# Proportion ample Size

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• #53650

chris reid
Participant

Hi, I’m working on a project in a contact centre and have a plan to reduce call handling time. I want to work out how many calls I should plan to sample to prove Ive made a difference.

The Existing data.
Standard deviation= 477
Mean 1014

I believe my project can reduce this handle time (in a good way) by 240 seconds. I’ve used a 2 sample t test (mini tab 15) and eneterd the following

Differences 240
power .9
Standard deviation = 477
Result = 84 samples required.

Have I done this correctly??? :unsure:

0
#190998

Christopher G. Connors
Participant

Sorry, no. I don’t think you have a normal distribution with cycle time data. A non-parametric test should be used (You don’t say what the sample size was to collect the baseline for your average and sd, but with ). Typically you would use a discrete sample size calculation.

Wiki
In probability theory and statistics, the Poisson distribution (pronounced [pwas??]) (or Poisson law of small numbers[1]) is a discrete probability distribution that expresses the probability of a number of events occurring in a fixed period of time if these events occur with a known average rate and independently of the time since the last event. (The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume.)

http://www.une.edu.au/WebStat/unit_materials/c6_common_statistical_tests/nonparametric_test.html

http://www.qualitydigest.com/dec99/html/nonnormal.html

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#190999

Robert Butler
Participant

The t-test is robust with respect to non-normality so there’s a good chance your approach is ok. One issue with a Poisson is that of the possible extreme nature of the tail. If it is too extreme the t-test will fail to detect a difference when one exists. If you have this situation a better test would be the two sample Wilxocon-Mann-Whitney test.

I’d recommend looking at your existing distribution – histogram as well as the summary statistics of minimum, maximum, median, and mean. If the median and mean are “reasonably close” ( a judgement call to be sure) then the t-test will probably suffice. If they aren’t or if you just want to make certain from the start – generate the sample size estimate for the Wilcoxon-Mann-Whitney instead.

However, given what you have provided, I’m getting the following for a one sample test:

Null Mean = 1014

Improved Mean = 774

Sample standard deviation = 477 (assumed to be the same as the initial )

One sided test (since we are only interested in improvement)

Power = .9 and alpha = .05

therefore n total = 36

Where the n is for the samples taken using the new procedure.

If I try a two sample mean calculation and choose the option of differences in means and assume both populations have the same standard deviation, the test is one sided, power = .9 and alpha = .05 I get an N total of 138 or 69 samples per group.

I don’t have Minitab but the fact that I can’t get your number bothers me. The only thing I can think of is that perhaps Minitab has a distribution option. If that’s the case and if you chose Poisson then I could believe this would explain the difference in our results. If this isn’t the case then I really don’t know what to offer in this regard.

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#191002

chris reid
Participant

:) Thanks a lot, really well explained and much appreciated. I’vee have another go.

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#191006

broderick
Participant

Do you mean you ran a Power and Sample Size calculation for a 2 sample T using MTB 15? If not, what are you comparing?

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