Answer:
a) The 90% CI for the true rate is [4.44%, 4.65%].
[tex]0.04437 \leq \pi \leq 0.04653[/tex]
b) The rate of 4.8% is not plausible.
Step-by-step explanation:
The question is incomplete:
A philanthropic organization sent free mailings labels to a random sample of 100,000 potential donors and received 4,545 donations.
Give a 90% confidence interval for the true proportion of those from their entire mailing list who may donate. (b) A staff member thinks that the true rate is 4.8%. Given the confidence interval you found, do you find that rate plausible?
The proportion of the sample is:
[tex]p=X/N=4,545/100,000=0.04545[/tex]
The estimated standard deviation is
[tex]\sigma=\sqrt{\frac{p(1-p)}{N}}=\sqrt{\frac{0.04545*0.95455}{100,000}}=0.00066[/tex]
The z-value for a 90% confidence interval is z=1.645.
Then, the CI can be constructed as:
[tex]p-z \sigma \leq \pi \leq p+z \sigma\\\\0.04545-1.645*0.00066 \leq \pi \leq 0.04545+1.645*0.00066\\\\0.04545-0.00108 \leq \pi \leq 0.04545+0.00108\\\\0.04437 \leq \pi \leq 0.04653[/tex]
The 90% CI for the true rate is [4.44%, 4.65%].
b) The value 4.8% lies outside the 90% confidence interval (more specifically, over the upper limit), so it has less than 5% of probability of being the true rate.
The rate of 4.8% is not plausible.
