Hypothesis Testing

Hypothesis Testing
An Introduction Using the Normal
Density for Testing Claims Against the
Population Mean

Objectives
• To introduce hypothesis testing components
when determining the accuracy of a claim
against a population mean.
• To demonstrate the four-step process for
answering questions using hypothesis testing
methods.

Materials To Review
• Normal density curves.
• z-scores
• Sampling distribution of means
• Central Limit Theorem

Terminology
• A hypothesis is a claim or statement about a
property of a population.
• A hypothesis test (or a test of significance) is
a procedure for testing a claim about a
property of a population.
• The null hypothesis, denoted as H0, is a
statement that the value of a population
parameter is equal to some claimed value.

Terminology
• The alternate hypothesis, denoted as Ha, is
the statement that the parameter has a value
that differs from the null hypothesis.
• The test statistic is a measure of the distance
the sample statistic is from the claim based on
the distribution.

Terminology
• The P-value (or probability value) is the
probability of getting a value of the test
statistic that is at least as extreme as the one
representing the sample data, assuming the
null hypothesis is accurate.

Terminology
• The critical region (or rejection region) is the
set of all values of the test statistic that cause
us to reject the null hypothesis.
• The significance level (denoted by a) is the
probability that the test statistic will fall in the
critical region when the null hypothesis is
accurate.

Terminology
• A critical value is any value that separates the
critical region from the values of the test
statistic that do not lead to rejection of the
null hypothesis.
• The critical value depends on:
1. Nature of the null hypothesis,
2. The sampling distribution that applies, and
3. The significance level, a.

Forming Hypotheses
• If you are conducting a study and want to use
a hypothesis test to support a claim. The claim
must be worded so that it becomes the
alternate hypothesis.
• In other words, we always test AGAINST the
null hypothesis. So if you wish to find evidence
in support of a claim, that claim must be the
ALTERNATE hypothesis.

Cautions
• When conducting hypothesis tests as
described in this course, be sure to consider:
1. The context of the data,
2. The source of the data,
3. The sampling method used to collect the
data.
• All of these can have an effect on the
decisions made for hypothesis tests.

The Rare Event Rule
• If, under a given assumption, the probability
of a particular observed event is smaller than
a significance level, we conclude that the
assumption is probably not accurate.
• Note that this decision criteria is based on the
significance level.
• Since the investigator usually chooses the
significance level, in order to be meaningful, it
is usually 10% or less.

Conditions for Using Normal Density
• To construct a confidence interval for the
purpose of estimating a population mean, the
normal density may be used under the
following conditions:
1. The sample is a simple random sample.
2. The value of the population standard
deviation is known.

Conditions for Using Normal Density
• As an alternative to condition 2, we have:
3. Either or both of these conditions are
satisfied:
a) The population is known to be normally
distributed, or
b) The sample size is greater than 30.

Components for 1-Sample Proportions
Test
• Distribution: Normal if conditions are met
• Significance Level: a (usually given), 0.05 if
not given.
• Null hypothesis: H0: m = m0
• Alternate hypothesis:
Type of Test Mathematical Translation
Left-Sided Test Ha: m < m0
Right-Sided Test Ha: m > m0
Two-Sided Test Ha: m ≠ m0

Components for 1-Sample Proportions
Test
• Test statistic: 𝑧 =
𝜇−𝜇0
𝜎/ 𝑛
.
• P-value: Area under the distribution from the
test statistic that describes the probability of
obtaining such a sample or a sample with a
more extreme value.

Components: Density with Null
Hypothesis
Claim

Components: Significance Level and
Alternate Hypothesis
• This is the type of diagram that would be seen for
a left sided test. The alternate hypothesis
suggests the true value is less than what is
claimed.
Claim
Significant Level = a
z*

Components: Significance Levels and
• This is the type of diagram that would be seen for
a right sided test. The alternate hypothesis
suggests the true value is greater than what is
claimed.
Claim
z*

• This is the type of diagram that would be seen
for a two-sided test. The alternate hypothesis
suggests the true value is different than what
is claimed.
Claim
Significant Level = a/2
z*
Significant Level = a/2
z*

Alternate Hypotheses
• In each of the last three diagrams, the area in
the red region corresponds to the significance
level.
• The region colored in red is referred to as the
region of rejection.

