ends. Estimates 30% of customer do so. Results should be accurate within 5%. Also 95% confident of the results. What sample size is necessary? |
| out of a random sample 167 students pass an exam out of the 300. how do you calculate an exact 99% confidence interval for the proportion of sudents who passed the exam? |
| Suppose that in NY State, 35% of voters are Democrats, 30% are Republicans, and 11% are independent. You are conducting a poll by randomly calling registered voters. In your first four calls, what is the probability that you talk to:
a)all Republicans?
b)exactly 2 Republicans?
c)no Independents?
d)all Independents?
e)at least one independent? |
| A random sample of 25 households finds that an average of 2.3 people reside in each house (the standard deviation is 0.35). With a 95% confidence level, what is your estimation of the population average? |
| A telescope manufacturer wants its telescopes to have standard deviations in resolution to be significantly below 2 when focusing on objects 500 light-years away. When a telescope is used to focus on an object 500 light years away 30 times, the sample standard deviation turns out to be 1.46.
a.State explicit null and alternate hypotheses
b.Test your hypothesis at the á=0.01 level. |
| When calculating confidence interval estimates...how is the x and standard deviation calculated? |
| The Web-based company Oh Baby! Gifts has a goal of processing 95 percent of its orders on the same day they are received. If 485 out of the next 500 orders are processed on the same day, would this prove that they are exceeding their goal, using á = .025? (See story.news.yahoo.com accessed June 25, 2004.) |
| A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured
(a display ad is a large block of multicolored illustrations, maps, and text). The data (in square millimeters) are shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
(a) Construct a 95 percent confidence interval for the true mean. (b) Why might normality be an
issue here? (c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence? (d) If this is not a reasonable requirement, suggest one that is. I am new at this and it would help if you could give me the formula and break it down step by step so I can understand. Thanks |
| Your company asks you to compare a new advertising campaign and the old one. Sample data on the accounts as follows:
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. |
| Administrative staff based at a business school in the UK are advised to take a 15 minute break from their personal computer after working on it for 90 minutes continuously. A random sample of 36 staff revealed that, one average a break was taken after 97.4 minutes of continuous use. The corresponding standard deviation was measured at 5.1 minutes.
a. Provide a 99% confidence interval for the population mean time for working on a PC continuously, prior to taking a break.
b. Conduct a suitable test at the 1% level of significance, to see whether staff, on average, are working longer than advised. |
| 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? |
| A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm. 1. State the null hypothese 2. state the alternative 3. Identify the test statistic 4. Find the P-value 5. What is the conclusion regarding the null hypothesis? 6. What is the final conclusion that addresses the original claim? |
| You are interested in whether men participate more than women in political activities, so you survey 632 men and 675 women and ask them how many political events they attended within the past year. Here are your results:
MEN WOMEN
mean = 1.58 mean = 1.42
s.d. = 0.10 s.d. = 0.23
N=432 N=375
With 95% confidence what do you conclude? |
Ask a Question
| June 3, 2009 | B Joseph wrote: |
| In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a 95 percent confidence interval for the population proportion of positive drug tests. (b) Why is the normality assumption not a problem, despite the very small value of p |
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| May 25, 2009 | Sujan Karki wrote: |
| I want to calculate confidence intervel of cluster sample. How to use this calculator for CI for cluster effect? any modification or can not use this calculator?
thanks
sujan |
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| April 4, 2009 | sammy wrote: |
| 4nWrDb vkoo7wvY5Xkfak7bf1Th |
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| April 4, 2009 | sammy wrote: |
| 4nWrDb vkoo7wvY5Xkfak7bf1Th |
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| February 8, 2009 | Alexandre miranda wrote: |
| cant find how to calculate the exercise on page 66 -confidence interval based on binomial distribution- (figure 4.1) |
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| May 20, 2008 | Charles Bedard wrote: |
| Not sure if my comment went throug. Instead of 2 as an answer to the question "What is 1+1", I entered 1.999999..... , which is mathematicaly equivilent. My joke. to repeat my comments.
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I find that all the estimators have one fatal flaw. A two sided confidence interval is specified with the presumtion that the error in each tail is alpha/2. When the number of successes is equal to zero or the number of trials, all the stated CI's take either 0 or 1 as one end of the CI and put ALL the error in the inside tail, making the CI a one sided confidence interval with alpha (not alpha/s) in the tail. I prefer a modified Agresti CI (using an unassumed prior to keep frequentests happy or a simple uniform over .5 to 1 (or .5 to 0) The modified Agresti CI is based on the Beta distribution since the distribution of the proportion is a continuous distribution. More honest, especialy in one-shot (non-production) situations. |
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| May 14, 2008 | Pieter Johnson wrote: |
| This website is excellent! Very helpful. |
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