Download Hypothesis Testing - Dixie State University :: Business Department

William Christensen, Ph.D.
What is a Hypothesis?

Hypothesis – (in statistics) a claim or
statement about a property of a population.

Example: a claim that the mean weight of men is
greater than 175 lbs.

Example: a claim that less than 50% of
households in the U.S. have internet access

Example: a claim that annual income levels in St.
George have a standard deviation greater than
$10,000

Note: Hypothesis is singular and hypotheses is
plural
Hypothesis Testing

In this section you will learn how to test
hypotheses related to means, proportions,
and standard deviations (or variances)

We use sample data to test claims about
population parameters (means,
proportions, and standard deviations)

Statistics is all about chance or probability,
so our hypothesis tests are always based
on some given chance of being right (or
wrong) in our answer
Null & Alternative Hypothesis
Since we do not know in advance
what the results are going to be of a
hypothesis test, we ALWAYS state the
hypothesis in two forms:
1. H0 (read H-oh or H-zero) is the sign to
represent the “Null Hypothesis” and
always contains an =, , or  sign
2. H1 (read H-one) is the sign to
represent the “Alternative
Hypothesis” and always contains the
opposite sign (≠, , )

Null & Alternative Hypothesis
Here are examples of all the kinds of
hypothesis testing we learn in this section,
shown in their proper format:
Hypothesis tests for µ (population Mean)



The claim being tested always determines the value that is used,
but in this case we use a value of 100 just as an example
Null Hypothesis Alternative Hypothesis
Case 1
H0: µ = 100
H1: µ ≠ 100
Case 2
H0: µ  100
H1: µ  100
Case 3
H0: µ  100
H1: µ  100
Null & Alternative Hypothesis
Here are examples of all the kinds of
hypothesis testing for p (population
Proportion)


The claim being tested always determines the value that is used,
but in this case we use a value of 0.50 (50%) just as an example.
Note that proportions like probabilities must be between 0.0 and
1.0
Null Hypothesis Alternative Hypothesis
Case 1
H0: p = 0.50
H1: p ≠ 0.50
Case 2
H0: p  0.50
H1: p  0.50
Case 3
H0: p  0.50
H1: p  0.50
Null & Alternative Hypothesis
Here are examples of all the kinds of hypothesis testing for σ and
σ2 (population Standard Deviation and Variance)


The claim being tested always determines the value that is used, but in this case we use
a value of 10 for standard deviation and 100 for variance just as an example
Null Hypothesis
Alternative Hypothesis
Case 1
H0: σ = 10
H1: σ ≠ 10
Case 2
H0: σ  10
H1: σ  10
Case 3
H0: σ  10
H1: σ  10
Null Hypothesis
Alternative Hypothesis
Case 1
H0: σ2 = 100
H1: σ2 ≠ 100
Case 2
H0: σ2  100
H1: σ2  100
Case 3
H0: σ2  100
H1: σ2  100
Null & Alternative Hypothesis

You may have noticed that every Null hypothesis
contains a “condition of equality” ( =, , or  sign)

And that every Alternative hypothesis has the
opposite sign as shown in the following table
Case 1
Case 2
Case 3
Null Hypothesis
Alternative
Hypothesis
=


≠


Doing a Hypothesis Test

The previous slides show every
possible combination of Null and
Alternative Hypotheses for population
Mean, Proportion, and Standard
Deviation / Variance.

Now that you know what hypotheses
look like, the next step is to learn how
to test them.
Doing a Hypothesis Test

It is always the “Null Hypothesis” that
is being tested.

If the test “fails” then we reject the
Null Hypothesis, or in other words,
accept the Alternative Hypothesis

If the test does not fail, then we accept
the Null Hypothesis
Doing a Hypothesis Test
Whenever we test hypotheses we are
testing a “claim” that has been made



Example: The mean weight of men is greater
than 175 lbs.
Whenever possible, it is a good practice to
state the claim in terms of the Alternative
Hypothesis (not equal, less than, or greater
than) rather than the Null Hypothesis
(equal, less than or equal, or greater than or
equal)
Doing a Hypothesis Test
Here are the steps in doing a hypothesis test
State the Null and Alternative hypotheses
Determine the “Critical Value(s)” and “Critical
Region(s)” – this is what we’ll learn next
Calculate the “Test Statistic” and compare it
against the Critical Value(s)

1.
2.
3.


