STA 6166 SAMPLE Exam 2
I. Short Answers (15).
(3) 1. State the Central Limit Theorem for the sample mean.
(3) 2. State three ways to decrease the width of a confidence interval.
(3) 3. We are 90% confident that the interval from 10 to 16 contains the population mean. Since the true value is either in or out of this interval, what does this mean?
(5) 4. In an experiment with 50 students chosen at random, you must assign half to the treatment group and half to the control group. Assuming you have labeled them from 01 to 50, start at row 178 and show the numbers of the first 10 students chosen for the treatment group.
II. Problems.
1. Randomly chosen married couples are asked separately to rate their spouse romantically on a scale of 1 to 3 where a 1 is "very romantic" and a 3 is "not at all romantic". The findings are:
Wife's Rating of Husband
1 2
3
Husband's
1 40 16
8
Rating of
2 8 20
4
Wife
3 6 8
10
(5) a)
Find the probability of choosing a couple at random where the wife's rating
of her husband is "not at all romantic".
(5) b)
Find the percent of couples who agree on each other's romantic ratings (gave
each other the same rating).
(6) c) Find the probability of
choosing a wife who rates her husband as "very romantic" given that her
husband rates her as "not at all romantic".
2. For the
following probability distribution of educational levels,
(4) a)
find the probability of choosing a person at random who has some college
credits.
(4) b)
what is the probability of choosing a person at random who has completed at
least high school
(6) c) Given the
a person has completed at least high school, what is the probability
that they have completed a bachelor’s degree or more?
Less than High School
Some
Bachelor's More than
Education
High School Completed
College Degree
Bachelor's
---------------------------------------------------------------------------------------------------------------------
Probability
.09
.32
?
.31
.05
3. Various methods can be used to
estimate the percent of body fat in male athletes. A recent study included a
sample of 19 male gymnasts on college teams whose mean percentage of body
fat was 6.5. For a normed population, the
variance was 5.76.
(12) a)
Find a 90% confidence interval for all male gymnasts' mean body fat
percentage.
(3) b )
Why can you use the probability distribution you used in a) to construct the
interval?
(5) c) Write a
statement interpreting the interval for this problem.
(5) d)
Based on the confidence interval results, would a body fat mean of 5% be
reasonable for college male gymnasts? Explain.
(10) e) How large a sample of
gymnasts would you have to take in order to estimate the true mean body fat
to within a 1/2 % margin of error (using a 90%
confidence coefficient and a pop. variance of 5.76)?
4. A national survey was conducted to
determine how the general public views government involvement in domestic
projects. They were also classified by income level. The frequencies are
given below.
Involvement with government
Income too little
just enough too much
low
155
78
67
medium
140
88
72
high 112 99 89
(3) b) What is
the probability of choosing a person at random who believes government is
involved just enough?
(3) c) Find the probability of choosing a
person at random who is in the low income category and thinks government is
involved too little?
(5) d) Are the events "government involved too much" and "high income" statistically independent?
(7) a)
Find the probability of buying a used home in 2002 within $10,000 of the
mean.
(7) b)
In a certain neighborhood, a realtor observed that in 2002 a random sample
of 49 used homes sold for an average of $203,000. What are the chances
of observing this large a sample mean or larger, if the Miami Realtors
Association figure is correct?
Answers to Sample Exam 2
Short Answer.
1. For a random variable with an unknown
distribution, xbar will be approximately
normally distributed with mean μ and standard deviation σ/√n, when n is
large.
2. 1) decrease confidence coefficient, 2)
increase sample size, 3) decrease std. dev if
possible.
3. 90% refers to the method of
construction. That is, in repeated sampling, we would expect 90% of the
confidence intervals constructed by this method to contain the pop. mean.
4. 24, 21, 14, 26, 22, 43, 20, 18, 09,
06.
Problems.
1. a) P(w rate
h=1) = 22/120 = .1833
b) P(same
rating) = P(1,1) + P(2,2) + P(3,3) = (40+20+10) /120 = .5833 or 58.33%
c) P(w rate h=1 | h rate w=3) = 6/24 = .25
b) P (at least HS) = P(HS)
+ P(some college) + P(Bach) + P(Masters) = 32 + .23 +.31 + .05 = .91
c) P(Bach, Masters|HS)
= (.31+.05)/.91 = .36/.91 = .3956
3. a) 6. 5+
1.645 √5.76/19 or 6.5 + .9057 or (5.594, 7.406)
b) Since n=19, CLT
does not hold, so must assume the variable, % body fat, is normally
distributed.
c) We are 90% sure the
interval from 5.59 to 7.41% contains the true mean % body fat for male
athletes.
d) No, 5% does not
fall in the confidence interval, so is not a reasonable estimate.
e)
1.645*2.4 2
2
n = [ ------------ ] = [7.896]
= 62.35 or n = 63
.5
4. a) P(too much | high income) = 89/(112+99+89) = .2967
b) P(just enough) = (78+88+99)/900= .2944
c) P(too little and low income) = 155/900 = .1722
d) P(too much | high income) = .2967 and P(too much) = (67+72+89)/900 = 228/900 = .2533. Since the conditional probability is not equal to the marginal (unconditional) probability, these events are DEpendent statistically.
203000 - 195500
b) P ( x > 203000) = P ( z > 1.16 ) = .1230 z = ---------------------- = 1.16
45300/% 49