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 < [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)]
[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,
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.