The best way to explain this chapter is to outline one of the examples.
In this example, a researcher tests the effectiveness of semi-personal letters against form letters. The results from a test were:
- Sent 1,018 semi-personal letters, 325 responses.
- Sent 1,022 form letters, 225 responses
How much of an improvement in response is gained by using semi-personal letters?
Consider two population models: semi-personal (1) and form letters (2)
Prior density for each population is beta (1,1).
Therefore, the posterior density is:
Semi-personal: beta (326, 694)
Form letter: beta (226, 798)
We can calculate the following:
Semi-Person | Form | |
r | 0.3196 | 0.2207 |
r+ | 0.3203 | 0.2215 |
t (std deviation) | 0.01459 | 0.01295 |
We then calculate the z-score to obtain the probability that difference d = p1-p2 is less than zero:
Z = ( 0 – (r1 – r2)) / √ (t2 + t2) = -5.07
The probability for this z-score is p < 0.0001, therefore the probability of that difference being at least zero = 1
Similiarly, PdAL (probabilities that d is at least as big as) 0.05:
Z = ( 0.05 – (r1 – r2)) / √ (t2 + t2) = -2.506 à therefore PdAL0.05 = 0.994
Z = ( 0.10 – (r1 – r2)) / √ (t2 + t2) = 0.056 à therefore PdAL0.10 = 0.52
In other words, the probability of additional response rate of 5% using semi-personal letters as opposed to form letters is 99%.
The z-score can also be used to obtain probability intervals for the difference :
r1 – r2 + / - z * √ (t2 + t2).
If the probability interval for d contains zero, then the null hypothesis is supported.
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I’m still thinking about the differences between Frequentist and Bayesian statistics:
Bayesian statistics probably provides more information than Frequentist statistics:-
Bayesian
- Probability / credible interval
- Null hypothesis testing (using probability interval)
- Probability that difference between treatment & control success proportions is at least x
- Probability of next success observation
Frequentist
- Significant / p-value / null hypothesis testing
- Effect size
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