Alternate Hypotheses
• So the significance level tells us the amount of
area under the curve should be covered.
• The alternate hypothesis tells us the direction
of covering this area.
• The critical values, z*, tell us where on the
curve these regions begin so the proper
amount of area is covered.

Components: Test Statistic
• The test statistic is a measure in terms of
standard deviations of how far our sample
statistic is from the claimed value.
• Since we are using information from a sample, we
must consider the standard deviation from the
sampling distribution, not the original
population.
• The Central Limit Theorem provides the
mathematical support for the methods used
here.

Components: Critical Values
• The critical value is the value that begins the
region of rejection. It is a measure of how far
away from the claim a sample statistic needs
to be in order to cast doubt on the accuracy of
the claim.
• Once a test statistic is beyond the critical
value, our sample is in the region of rejection
and we would have sufficient evidence to
reject the claim.

More on Critical Values
• Before computers the critical values were used to
determine decisions based on these tests. This is
sometimes referred to as the traditional method.
• With computers available, the P-value is
computed as to give one single criteria for all
testing schemes as well provide information
about the chances of obtaining such samples.
This is referred to as the P-value method and this
is the method we shall use in our class.

Components: P-value (Rejection of
Null Hypothesis)
• The P-value is the area under the curve that
begins from the test statistic and is shaded in
the direction of the alternate hypothesis.
Claim
z* z
P-value

Components: P-value (Fail to Reject
Null Hypothesis)
• The P-value is the area under the curve that
begins from the test statistic and is shaded in
the direction of the alternate hypothesis.
Claim
z*z
P-value

Four-Step Process
• State: What is the practical question that
requires a statistical test?
• Plan: Before you devise your plan, do the
following:
Test to see if the conditions are satisfied.
If the conditions are satisfied, you may then
write out the hypotheses. Also identify the
parameter and choose the type of test you are
conducting.

Four-Step Process
• Solve: Compute the test statistic and
determine the p-value.
• Conclude: In this step we do the following:
1. Compare the p-value with the significance
level.
2. Write the conclusion in the context of the
problem.

EXAMPLES OF HYPOTHESIS TESTING
In this section we demonstrate how to perform a hypothesis test for
population proportions. Each type of alternative is demonstrated.

Example 1: Left-Sided Test
• Management at Ford Motor Company are trying
to determine if the average age of a car is now
less than what it was in 1995. The age of cars in
1995 was 8.33 years. Based on a random sample
of 18 automobiles owners, you obtain the ages of
their cars:
{8, 12, 1, 2, 13, 3, 5, 9, 12, 6, 5, 6, 10, 7, 10, 11, 6,
10}.
• Assuming that s = 3.8 years, test this claim at the
10% significance level.

Example 1: Left-Sided Tests
• Because of the small sample size, we need to
test to see if data are normally distributed
with no outliers. This is done with a normal
probability plot and a boxplot.
• The graphs are given on the next slide.

• Boxplot shown below
shows a roughly
symmetric shape with
no outliers.
• Normal probability plot
below shows the data is
roughly linear with no
outliers. This gives
evidence to suggest the
data are normally
distributed.

• First we identify the components:
1. Distribution: Since the boxplot and normal
probability plot support the use of a normal
density, we may use a normal density for our
test.
2. Significance level: We have a significance
level of 10%, so a = 0.10.

3. Null hypothesis: The null hypothesis would
be there is no difference in the age of cars
today compared to the age of cars in 1995,
which was 8.33 years. Thus Ho: m = 8.33
4. Alternate hypothesis: Since Ford Motor
Company’s management believes the age is
less than the age in 1995, we have Ha: m <
8.33.

Example 1: Distribution
Significance Level = region of rejection
Significance Level = 10%
Area to the left of critical value = a.
Critical Value, z* = 7.18
z*

• Test statistic: If we compute the test statistic
using the formula we have:
𝑧 =
𝑥−𝜇
𝜎/ 𝑛
=
7.56−8.33
3.8/ 18
= −0.86
• So our particular sample’s average is 0.86
standard deviations (on the sampling
distribution of means curve) to the left of the
mean.

