Basic "with confidence" Method Question

pterry

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Sep 19, 2013
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Hi all,

I am trying to answer a confidence question and want to choose the most appropriate methodology to do so. Here is a small mock table of the type of question:

Day of the weekSuccess Rate
Sunday5.01%
Monday5.42%
Tuesday5.71%
Wednesday5.86%
Thursday4.78%
Friday7.74%
Saturday5.31%
The "Success Rate" figures come from a calculation of # captured/# seen for each given day across one month. Is there a good way to say "with confidence" at any alpha-level that, for example, Friday performs better than the other days? It outperforms the mean and median but I'm wondering if I can say anything more conclusive. Thanks in advance!

Terry
 
Last edited:
Hi all,

I am trying to answer a confidence question and want to choose the most appropriate methodology to do so. Here is a small mock table of the type of question:

Day of the weekSuccess Rate
Sunday5.01%
Monday5.42%
Tuesday5.71%
Wednesday5.86%
Thursday4.78%
Friday7.74%
Saturday5.31%
The "Success Rate" figures come from a calculation of # captured/# seen for each given day across one month. Is there a good way to say "with confidence" at any alpha-level that, for example, Friday performs better than the other days? It outperforms the mean and median but I'm wondering if I can say anything more conclusive. Thanks in advance!

Terry
I would like to assign an uncertainty to each of the calculated percentages. To do that I have to know the actual numbers captured, not just the percent. One simple procedure would be to assume Poisson statistics, such that the standard deviation to be attached to a number N is sqrt(N). For instance if 200 were captured, record that as \(\displaystyle N = 200 \pm 14\). If the total seen for that same time was 4000, the resulting percentage with its "confidence" expressed as standard deviation would be

success rate \(\displaystyle = \dfrac{200 \pm 14}{4000} = (5.00 \pm 0.35)\% \)

That should be close enough for government work. If you also include the Poisson statistics on the number seen, then

success rate \(\displaystyle = \dfrac{200 \pm 14}{4000 \pm 63} = (5.00 \pm 0.36)\% \) .. showing that the preponderance of the uncertainty is due to the smaller number in the ratio.

Once you see the "error bars" for the individual days, you will have a better sense of what is (most likely) random and what may be statistically significant.
 
Hi, thanks for the speedy response. Attaching a chart of my actual counts, for a single given month. I'm afraid I don't follow your answer (fairly new to stats)--If I wanted to provide evidence that Friday's high success rate is variance due to the actual day and not randomness, how can I utilize that formula to that end? It would be great if you could also provide an example on one of the day's calculations being done. Thanks again!

Untitled.jpg
 
Hi, thanks for the speedy response. Attaching a chart of my actual counts, for a single given month. I'm afraid I don't follow your answer (fairly new to stats)--If I wanted to provide evidence that Friday's high success rate is variance due to the actual day and not randomness, how can I utilize that formula to that end? It would be great if you could also provide an example on one of the day's calculations being done. Thanks again!

View attachment 3238
Unfortunately the "enrolled" numbers are so small that random fluctuations are probably masking and "real" differences. Lets go through and find the standard deviations, based on the square roots of the data. I am taking the percentage of enrolled/total.

Sunday, (23 ± 4.8)/(459 + 23) = (4.77 ± 0.99)%
Monday, (30 ± 5.5)/(554 + 30) = (5.14 ± 0.94)%
Tuesday, (41 ± 6.4)/(718 + 41) = (5.40 ± 0.84)%
Wednesday, (42 ± 6.5)/(717 + 42) = (5.53 ± 0.85)%
Thursday, (32 ± 5.8)/(712 + 32) = (4.30 ± 0.78)%
Friday, (38 ± 6.2)/(491 + 38) = (7.18 ± 1.17)%
Saturday, (22 ± 4.7)/(414 + 22) = (5.05 ± 1.08)%

TOTAL,.....(196 ± 14.0)/3549 = (5.52 ± 0.39)%

The "null hypothesis" would be that all 7 samples are drawn from the same population distribution with mean 5.52%.
Your "alternate hypothesis" would be that the Friday number is greater than that.
However, the Friday number is only 1.4 standard deviations higher than 5.52, which would happen at random 7.8% of the time.
Thus there is an indication that Friday might be higher, but the statistical evidence is not strong enough to reject the null hypothesis - that is, "all days are equal" has not been disproved.

Sorry about using jargon, but I hope the words make sense in the context of the problem.
 
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