ECO391 Exam I Review and Preparation Information
Exam I will take place on Monday, February 24, regular time and location.
>>> Bring a calculator!
The Exam covers the following sections from the text:
Chapter 9
Sampling Distributions
Sections 9.1, 9.2, and 9.3
Chapter 10 Confidence Intervals and the t-distribution
Sections 10.1, 10.2, and 10.3
Chapter 11 Hypothesis Testing
Sections 11.1, 11.2, 11.4, 11.5, 11.6, and 11.8
Chapter 12 Tests of Hypothesis Involving Two Populations
Section 12.1
Chapter 13 Chi-Squared Tests
Sections 13.1 (omit pages 514-517) and 13.2
In preparing for the exam, you should review lecture notes, assigned readings, and the homework sets that have been
assigned thus far. I will ask 16 multiple choice questions, but the nature of the majority of questions on the test will be
comparable to the homework assignments.
Attached to your exam you will find the following tables: z, t, and Chi-squared.
Topical Review List:
Chapter 9: Sampling Theory and Sampling Distributions
___population
___population parameter
___sample
___statistical inference
___N
___n
___population mean, 
___sample mean, X(bar)
___Population variance, 2
___Population standard deviation, 
___Sample variance, S2
___Sample standard deviation, S
___How a Sampling Distribution is formed and what it is
___Estimator
___Estimate
___True sampling distribution (9.1)
___Approximate sampling distribution (9.2)
___Sampling Error
___Biased and unbiased
___Efficiency
___The expected value of an unbiased estimator, E(Xbar) = 
___A summary of some facts about sampling distributions (pp.303 and 306)
1. The shape of the sampling distribution of X(bar) is bell-shaped and approximately normal (even if the actual
population data are not.)
2. The larger the sample size, the closer to be normal the X(bar) sampling distribution becomes.
3. The mean of the sampling distribution, E(Xbar) is the same as the population mean x.
4. VAR(Xbar) = 2x/n The variance of the sampling distribution is the variance of the population data divided by the
sample size. (So the standard deviation is the square root of that
or /square root of n)
5. Given #4, as the sample size rises, the VAR(Xbar) falls. The estimator becomes more efficient. This is because as
the sample size rises the sample becomes a more precise approximation of the true population.
Chapter 10: Confidence Intervals
___Define confidence interval
___For normally distributed population with mean, , and population standard deviation, , which is known, Steps:
1)Determine the sample mean X(bar).
2)Record the sample size, n, the level of confidence (1-), and the population standard deviation, . Record the
standard deviation of the sampling distribution as X(bar). This is x(bar) = /square root of n.
3) Find the values + and - Z/2 from table A.5.
4)The two-sided confidence interval is
_
 _
 
 x  z / 2
, x  z / 2

n
n

5)Interpret your result.
___Steps for calculating a confidence interval if the population variance is unknown and sample size sufficiently large,
(n > 30) Same steps as previous test, use S as proxy for population variance.
_
S _
S 
 x  z / 2
, x  z / 2

n
n

___Two desirable features of a confidence interval
1) The interval should have a high level of confidence (1 - ).
2) The confidence interval should have a narrow width.
___Factors affecting the width of a confidence interval
1) As (1- ) rises, the size of the confidence interval rises.
2) As the standard deviation of the population, , rises, the width of the confidence interval rises.
3) As the sample size, n, rises, the width of the confidence interval will fall.
Chapter 11 Hypothesis Testing
___Define hypothesis testing
___Null and alternative hypotheses
___One-tailed (one sided) tests (negative and positive)
___Two-tailed (two-sided ) tests
___Type I and Type II errors
___General steps in a hypothesis test.
___Hypothesis tests for  when population is normal and population variance is known (11.2)
Steps 1-4 with test statistic
Z
X  o

and apply the z-distribution
n
___Hypothesis tests for  when population variance in not known, but sample is large (n>30) (11.4)
Still use the standard normal distribution - z - values
Still follow the same steps and formulas from section 11.2
The only difference - replace  (population standard deviation) with S (sample standard deviation)
___Hypothesis tests for  when population variance in not known, the population is normal, and n < 30. (11.5)
If the population variance is unknown and the sample size is small, the general steps of hypothesis testing are the
same except.
Instead of applying the z-distribution, we apply a t-distribution
Test statistic - calculated t value:
t
X  o
S
n
___Tests of the population proportion, p (11.6)
___What is a proportion?
___p
___q = (1-p)
___provided that np > 5 and nq > 5
___Hypothesis tests involving the population proportion
Follow the standard steps for testing with the following distinctions.
*For these tests, we will apply the z-distribution
*For our calculated z score (observed) we will use the following formula:

Z
p  po
po qo n
Chapter 12 Hypothesis Testing to Compare Two Population Means
___Hypothesis Tests for the Difference Between Two Population Means when the sample is large, the population
variances are known, and the samples are independent of each other.
In general - apply the seven standard hypothesis test steps we discussed earlier.
Case I: (Positive, one-tailed)
Ho: 1 - 2 = Do or 1 - 2 < Do
H1: 1 - 2 > Do
The test statistic:
Z
( X 1  X 2 )  D0

2
1
 2
2
n1
n2
Case II: (Negative, One-tailed test)
Ho: 1 - 2 = Do or 1 - 2 > Do H1: 1 - 2 < Do
Case III: (Two-Tailed Test)
Ho: 1 - 2 = Do H1: 1 - 2  Do
___If the population variances are unknown, but the sample sizes are large the above tests can be done by replacing
the population variances with the sample variances.
Chapter 13 Chi-Squared Tests
___The Chi-Squared Goodness of Fit Test
General Steps for the Chi-Squared Goodness of Fit Test
We observe a random sample of n observations, where each observation falls into exactly one of K categories. In the
population, the hypothesized proportion of observations in category i is denoted by pi.
1) Determine the null and alternative hypotheses:
Ho: p1, p2, …..pk are equal to a set of pre-specified values
H1: At least one of the proportions if not correct
2) Assume a level of significance, 
o1, o2, …….ok refer to the observed frequencies
e1, e2,, ….., ek refer to the expected frequencies where ei = npi
3) The test statistic:
k
2  
i 1
( o i  ei ) 2
This test statistic is distributed along a chi-squared distribution with degrees of freedom v = (k-1)
ei
4) Look up critical value:
 2 ,
5) Decision Rule: Reject Ho if:
 2   2,
6) Rejection region:
7) State conclusion:
___Tests of Independence and Contingency Tables (13.2)
___Contingency Table
The following are the steps for a test that will let us to see if these two factors are independent of each other.
General Steps for the Chi-Squared Test of Independence
We observe a random sample of n observations, where each observation may be classified according to two
variables. For one variable there will be H classifications divided into rows. For the other variable there will be K
classifications divided into columns. This will create a contingency table with dimensions KXH. Any given cell in
this table would be denoted by ricj where r refers to the row and c refers to the column. For instance r 1c2 in the
table above refers to men who are dating.
Determine the null and alternative hypotheses:
Ho: Two variables are independent of each other
H1: Two variables are dependent upon each other.
Assume a level of significance, 
This test statistic is distributed along a chi-squared distribution with degrees of freedom v = (H-1)(K-1)
Look up critical value:   ,( k 1)( H 1)
2
Decision Rule: Reject Ho if: 
2
  2,
Rejection region:
oij refer to the observed frequencies
eij refer to the expected frequencies where ei = ricj/n
The test statistic:
H
 
k
(oij  eij ) 2
j 1
eij

2
i 1
state conclusion:
Formula sheet that will be attached to the exam

x 
n
_
 _
 
 x  z / 2
, x  z / 2

n
n

Z
t
X  o

n
X  o
S
n

p  po
Z
Z
po qo n
( X 1  X 2 )  D0

k
2  
i 1
H
 
2
i 1
2
1
n1
2
2
n2
( o i  ei ) 2
, with degrees of freedom v = k-1
ei
k
(oij  eij ) 2
j 1
eij

k
njxj
j 1
n
X 

, with degrees of freedom v = (K-1)(H-1)