• P-value: 0.1936
• So the probability of obtaining such a sample
average or one more extreme than this is
approximately 19.4%.

Example 1: Test Statistic and P-Value
Critical Value, z*
P-value = hatched shading
P-value = 0.1936 or 19.4%
Test Statistic
Significance Level
a = 0.10 or 10%

Example 1: Analyzing the Diagram
• We can see that the test statistic is not further
away from the claim than the critical value.
• We can see the amount of area in red is less
than the amount of area that is brown.
• The test statistic, 𝑥 is not less than the critical
value z*.

• State: Are cars younger today than they were
in 1995?
• Plan: We shall use a Left-Sided Z-Test for this.
We have checked the conditions, since the
sample size is small and have concluded that
the conditions are satisfied. Our hypotheses
are:
Ho: m = 8.33
Ha: m < 8.33

• Solve: Our test statistic is -0.86. Our P-value is
0.1936.
• Conclude: The P-value > significance level.
This implies we fail to reject the null
hypothesis. We do not have sufficient
evidence to suggest that the age of cars are in
fact younger than they were in 1995.

Example 2: Right-Sided Test
• A researcher claims that the average age of a
woman before she has her first child is greater
than the 1990 mean age of 26.4 years, on the
basis of data obtained from the National Vital
Statistics Report, Vol. 48, No. 14. She obtains a
simple random sample of 40 women who gave
birth to their first child in 1999 and finds the
sample mean age to be 27.1 years. Assuming the
population’s standard deviation is s = 6.4 years,
test the researcher’s claim using a significance
level of 5%.

1. Distribution: Normal density, we have a large
enough sample of 40 individuals and the
population standard deviation is given at s =
6.4 years.
2. Significance level: Given at 0.05 or 5%.

3. Null hypothesis: The claim we wish to find
evidence against is there is no difference
between ages of women between the years
1990 and 1999. According to the National
Vital Statistics Report, the average age was
26.4 years. So H0: m = 26.4.
4. Alternate hypothesis: The researcher
believes that the age is now older, so we
have: Ha: m > 26.4.

Significance Level = region of rejection
Significance Level = 5%
Area to the left of critical value = a.
Critical Value, z* = 28.06
z*

• Test statistic:
𝑧 =
𝑥 − 𝜇0
𝜎/ 𝑛
=
27.1 − 26.4
6.4/ 40
= 0.6917
• This means that my particular proportion from
our sample is about 0.69 standard deviations
from the claim under the sampling
distribution of the means.

• P-value: 0.2445.
• This means that we have approximately a
24.45% chance of seeing such a sample
average or a sample average that is greater.

Example 2: Test Statistic and P-value
Critical Value, z*
P-value = hatched shading
P-value = 0.2445 or 24.45%
Test Statistic
Significance Level
a = 0.05 or 5%

Example 2: Analyzing the Diagram
• We can see that the test statistic is not further
away from the claim as the critical value is.
• We can see the amount of area hatched is
greater than the amount of area shaded in
red.
• Both of these are indicators that we fail to
reject the null hypothesis. We did not find
sufficient evidence to suggest that women are
in fact having children later in life.

• State: Are women waiting longer before
having their first child?
• Plan: I will conduct a Z-Test using a sample of
40 mothers who had their first child in 1999. I
am comparing the results of this new data to
the results obtained and published in National
Vital Statistics Report, Vol. 48, No. 14. in 1995.
The condition for using the normal
approximation is satisfied.

• Solve: The test statistic is approximately
0.6917 and the P-value is approximately
0.2445.
• Conclude: Since the P-value > 0.05, we fail to
reject the null hypothesis. We do not have
sufficient evidence to suggest that women are
waiting longer to have their first child.

Example 3: Two-Sided Test
• Suppose I am in the market to purchase a
2010 Ford Mustang. Before shopping around, I
want to determine what I should expect to
pay.

