Use your calculator's invNorm function to determine the z-score, accurate to two decimal places, for confidence intervals of these percents, the blue area in the graph.
These z-values are called critical values.

99.7%
95%
90%
Exact
General
Usage answer: 2.9677
answer: 1.9599
answer: 1.6448
3.000
1.960
1.645
invNorm (A) = z
Practice on these (great test question):
62.5% answer: 0.88714
87.4 %
50%
100% answer: 1.530067
answer: 0.6744897
answer: infinity
Check your answers using normalcdf function on your calculator.
normalcdf (-z, z) should give you the percent you started with.

#28, page 506
In 2004 ACT, Inc. reported that 74% of 1644 randomly selected college freshmen returned to college the next year.
Estimate the national freshman-to-sophomore retention rate.
a) Verify that the conditions are met.
b) Construct a 98% confidence interval.
c) Interpret your interval.
d) Explain what "98% confidence" means.

a) It seems likely to me that one's decision to continue to sophomore year is independent of other's decisions.
10*1644 = 16440 < total population of college freshmen n*p = 1644*74% = 1216.56 > 10 n*q = 1644*26% = 427.44 > 10 b) (1 - 0.98)/2 = 0.01
invNorm(0.01) = -2.32634 ==> z* = 2.326
p-hat = 0.74
ME = z*SD = 2.326*sq-rt(0.74*0.26/1644)
= 2.326*0.010818
= 0.02516
CI = p-hat +/- ME
= 0.74 +/- 0.02516
= 0.715, 0.765
c) We are 98% confident that between 71.5% and 76.5% of freshmen return for their sophomore year.
d) If we were to select repeated samples like this we'd expect about 98% of the confidence intervals we created to contain the true proportion of freshmen who return for their sophomore year.

#35, page 507
It's believed that as many as 25% of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group.
a) How many of this younger age group must we survey in order to estimate the proportion of non-grads to within 6% with 90% confidence?
b) Same question within 4%.
c) Same question within 3%.

ME = z*sq-rt(p*q/n) OR n = pq(z*/ME) 2
First, find z* for 90% confidence interval. It is 1.645.
p = 0.25; q = 0.75; z* = 1.645
n = 0.25*0.75*(1.645/ME) 2 n = 141 for ME = 6% n = 318 for ME = 4% n = 564 for ME = 3%
ALWAYS round up!

#35 modified, page 507
The modification: we don't know this 25%!!
It's believed that as many as 25% of adults over 50 never graduated from high school.
We wish to see if this percentage is the same among the 25 to
30 age group.
a) How many of this younger age group must we survey in order to estimate the proportion of non-grads to within 6% with 90% confidence?
b) Same question within 4%.
c) Same question within 3%.

A larger value for n will shrink the ME.
The formula for n is n = pq(z*/ME) 2 .
The z* and ME are fixed, so we look at pq.
Calculus: use x for p and (1-x) for q and y for pq.
y = x(1-x) = -x 2 + x for 0<= x <= 1.
It is a parabola opening down with max value when x = 0.5.
So, if we have no idea what the value for p is then use 0.5 to calculate n. This may give you a much larger n than is necessary. But it is the best we can do.