Homework 7 Solution

5.2.32
X represents the score of a single randomly selected student.  X is roughly  Normal, mean =21, SD = 4.7.
a) P(X > 23) = P(Z > (23-21)/4.7) = P(Z > .4255) = .335 or 33.5%
b) Xbar is an average of 50 randomly selected students.  Xbar has an approximately normal distribution.  The mean is still 21, the standard error (standard deviation) is 4.7/sqrt(50) = .664
c) P(Xbar > 23) = P(Z > (23-21)/.664) = P(Z > 3.01) = .001
d) The normal probability calculation in (c) is more accurate because it is based on  a larger sample size.  Because the distribution is not normal, all calculations are approximate.  But the Central Limit Theorem says the larger the sample size, the better the approximation.  The sample size in (a) is n = 1 and in (b) it's n = 50.

#3) The Pew Reserach Center Poll held on May 14 claims that 60% of Americans favor affirmative action (at least within the context of the question asked on the poll. )  They surveyed 1201 adults.  Assume that p-hat, the proportion in a random sample of size 1201 that would say they favor affirmative action is normally distributed.  Now a politician claims that the 60% represented by the Pew Center is a statistical fluke.  In fact, he claims, the true percentage of Americans favoring affirmative action is no more than 50%.   Find a 95% confidence interval for the percent of all Americans who would say they favor affirmative action.  Interpret this confidence interval with respect to the politician's claim.
HINT:  The standard error of p-hat is the squareroot of (p*(1-p)/n).  But you don't know p.  Set p = 1/2  and you'll get a "worse case" scenario that provides a reasonably conservative CI.  Then assume that phat follows a normal distribution (since np > 10 and n(1-p) > 10)

The proportion of people who favor A.A. within the population is unknown.  The proportion in the sample (which had only 1201 of the millions or so in the population) had 60% who favored.  A proportion based on a random sample, which we call p-hat, follows an approximately normal distribution with mean p and standard deviation sqrt(p*(1-p)/n).  An approximate 95% confidence interval would be
p-hat +/- 1.96*sqrt(p*(1-p)/n)
Problem is that we don't know p, and so to get a conservive margin of error, we use p = 1/2

95% CI is .60 +/ 1.96*sqrt(.5*.5/1201)   or .60 +/-  .028  or (.57, .63).

This makes the politician's claim that the mean is really 50% seem implausible.