• According to the Kelly Blue Book value, Using
the following criteria:
1. Very Good Condition
2. 2-Door Coupe
3. No Added Options
4. For Private Party
5. 43,500 miles
• We have a price of $12,972.

• After doing some shopping on carsforsale.com, I
found the following prices for fourteen 2010 Ford
Mustangs:
{15950, 17995, 16999, 16995, 17995, 15945,
17985, 18977, 14259, 18689, 14500, 21995, 15900,
18995}
• After seeing these values, I think I’ll be paying
something different than what Kelley Blue Book
published. Run a test to determine if the Kelley
Blue Book value is an accurate estimate to what I
would be paying. Assume s = $2100.

• Since our sample is small, we need to
determine if the conditions are upheld to
conduct such a test.
• We use a boxplot and a normal probability
plot to determine if the assumption of
normality is upheld.
• These pictures are on the next slide.

• Boxplot is roughly
symmetric with no
outliers.
• Normal Probability Plot
shows the data is
roughly linear giving
evidence that a normal
density may be used.

1. Distribution: Using the normal probability
plot and the boxplot we conclude that the
normality condition is upheld. Hence we may
use a normal density.
2. Significance level: Not given so we shall set it
to be 0.05 or 5%.

3. Null hypothesis: The claim is there is no
statistically significant difference in the
published value of the Mustang in Kelley Blue
Book and the sample average I found, so H0:
m = 12,972.
4. Alternate hypothesis: Since we are testing to
see if there is any difference in the averages,
we have: Ha: m ≠ 12,972.

1. Shading starts at the critical
values z* after we divide the
significance level in half.
2. Direction of shading is
based on the alternate
hypothesis.
𝛼
2
𝛼
2
Critical Value Critical Value

• Test statistic:
𝑧 =
𝑥 − 𝜇0
𝜎/ 𝑛
=
17639.93 − 12972
2100/ 14
= 7.836
• This means that my particular proportion from
our sample is about 7.836 standard deviations
from the claim using the sampling distribution
of means.

• P-value: 4.697 × 10-15.
• This means that we have approximately a
0.0000000000004697% chance of seeing such
a sample proportion or a sample proportion
that is greater.

Significance Level = a/2 Significance Level = a/2
Critical Value Critical Value
Half of P-value Half of P-value

Right Side of Claim
P-value = hatched area
P-value = 2.35 × 10−15
Significance Level = 0.025
Critical value Test Statistic

Left Side of Claim
P-value = hatched area
P-value = 2.35 × 10−15
Significance Level = 0.025
Critical valueTest Statistic

Example 3: Analyzing the Diagrams
• We can see that the test statistic is further
away from the claim than the critical value is.
• We can see the amount of area hatched is less
than the amount of area shaded in red.
• Both of these are indicators that we reject the
null hypothesis. We did find sufficient
evidence to suggest the Kelley Blue Book value
is not accurate to what I would be paying for a
2010 Ford Mustang.

Example 3: Analyzing The Diagrams
• We have two detailed diagrams because we have
two sides to our test.
• Since we are testing the claim of no difference we
must account for the possibility that the test
statistic could be on either side of the claim.
• This also accounts for the significance level being
halved as well. The significance level corresponds
to the total area under the curve that represents
what we describe as “a rare event”.

• State: Is Kelley Blue Book’s average value of a
2010 Ford Mustang accurate to what people
are selling?
• Plan: I will conduct a Z-Test using a sample of
14 prices. The condition for using the normal
approximation is satisfied. Therefore our
hypotheses are:
H0: m = 12,972
Ha: m ≠ 12,972

• Solve: The test statistic is approximately 7.836
and the P-value is approximately 0.
• Conclude: Since the P-value < 0.05, we reject
the null hypothesis. We have sufficient
evidence to suggest the Kelley Blue Book value
and the price I’ll be paying for a 2010 Mustang
are different.

Lurking Variables
• What are some lurking variables that could
have an effect on my conclusion?
• Individuals will try to make a profit and
therefore may set the price of their cars
higher?
• Some cars may have additional features, like
being convertible. This influences the price.
• Any others?

Hypothesis Testing

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