If the Test Statistic lies within the Critical Region then
reject the null hypothesis (accept the alternative
hypothesis)
If the Test Statistic does not lie within the Critical Region
then fail to reject the null hypothesis (accept the null
hypothesis)
Critical Region(s) and Value(s)



Critical values are the z-scores or x values associated with the standard
normal distribution. These values are established based on the level of
confidence or alpha value we choose.
Critical regions are bordered by a critical value and include all the area
outside (toward the tail).
Whenever we do hypothesis testing we do it based on some level of
“confidence” or α (alpha) value

For example, an α = 0.05 represents a 95% confidence level or 95% chance
that our answer will be truly correct and a 5% chance that our answer will
not be correct. The total area contained in the critical region(s)
equals alpha. E.g., if α = 0.05 then the total
combined area in these critical regions equals
5% of the total area (0.05/2 = 0.025 on each side)
Critical
Values
Critical
Regions
Hypothesis Test:
population Mean
using large
sample (n  30)
Assumptions

When testing claims about population
means from large samples we assume
the following conditions exist:
1. The sample is a random sample
2. The sample is large (n  30)
Critical Value(s)


You actually already know how to find the critical value for hypothesis tests of
population Mean using large samples. Remember the Excel function
NORMSINV(probability)? It allows us to determine the x value (same as zscore) for a standard normal distribution where we know the area (probability)
as measured from the extreme left out to the critical value.
For example, if α = 0.05 (total combined area) then the area from the extreme
left to the critical value on the left can be found by NORMSINV(0.025) and
the critical value of the right can be found by NORMSINV(0.975)
Critical
Values
Area from left = 0.975
Area from left = 0.025
0.025
0.025
µ Critical Values
Remember the 3 types of null / alternative hypothesis
combinations we have:


This is very important in order to properly determine the critical
value(s) for nearly ALL HYPOTHESIS TESTS
Null
Hypothesis
Alternative
Hypothesis
Two-tail test (the area represented
by alpha is split equally between the
left and right tails)
=
≠
Right-tail test (the area represented
by alpha is entirely in the right tail
with no left tail at all)


Left-tail test (the area represented
by alpha is entirely in the left tail with
no right tail at all)


The easy way to
remember this is that
the sign of the
Alternative hypothesis
points the direction of
the test. A ≠ sign kind
of points both ways
and represents a twotail test. A  sign points
right, and a  sign
points left.
µ Critical Value Example
α = 0.05
CV = -1.96
Null
Hypothesis
Alternative
Hypothesis
=


≠


Two-tail test
CV = 1.96
Area = α/2 =
0.025
Area = α/2 =
0.025
Right-tail test
Left-tail test
CV = -1.645
CV = 1.645
Area = α =
0.05
Area = α =
0.05
Notice how critical values for the left-tail are ALWAYS
negative and critical values for the right-tail are
ALWAYS positive
µ Critical Value Exercise
α = 0.01
CV = -2.576
Null
Hypothesis
Alternative
Hypothesis
=


≠


Two-tail test
CV = 2.576
Area = α/2 =
0.005
Area = α/2 =
0.005
Right-tail test
Left-tail test
Use Excel function
NORMSINV to confirm
these critical values are
correct
CV = -2.326
Area = α =
0.01
CV = 2.326
Area = α =
0.01
Test Statistic



A “Test Statistic” is a value we calculate from the sample
data
We use the Test Statistic to test our hypothesis by
comparing the Test Statistic against the critical value(s)
The following graph shows how the position of the Test
Statistic determines whether we accept the Null
hypothesis or the Alternative hypothesis
If the Test Statistic falls in the critical region(s) then reject the
null hypothesis (accept the alternative hypothesis)
If the Test Statistic does not
fall in the critical region(s)
then “fail to reject H0 (accept
the null hypothesis)
Test Statistic for µ

