mm
The number of respondents to test mailing of 1000 is 123, both these
numbers are large enough that we may assume that the Normal
approximation to the Binomial distribution is appropriate.
Then (N is the sample size in this case 1000, and p is the proportion of
recipients that will respond):
z = [123 - pN]/sqrt[N p (1-p)]
has a standard normal distribution, and so in 90% of cases lies between
+/-1.645. So:
-1.645 < [123 - pN]/sqrt[N p (1-p)] < 1.645
90% of the time, rearranging these inequalities:
-1.645 sqrt[N p (1-p)] < 123 -pN < 1.645 sqrt[N p (1-p)]
or:
[123-1.645 sqrt[N p (1-p)]]/N < p < [123+1.645 sqrt[N p (1-p)]]/N
90% of the time.
The problem now is that we don't know the value of sqrt[N p (1-p)], so
we have to estimate it from out test sample, for which N=1000, and our
estimate for p is pest=123/1000=0.123. In which case our estimate of
sqrt[N p (1-p)] is 10.39, and our inequality is now:
[123-1.645*10.39]/1000 < p < [123+1.645*10.39]/1000,
or:
0.106 < p < 0.140.
Which gives a 90% confidence interval for p of approximately (0.106, 0.140).
This interval is a random variable which 90% of the time contains theb) Explain what this interval means.
actual proportion of respondents.
see abovec) Explain what "90% confidence" means.
The confidence interval suggests that the results obtained from the testd) The company must decide whether to now do a mass mailing. The mailing won't be cost-effective unless it produces at least 5% return. What does your confidence interval suggest? Explain.
shot are unlikely to have been the result of a test on a population with a
response rate as low as 5%, but only suggests this as this is the wrong
procedure to apply for this last part.
RonL