Collection and Use of Data in Economics
Hypothesis
Testing
Concepts of Hypothesis Testing Hypothesis: a claim about a population parameter.
The proportion of adults in this city with cell phones is P = 88%
Null Hypothesis We assume the null hypothesis is true (e.g. H0: P=.88).
Always contains an equal sign (“=” , “≤” or “” )
Status quo
May or may not be rejected.
Alternative Hypothesis Opposite of the null hypothesis (e.g. H1: P≠.88)
Never contains an equal sign (< or >)
Challenges the status quo
May or may not be accepted.
Hypothesis Testing Process
Level of Significance and Rejection Region Level of Significance defines the unlikely values of the sample
statistic if the null hypothesis is true.
Defines rejection region of the sampling distribution.
Selected by the researcher at the beginning.
Provides the critical value(s) of the test.
Testing the Mean of a Normal Distribution (σ Known)
Calculate the critical value Zα based on α
Convert sample result ( ) to a computed Z value Zc
The decision rule is:
α 0
c0 z
n
σ
μx z if HReject
σ Known σ Unknown
Hypothesis
Tests for
Consider the test
00 μμ:H
01 μμ:H
(Assume the population is normal)
x
Hypothesis Testing Steps
Reject H0 Do not reject H0
zα 0
μ0
H0: μ ≤ μ0
H1: μ > μ0
Critical value
n
σ zμ α0
α 0
c0 z
n
σ
μx z if HReject
1. State the null and alternative hypotheses.
2. Specify the desired level of significance (.05, .10, .01).
3. Choose a sample size (n=?).
4. Determine statistics.
5. Determine Critical Value.
6. Calculate Computed Value.
7. Decision Rule.
1. A phone industry manager thinks that customer monthly cell phone bills have increased, and
now average over $52 per month. The company wishes to test this claim. (Assume =10).
Suppose a sample is taken with the following results: n = 64, x = 53.1. Calculate the critical
and computed value at a .05 level of significance and determine whether to accept or reject
the hypothesis.
2. Test the hypothesis that the average number of T.Vs in U.S. households is 3. Your sample
consists of 100 households with a mean of 2.84 T.Vs. You know the population standard
deviation to be .8. Your desire a level of significance of 0.05
3. A national perfume manufacturer, claims that the cost of processing a sales order is $12.50.
The company accountant suspects that the average cost of processing is more than $12.50.
In order to test her beliefs she obtains a random sample of 40 orders and finds the mean
processing cost is $12.75 with a population standard deviation of $0.50. Test her suspicions
at the 0.01 level of significance.
Application
Tests of the Mean of a Normal Population (σ Unknown)
σ Known σ Unknown
Hypothesis
Tests for
The decision rule is:
α 0
c0 t
n
s
μx tif HReject
Consider the test
(Assume the population is normal)
00 μμ:H
01 μμ:H
1. Ralphs supermarket claims its low fat ground beef is 15% fat by weight. The
market’s quality assurance department would like to know if the 15% goal is
being met. Too much fat could bring on complaints from consumer groups
while not enough fat could effect the flavor of the product. What can the quality
control people conclude at the 0.05 level of significance if a random sample
has values of 16.1, 16.7, 14.5, 16.4, 17.3, and 15.8%
2. The salary data for college graduates in entry level management positions in
CA are normally distributed with a population mean of 63,200. In San Diego a
random sample of 10 women who were college graduates in these positions
had a mean salary of 59,870. The sample standard deviation was 2,590. Do
the sample results suggest the women earn less than 63,200 annually. Use a
level of significance of 0.05
Application
Tests of the Population Proportion
The sample proportion in the success category is
denoted by
When nP & n(1 – P) > 5, can be approximated by
a normal distribution with mean and standard
deviation:
The sampling distribution of is approximately
normal, so the test statistic is a z value:
Hypothesis test:
sizesample
sampleinsuccessesofnumber
n
x p ˆ
Pμ p̂ n
P)P(1 σ
p̂
p̂
p̂
n
)P(1P
Pp̂ z
00
0 c
p̂
00 PP:H
01 PP:H
1. A marketing company claims that it receives 8% responses from its mailing. To test this
claim, a random sample of 500 were surveyed with 25 responses. Test at the = .05
significance level.
2. The Scenic Blight sign company claims that Alliance Medical Group gets at least 25% of
its new patients from the sign company’s bill board on highway 101. To test this claim
the medical group randomly selects 120 patient information forms to determine the
answers given to the question “Where did you hear about our service?” If 26 of the
forms had the response: “The billboard on highway 101” what should the medical group
conclude at α = 0.05?
Application