For large samples, testing claims about
population means we calculate the Test
Statistic using the following formula (you
must remember this one)
Test Statistic =
x-µ
σ
n
Test Statistic for µ
For large samples
This is the sample
average/mean
which we calculate
from our sample
data
Test Statistic =
n is simply the size of our
sample. E.g., if our
sample consists of 50
men, then n=50
x-µ
σ
n
We never really
know the population
mean (µ) but we do
have a value
included in the
hypothesis (e.g., H0:
µ=100) and it is this
value that we put
here – the value
from the hypothesis
If we know σ (population
std. deviation) then we
put it here. However, we
usually only have sample
data, so we use s
(sample std. deviation)
here most of the time
Hypothesis Test Example I
testing a claim about µ from a large sample
We generally assume that the average or normal human
body temperature is 98.6 degrees F. To test this
assumption or claim, data was collected from 106 healthy
adults (n = 106). From this sample, the following
statistics were obtained.
Mean temperature = 98.2o
Standard deviation (s) = 0.62o
Using an alpha of 0.05 (a 95% confidence level that our
solution will be correct, with a 5% chance of being wrong)
test the claim that the mean body temperature of all
healthy adults is equal to 98.6o.
Hypothesis Test Example I
testing a claim about µ from a large sample
1. Write out the claim and the hypotheses
• The claim is that the mean body temperature
of healthy adults is equal to 98.6o or µ = 98.6o
•
Notice that we are making a claim about the population of
all healthy adults, but we only have a sample of 106 adults
upon which to test this claim
• Thus, the null hypothesis H0: µ = 98.6o
• The alternative hypothesis H1: µ ≠ 98.6o
• You must be able to read ‘word problems’
and be able to translate them into null and
alternative hypotheses
Hypothesis Test Example I
testing a claim about µ from a large sample
2. Draw a picture of the normal distribution and
determine the critical values
•
Note that the sign of the alternative hypothesis is ≠ so we have
a two-tail test
Note: since the standard normal
distribution is perfectly
symmetrical, you can simply take
the negative critical value on the
left and change it to a positive
value for the right side – or go
ahead and use NORMSINV again
to solve for the right side
α = 0.05, α/2 = 0.025
0.4750
0.025
0.4750
0.025
-1.96
1.96
Critical values are determined using Excel NORMSINV
Hypothesis Test Example I
testing a claim about µ from a large sample
3. Calculate the Test Statistic and form your final
conclusion
•
•
Remember we were given the following information:
• n = 106, s = 0.62o, sample mean = 98.2o, α = 0.05, and our
null hypothesis is µ = 98.6o
We can now calculate the Test Statistic using the formula we
learned:
x
µ
zTS = s
n
=
98.2 - 98.6
0.62 106
= - 6.64
Note: If the test statistic falls inside the critical values then we always
accept the null hypotheses (mean body temperature equals 98.6), and if
the test statistic falls outside the critical values we always accept the
alternative hypothesis (mean body temperature does not equal 98.6).
Hypothesis Test Example I
testing a claim about µ from a large sample
3. Calculate the Test Statistic and form your final
conclusion
We calculated the test statistic zTS = -6.64
CONCLUSION: Accept the alternative hypothesis H1: µ ≠ 98.6o
The mean body temperature of healthy adults is not 98.6o F
Accept the alternative
hypothesis
Accept the alternative
hypothesis
Accept the null hypothesis
0.025
Test statistic falls
outside the critical
value
-1.96
0.025
1.96
Critical values determined using Excel NORMSINV
Hypothesis Test Example II
testing a claim about µ from a large sample
Men are often accused of channel-surfing (constantly
changing channels using the remote control). With
this in mind, a study was done to test the claim that,
on average, men change channels every 5.00
seconds or less. To investigate this claim, sample
data was collected and the following statistics were
obtained:
n = 80 (80 men were observed)
Sample mean = 5.25 seconds
standard deviation = 2.50 seconds
Use an alpha = 0.10 to test this claim.
Hypothesis Test Example II
testing a claim about µ from a large sample
1. Write out the claim and the hypotheses
• The claim is that men change channels every
5 seconds or less or µ  5.00 seconds
•
Notice that we are making a claim about the population of
all TV-watching men, but we only have a sample of 80 men
upon which to test this claim
• Thus, the null hypothesis H0: µ  5.0 seconds
• The alternative hypothesis H1: µ  5.0 seconds
•
REMEMBER that the sign for the null hypothesis MUST ALWAYS
be =, , or  and the sign for the alternative hypothesis is always ≠,
, or  (the opposite of the null hypothesis)
Hypothesis Test Example II
testing a claim about µ from a large sample
2. Draw a picture of the normal distribution and
determine the critical values
•
Note that the sign of the alternative hypothesis is  so we have
a right-tail test
Note: since we have a one-tail test to
the right, the entire critical region is on
the right side and has an area equal to
alpha. We can either use
NORMSINV(0.10) and change the
answer to a positive value, or use
NORMSINV(0.90) which will return the
same positive value.
α = 0.10
0.500
0.400
0.100
1.28
Critical value is determined using Excel NORMSINV
Hypothesis Test Example II
testing a claim about µ from a large sample
3. Calculate the Test Statistic and form your final
conclusion
•
•
Remember we were given the following information:
• n = 80, s = 2.50 seconds, sample mean = 5.25 seconds, α =
0.10, and our null hypothesis is µ  5.0 seconds
We can now calculate the Test Statistic using the formula we
learned:
x
µ
zTS = s
n
=
5.25 – 5.00
2.50 80
= 0.89
Note: If the test statistic falls inside the critical value then we always accept the
null hypotheses (men switch channels every 5 seconds or less), and if the test
statistic falls outside the critical value we always accept the alternative
hypothesis (men wait more than 5 seconds before switching channels).
Hypothesis Test Example II
testing a claim about µ from a large sample
3. Calculate the Test Statistic and form your final
conclusion
We calculated the test statistic zTS = 0.89
CONCLUSION: Accept the null hypothesis H0: µ  5.0 seconds
Men switch channels every 5 seconds or less
Accept the alternative
hypothesis
Accept the null hypothesis
0.100
Test statistic falls inside the critical value
1.28
Critical value determined using Excel =NORMSINV(0.90)
Hypothesis Test Example II
testing a claim about µ from a large sample
•
•
•
•
•
Discussion: Perhaps you noticed in the last exercise that the sample mean was
5.25 seconds and yet we concluded that men change channels every 5 seconds or
less. How is this possible?
You must remember that we are talking about probabilities here. We did this
problem based on a 90% confidence of being correct (10% chance of being
wrong). Therefore, we are not certain about our conclusion, it is just our best
estimate based on the info we were given.
We do know that a sample of 80 men does NOT give us exactly the same data as
we would have if we knew the channel-surfing habits of all men in our population.
Certainly if the mean channel-surfing interval for all men was 5.25 seconds then it
would be ridiculous to make any other claim. However, in this case we must make
our conclusion about the entire population of men based on our sample of only 80
men.
Since our sample mean of 5.25 seconds was pretty close to our claimed
population mean of 5.00 seconds, and since we had only 80 men in our sample,
statistically (at alpha = 0.10) we did not have enough evidence to reject the null
hypothesis that the population mean could actually be 5 seconds or less even
though the sample mean was more than 5 seconds.
On your own, try doing the problem again assuming the sample size was 200 men
instead of 80 men. Does it change the answer?
Hypothesis Test:
population Mean
using small
sample (n  30)
Assumptions

