STAT 301 – BUSINESS STATISTICS
FALL SEMESTER 2010
“Knowledge Festival” #1 – Part 1 (Computational)
This “knowledge festival” is intended to assess your mastery of the computational aspects of this
course. This test is open-book, open-notes. That is, you may use any written resource material
during the examination (textbook, other reference books, course notes, lecture review
assignments, tables, etc.). You may not use internet-based sources, and may not confer with
others. You may, of course, use a calculator. You must show your work on all problems.
There are 100 points worth of problems on this exam; relative problem weights are given in
brackets. Unless the problem specifically indicates otherwise, use the traditional confidence
level of 95% and the traditional significance level of =0.05.
This “knowledge festival” is given under provisions of the Honor System of Stetson University.
The word “pledged” written before your name at the top of this page is a symbol of your ongoing
commitment to the Honor System.
Enjoy!!
Question 0 [4 points]:
Spell your name (correctly and legibly) at the top of the page.
Question 1 [18 points; 6 each part]:
Wildlife biologists studying dwarf pygmy wombats on the island of Kafoonistan have
observed that the weight of adult wombats is normally distributed with a mean of 42 ounces and
a standard deviation of 6 ounces.
a) What percentage of adult dwarf pygmy wombats weigh 46 ounces or more?
b) How heavy are the heaviest ten percent of all adult dwarf pygmy wombats?
c) What percentage of adult dwarf pygmy wombats weigh 20 ounces or less?
Question 2 [10 points]:
Find the sample mean, median, variance, and standard deviation for the following data.
(Show your work!)
4
3
8
4
1
Question 3 [12 points]:
The Gallup Poll website reports on a recent survey of 3527 adults nationwide. One
question on the survey was whether the respondent generally approved or disapproved of the job
President Obama was doing. Of those surveyed, 1552 reported that they generally approved.
Give a 95% confidence interval for the percentage of all U.S. adults who approve of Obama’s
job performance.
Question 4 [6 points]:
Monthly stock returns on the Boravian Stock Exchange are normally distributed with a
mean of +0.04% and a standard deviation of +1.2%. Professional market analysts at the
investment firm of Dewey, Cheatham, and Howe construct their mutual fund of Boravian stocks
simply by selecting companies at random. (This cuts down on research costs, leaving more
money for analyst paychecks.) What is the probability that a randomly selected group of thirty
Boravian stocks will have an average positive monthly return (that is, 0.0% or higher)?
Question 5 [20 points, divided as indicated]:
Ismerelda Tempusfugit knows that 42 is The Answer to Life, the Universe, and
Everything. She wonders whether it is also the Secret to winning the lottery. In other words, she
wonders whether 42 is more likely to be a winning lottery number than would be expected from
chance alone.
She knows that in the Florida lottery, six out of 53 numbers are selected as “winners”
each week. She obtains historical data from the Florida Lottery Commission. From those data,
she notes that out of the past 500 drawings in the Florida Lottery, 42 has been one of the winning
numbers fifty times.
a) [6] State Ismerelda’s null and alternative hypotheses, in words and in symbols.
b) [8] Compute an appropriate test statistic. Give the p-value.
c) [6] State an appropriate conclusion, both in statistical terms (reject/don’t reject) and in the
context of the problem.
Question 6 [8 points]:
The data below give annual tuition amounts (in Boravian dollars) for all twenty private
universities in the Kingdom of Boravia.
$10,000
$10,100
$10,250
$10,350
$10,475
$10,550
$10,750
$10,950
$10,950
$11,125
$11,375
$11,570
$11,750
$11,900
$12,350
$12,500
$13,500
$13,750
$14,150
$18,500
Sketch a histogram of these data. What does your graph tell you about tuition figures in
Boravia?
Question 7 [22 points, divided as indicated]:
Recall that an American roulette wheel has 38 numbers—1 through 36 plus 0 and 00.
We noted in class that any bet you place at roulette has a payoff of -$2/38 per dollar bet. There
is, however, one exception to that rule. One of the many ways you can gamble at roulette is to
bet on a block of five numbers (namely: 0, 00, 1, 2, 3). If you place this bet and win (that is, one
of those five numbers comes up), you win $6 net for every dollar bet. (You get your $1 back
plus $6 more.) Otherwise, you lose your bet.
a) [12] Find the expected value and variance of the net return from a single $1 bet on the block of
five.
b) [10] Suppose the bet is made 2000 separate times. What is the probability that you win
money, on average?
PARTIAL SOLUTIONS
1a) Pr(46 or more): z = (46-42)/6 = .67. From table: .2486. So prob = .5 - .2486 = .2514
1b) Top 10%: Look up 40% in table: 1.28. So 1.28 SD above mean: 42+(1.28*6)=49.68
1c) Pr(20 or less): z = (20-42)/6 = -3.67. Off the table. Probability is approx. 0
2) Mean = (4+3+8+4+1)/5 = 4
Median: 1 3 4 4
Variance = [ (4-4)2 + (3-4)2 +(8-4)2 +(4-4)2 +(1-4)2 ]/4 = 26/4 = 6.5
Standard deviation = square root of 6.5 = 2.55
8
3) We haven’t gotten to confidence intervals yet; they’ll be on the next knowledge festival.
Solution: [ Best Guess ] + [ # ]*[ st dev ]
Best guess = 1552/3527 = .44
( .44 ) + (1.96)*sqrt(.44*.56/3527)  .44 + .016
4) We haven’t gotten to this yet either.
Solution: z = (obs-exp)/sd = (0.0-0.04)/(1.2/sqrt(30)) = 0.18  table = .0714
 prob = .5+.0714 = .5714
5) We haven’t gotten to this yet earlier.
(Looks like I should have scheduled this knowledge festival for later!)
5a) H0: 42 appears randomly. =6/53 = .113. H0: 42 appears more than expected. >6/53.
5b) p=50/500 = .1. z = (obs-exp)/sd = (.1-.113)/sqrt(.113*.887/500) = .92
p-value = .5-.3212 [table] = .1788
5c) Do not reject the null hypothesis.
There is not enough evidence to conclude that 42 appears more than would be expected by
chance.
6) You should do a histogram of these data. Any (reasonable) scale is fine.
While sketching a histogram is simple, putting together a graphic that will illustrate one will
take a ton of time (far more than it takes to type up the rest of these solutions).
If you’re fuzzy on what a histogram is, check lecture notes from 1/24, or check chapter 1.5 of
the text. Or look up “histogram” in any intro textbook … or online for that matter.
7a) possible outcome
+6
-1
| probability
|
5/38
|
33/38
E(X) = (6)(5/38) + (-1)(33/38) = -3/38 = -$0.0789
V(X) = (6-[-3/38])2(5/38) + (-1-[-3/38])2(33/38) = 5.599
OR
V(X) = [ (6)2(5/38) + (-1)2(33/38) ] – (-3/38)2 = 5.599
7b) We haven’t covered this yet.
Solution: z = (obs-exp)/sd = (0 – [-3/38])/(sqrt(5.599)/sqrt(2000)) = 1.49
So prob = .5 - .4319 = .0681