Problem Type: Confidence Interval Estimation of a Population Mean, , with a known or
given population standard deviation, .
Find the 98% confidence interval for the mean annual cost of auto insurance, . AAA found
that for 100 randomly selected clients who bought insurance on a given day paid a mean
premium of $750/yr with a standard deviation of $48/yr.
x  750
 x  48 / 100  4.80
98% confidence = .99 left tail area → z = 2.33
E = 2.33 * 4.80 = 11.18
Therefore there is a 98% probability that the population mean insurance costs for autos is within
$11.18 of $750, somewhere between $738.82 and $761.18.
Problem Type: Confidence Interval Estimation of a Population Proportion, p, with a large
sample (np>5 and nq>5)
A randomly selected sample of 400 potential customers were asked if they would commit
to buying a recently improved product they consume frequently. 65% of the sampled customers
stated their commitment to buy the recently improved product. Construct the 92%-confidence
interval-estimate for the proportion of all potential customers who will choose to commit to
buying the recently improved product, p.
pˆ  E  .65  .0419  from .6081 to .6919
E  z *  pˆ
.92% confidence means a left tail area of .9600  z  1.76
p * (1  p )
.65 * (.35)

 .0238
n
400
E  1.76 * .0038  .0419
 pˆ 