When testing claims about population
means from small samples we assume
the following conditions exist:
1. The sample is a random sample
2. The sample is small (n  30)
Critical Value(s)




Finding the critical value(s) when doing hypothesis testing of population
means using small samples is similar to the method we learned when we
have large samples, but with some important differences. For one thing,
rather than using the Excel function NORMSINV(probability) as we did with
large samples, we must now use the Excel function
TINV(probability,deg_freedom) where degrees of freedom equals n-1.
Also, and this is really important, this function assumes we are doing a
two-tail test, so we enter alpha (not alpha/2) when doing a two-tail test, and
we must enter 2 times alpha (alpha x 2) if we are doing a one-tail test.
Finally, TINV always returns a positive (right-side) value, so we must know
and remember to make the value negative if using it on the left-side
For example, if α = 0.05 (total combined area), and n=20, then a left-tailed
test critical value would be found by negating the value we get with
TINV(0.10,19). For a right-tailed test critical value we can simply use
TINV(0.10,19). For a two-tailed test we take the positive and negative value
of TINV(0.05,19).
Critical Value(s)
two-tail hypothesis test

For example, if α = 0.05 (total combined area), and
n=20, then the right-side critical values is found by
TINV(0.05,19) and for the left-side critical value
we simply take this same value and make it
negative (-2.09).
Critical
Values
0.025
-2.09
0.025
2.09
Critical Value(s)
right-tail hypothesis test

For example, if α = 0.05 (total area), and n=20,
then, for a right-tail test, the critical value is found
by TINV(0.10,19) (note we must double alpha). If
this was a left-tail test we would do the same thing except take the
value and make it negative.
Critical
Value
0.05
1.729
Critical Value(s)
small samples
The smaller the sample size, the larger the
critical value. The t-distribution (TINV) we
use takes into account smaller samples
and requires the sample data to evidence
greater differences from the claimed µ
(population mean) before the results will
reject the null hypothesis (i.e., accept the
alternative hypothesis).
Test Statistic for µ

For small samples, when testing claims
about population means we use the same
formula we used with large samples to
calculate the Test Statistic (you must
remember this)
Test Statistic =
x-µ
σ
n
Test Statistic for µ

For small or
large samples
This is the sample
average/mean
which we calculate
from our sample
data
Test Statistic =
n is simply the size of our
sample. E.g., if our
sample consists of 50
men, then n=50
x-µ
σ
n
We never really
know the population
mean (µ) but we do
have a value
included in the
hypothesis (e.g., H0:
µ=100) and it is this
value that we put
here – the value
from the hypothesis
If we know σ (population
std. deviation) then we
put it here. However, we
usually only have sample
data, so we use s
(sample std. deviation)
here most of the time
Hypothesis Test Example
testing a claim about µ from a small sample
Because of the expense involved, car crash tests
often use small samples. When 5 BMW cars are
crashed under standard conditions, the repair costs
(in dollars) are: $797, $571, $904, $1147, and $418
(mean=$767.40, s=$284.73). Use a 0.05
significance level to test the claim that the mean
repair cost for all BMW cars is less than $1,000.
Would BMW be justified in advertising that the
average repair cost is less than $1,000?
Hypothesis Test Example
testing a claim about µ from a small sample
1. Write out the claim and the hypotheses
• The claim is that the average repair cost for
a BMS is less than $1,000 or µ  $1,000
•
Notice that we are making a claim about the population of
all BMWs, but we only have a sample of 5 crashed BMWs
upon which to test this claim
• Thus, the null hypothesis H0: µ  $1,000
• The alternative hypothesis H1: µ  $1,000
•
REMEMBER that our claim may be the null hypothesis or the
alternative hypothesis, but the sign for the null hypothesis MUST
ALWAYS be =, , or  and the sign for the alternative hypothesis is
always ≠, , or  (the opposite of the null hypothesis)
Hypothesis Test Example
testing a claim about µ from a small sample
2. Draw a picture of the normal distribution and
determine the critical values
•
Note that the sign of the alternative hypothesis is  so we have
a left-tail test
α = 0.05
Note: since we have a one-tail
test we must enter the
probability in TINV as 2 x alpha
(0.05 x 2 = 0.10). Degrees of
freedom = n-1 = 5 – 1 = 4
0.450
0.500
0.05
-2.132
Change sign to negative
since value is on left-side
Critical value is determined using Excel TINV
Hypothesis Test Example
testing a claim about µ from a small sample
3. Calculate the Test Statistic and form your final
conclusion
•
•
Remember we were given the following information:
• n = 5, sample mean = $767.40 and sample standard
deviation (s) = $284.73
• α = 0.05, and our null hypothesis is µ  $1,000
We can now calculate the Test Statistic using the formula we
learned:
767.4 – 1000
tTS = xs - µ =284.73
= -1.83
n
5
Note: If the test statistic falls inside the critical value then we always accept the
null hypotheses, and if the test statistic falls outside the critical value we always
accept the alternative hypothesis.
Hypothesis Test Example
testing a claim about µ from a small sample
3. Calculate the Test Statistic and form your final
conclusion
We calculated the test statistic tTS = -1.83
CONCLUSION: Accept the null hypothesis H0: µ  $1,000
The average BMW repair cost is greater than or equal to $1,000
Accept the alternative
hypothesis
Accept the null hypothesis
0.05
Test statistic falls inside the critical value
-2.132
Critical value determined using Excel =TINV(0.10,19)
Hypothesis Test Example
testing a claim about µ from a small sample
•
•
•
•
•
Discussion: Perhaps you noticed in the last exercise that the sample mean was
$767.40 and yet we concluded that the mean BMW repair cost is greater than or
equal to $1,000. How is this possible?
You must remember that we are talking about probabilities here. We did this
problem based on a 95% confidence of being correct (5% chance of being wrong).
Therefore, we are not certain about our conclusion, it is just our best estimate
based on the info we were given.
We do know that a sample of 5 BMWs does NOT give us a very good
understanding of ALL BMWs (the population). Certainly if the mean repair cost for
all BMWs involved in crashes was less than $1,000 then we could make that claim
without equivocation. However, in this case we must make our conclusion about
the entire population of BMWs involved in crashes based on our sample of only 5
crashed BMWs.
Since our sample mean of $767.40 is not that far from $1,000 (our claimed
population mean), and since we had only 5 BMWs in our sample, statistically (at
alpha = 0.05) we did not have enough evidence to reject the null hypothesis that
the population mean could actually be $1,000 or more, even though the sample
mean was less than $1,000.
On your own, try doing the problem again using an alpha of 0.10. Does it change
the answer? Can you see how statistics can be used to sell almost anything?
Hypothesis Test:
population
Proportions
Assumptions

When testing claims about population
proportions we assume the following
conditions exist:
1. The sample is a random sample
2. The conditions for a binomial experiment are satisfied

The are only two possible outcomes (e.g., yes/no, is/isn’t, etc.)
3. Assuming the n*p  5 and n*q  5, we can use the
standard normal distribution to determine our critical
values with µ = n*p and σ = sqrt(n*p*q)
Population Proportions:
Notation
n = number of trials
p̂ (p-hat) = sample proportion
p-hat is sometimes given directly
example: 10% of observed sports cars are red, thus p-hat =
0.10 remember p-values must always be between 0 and 1
other times p-hat must be calculated
(example: 96 households have cable TV and 54 do not, thus the
proportion of households with cable TV = 96 / (96 + 54), so p-hat
= 0.64
p = population proportion (used in the
null hypothesis)
q= 1-p
Critical Value(s)


Finding the critical value(s) when doing
hypothesis testing of population proportions
is exactly the same as finding critical
value(s) for population means using large
samples. That is, we use the Excel
function NORMSINV(probability)
Please review finding critical values for
population means using large samples if
you need to review this
Critical Value(s)
two-tail hypothesis test

For example, if α = 0.05 (total combined area) then
the critical values are NORMSINV(0.025) for the leftside and NORMSINV(0.975) for the right-side (or
simply change the left-side critical value to a positive
number and use it for the right-side critical value)
Critical
Values
0.025
-1.96
0.025
1.96
Critical Value(s)
right-tail hypothesis test

For example, if α = 0.05 (total area), for a right-tail
test the critical value is NORMSINV(0.95). Or, we
could take the left-tail critical value NORMSINV(0.05) and simply
change it to a positive value.
Critical
Value
0.05
1.645
Test Statistic for p

For testing claims about population
proportion we use the following
formula for the Test Statistic (you
must remember this)
Test Statistic (zTS) =
p̂  p
pq
n
Test Statistic for p
This is the sample
proportion which
we are given or
can calculate from
the sample data
We never really know the population
proportion (p) but we do have a value
included in the hypothesis (e.g., H0:
p=0.50) and it is this value that we put
here – the value from the hypothesis
Test Statistic (zTS) =
This is the same value (p) used in the
numerator and found in the hypotheses.
The value of q is simply 1 – p (p+q=1)
p̂  p
pq
n
n is simply the size of our
sample. E.g., if our
sample consists of 150
households then n=150
Test Statistic for p
CAUTION
 When
calculating the test statistic or
p-hat, do not round until the very end.
Rounding during calculation can
cause large errors in the final answer.
Remember, p-values are always
between 0 and 1 and so they are often
very small numbers.
Hypothesis Test Example
testing a claim about p
Geico Insurance company obtains a simple random
sample of 850 drivers and finds that 544 of them use
cell phones while driving. Use a 0.05 significance
level (α = 0.05) to test the claim that most drivers
(that means more than 50% or p  0.50) use cell
phones while driving.
Hypothesis Test Example
testing a claim about p
1. Write out the claim and the hypotheses
• The claim is that most drivers use cell phone
while driving or p  0.50
•
Notice that we are making a claim about the population of
all drivers, but we only have a sample of 850 drivers upon
which to test this claim
• Thus, the null hypothesis H0: µ  0.50
• The alternative hypothesis H1: p  0.50
•
REMEMBER that our claim may be the null hypothesis or the
alternative hypothesis, but the sign for the null hypothesis MUST
ALWAYS be =, , or  and the sign for the alternative hypothesis is
always ≠, , or  (the opposite of the null hypothesis)
Hypothesis Test Example
testing a claim about p
2. Draw a picture of the normal distribution and
determine the critical value(s)
•
Note that the sign of the alternative hypothesis is  so we have
a right-tail test
α = 0.05
Critical
Value
0.05
1.645
Critical value is determined using Excel NORMSINV
Hypothesis Test Example
testing a claim about p
3. Calculate the Test Statistic and form your final
conclusion
•
•
p̂  p
pq
n
Remember we were given the information: n = 850 and 544 of
these 850 drivers use cell phones while driving. Thus we can
calculate p-hat (drivers in our sample who whose cell phones
while driving) as 544 / 850 = 0.64. Remember, p-values MUST
ALWAYS and CAN ONLY be between 0 and 1.
We can now calculate the Test Statistic using the formula:
We know that p-hat=0.64, and p=0.50 (from the hypotheses), and
since q=1-p then q=1-0.5=0.5, and from the problem we know n=850
So, we know all the parts and can calculate the Test Statistic =
(0.64-0.50) / sqrt ((0.50*0.50) / 850) = 8.16
Hypothesis Test Example
testing a claim about p
3. Calculate the Test Statistic and form your final
conclusion
We calculated the test statistic = 8.16
CONCLUSION: Accept the alternative hypothesis H1: p  0.50
Most (more than 50%) of all drivers use cell phone while driving
Accept the alternative
hypothesis
Test statistic
falls outside the
critical value
Accept the null hypothesis
0.05
CV =1.645
Critical value determined using Excel =NORMSINV(0.95)
Hypothesis Test:
population
Standard Deviation
or Variance
Assumptions

When testing claims about population
standard deviation or variance we
assume the following conditions exist:
1. The sample is a random sample
2. The population has values that are normally
distributed (this is a strict requirement but
confirming this assumption is beyond the
scope of this class so see me if you need to
know how to do this)
2

Critical Value(s)

Finding the critical value(s) when doing hypothesis testing
of population standard deviation or variance is a little
different than anything we have done so far in this section.

Here we use the
Chi-squared (2)
distribution and the
Excel function
CHIINV where
degrees of freedom
(df) = n-1
2 Critical Value(s)


The 2 (chi-squared) distribution looks like this.
Notice how it starts at 0. That means there are NO
NEGATIVE values in a chi-squared distribution. That
also means we have NO NEGATIVE critical values
when using a chi-squared distribution.
In addition, since the distribution is NOT
SYMMETRICAL, we must determine left-side values
separately from right-side values
0
2 Critical Value(s)
two-tail hypothesis test


For example, if α = 0.05 (total combined area), and n=11
(df=n-1=10) then the critical values are CHIINV(0.975,10)
for the left-side and CHIINV(0.025,10) for the right-side
No, I did not get that backwards, in the Chi-squared
distribution the “probability” you input is the area to-theright of the critical value. If you get confused about this,
just remember that the smaller number goes on the left
and the larger number goes on the right.
Left-side 2 critical value
0.025
CV=3.247
Right-side 2 critical value
0.025
CV=20.483
2 Critical Value(s)
left-tail hypothesis test

If α = 0.05 and n=11 for a left-tail test (i.e., the
alternative hypothesis has the  sign) then the
critical value is CHIINV(0.95,10) since the areato-the-right of 0.05 in the left-tail is 0.95
Left-side 2 critical value
Area-to-the-right = 0.95
0.05
CV = 3.940
2 Critical Value(s)
right-tail hypothesis test

If α = 0.05 and n=11 for a right-tail test (i.e., the
alternative hypothesis has the  sign) then the
critical value is CHIINV(0.05,10) since the areato-the-right of 0.05 in the right-tail is simply 0.05
Right-side 2 critical value
Area-to-the-right = 0.05
0.05
CV = 18.307
Test Statistic for 

2
For testing claims about population
standard deviation or variance we use
the following formula for the Test
Statistic (you must remember this)
Test Statistic
2
( )
=
(n  1) s

2
2
Test Statistic for 2
n is simply the size of our
sample. E.g., if our
sample consists of 81
altimeters then n=81
s is the sample standard
deviation, which we must
square. Note, if given sample
variance (s2) then do not
square it again.
Test Statistic (2) =
σ is population standard deviation, which we don’t
really know, but a value for σ is always included in the
hypotheses. Therefore, simply take this value from
the hypotheses. Note, if the hypotheses relate to
variance (σ2) then put that value directly in the
equation without squaring it. However, if the
hypotheses relate to standard deviation (σ) then make
sure to square that value in the equation.
(n  1) s

2
2
Test Statistic for 2
CAUTION
 You
cannot have a negative chi-squared
value. In a two-tail chi-square
hypothesis test, the smaller value always
goes on the left.
square chi-squared (2). Chisquared is just a name or label, don’t
think there is a chi-value that you then
must square
 NEVER
Hypothesis Test Example
testing a claim about 2
Aircraft altimeters (the instrument that tells a pilot
how high the plane is) made by XYZ Company in the
past have a standard deviation of 43.7 ft. (that means
that on average, they are off by about 43.7 ft.). XYZ
recently purchased new new production equipment
and a sample of 81 new altimeters have a standard
deviation of 52.3 ft. Using an alpha of 0.05 test the
claim that the new altimeters have a standard
deviation different from the old altimeters. Note:
different means “not equal to”.
Hypothesis Test Example
testing a claim about 2
1. Write out the claim and the hypotheses
• The claim is that the population of
altimeters made on the new equipment
have a standard deviation different from
the old altimeters (43.7 feet) or σ ≠ 43.7 ft.
•
Notice that we are making a claim about the population of
new altimeters, but we only have a sample of 81 new
altimeters upon which to test this claim
• Thus, the null hypothesis H0: σ = 43.7 ft.
• The alternative hypothesis H1: σ ≠ 43.7 ft.
Hypothesis Test Example
testing a claim about 2
2. Draw a picture of the chi-squared distribution
and determine the critical value(s)
•
Note that the sign of the alternative hypothesis is ≠ so we have
a two-tail test
α = 0.05, df = n-1 = 81-1 = 80
0.025
57.153
0.025
106.629
Critical values are determined using Excel CHIINV
Hypothesis Test Example
testing a claim about 2
3. Calculate the Test Statistic and form your final
conclusion
•
•
From the problem we know that the sample standard deviation
(s) = 52.3 ft. and the population standard deviation from the
hypotheses (σ) = 43.7 ft.
Now we can use our Test Statistic formula:
Test Statistic (2) =
2 = (81-1)*52.32 / 43.72 = 114.586
(n  1) s

2
2
Hypothesis Test Example
testing a claim about 2
3. Calculate the Test Statistic and form your
conclusion
We calculated the test statistic = 114.586
CONCLUSION: Accept the alternative hypothesis H1: σ ≠ 43.7
ft. It appears the new production equipment does a worse job at
controlling standard deviation than the old equipment
Test statistic falls outside
the critical value – accept
the alternative hypothesis
Accept the alternative
hypothesis
Accept the alternative
hypothesis
Accept the null hypothesis
0.025
57.153
0.025
106.629
Critical values determined using Excel = CHIINV
Hypothesis Tests:
End Notes
Type I Error


A Type I Error is made when we accept the
alternative hypothesis when, in fact, we
should have accepted the null hypothesis
Type I Error is represented by α (alpha)
Type II Error


A Type II Error is made when we accept the
null hypothesis when, in fact, we should
have accepted the alternative hypothesis
Type II Error is represented by  (beta)

This is the first and last time that I will mention
beta in this class – there is nothing you need to
remember about beta
Controlling Type I and Type II Errors

The bottom-line when it comes to
controlling for Type I and Type II errors is to
have as large a sample size as possible.

Larger sample size means less Type I and
less Type II error
Hint for Test

How to correctly state the conclusion of
a hypothesis test! (MUST KNOW FOR TEST –
see practice exam for more examples)
A.
B.
Remember we are testing a claim, so the
correct wording of a conclusion requires us
to correctly state the claim
One of two outcomes results from any
hypothesis test – you must pick the
conclusion that matches the test results
1)
2)
We find sufficient evidence to support the claim
We do not find sufficient evidence to support the
claim
Hint for Test

How to correctly state the conclusion of a hypothesis test! (Example)
Question: A researcher claims that more than 44% of voters favor a certain initiative. A
test of the hypothesis is conducted and the null hypothesis cannot be rejected.
Which of the following sentences most correctly states the conclusion? Hint: first
note that the claim (p > 0.44) is H1, so H0 must be p ≤ 0.44. Since we cannot reject
the null hypothesis (per the test results), we are sort of accepting the null, which
means there is NOT sufficient evidence to support the alternative hypothesis (our
claim). Thus, we can analyze each of the following sentences to see which is most
correct. First, see if the claim is correct, then see if the conclusion is correct.
A. There is sufficient evidence to support the claim that the true proportion is less
than 44% (this sentence incorrectly states both the claim and the conclusion)
B. There is sufficient evidence to support the claim that the true proportion is
greater than 44% (this sentence correctly states the claim, but incorrectly states the
conclusion)
C. There is not sufficient evidence to support the claim that the true proportion is
less than 44% (this sentence incorrectly states the claim, although the conclusion is
correct)
D. There is not sufficient evidence to support the claim that the true proportion is
greater than 44% (this sentence correctly states both the claim and the conclusion)
William Christensen, Ph.D.