Hypothesis Testing for Large and
Small Sample Mean
Developing Null and Alternative
Hypotheses
 Hypothesis testing can be used to determine whether
a statement about the value of a population parameter
should or should not be rejected.

The null hypothesis, denoted by H0 , is a tentative
assumption about a population parameter.
 The alternative hypothesis, denoted by Ha, is the
opposite of what is stated in the null hypothesis.
 The alternative hypothesis is what the test is
attempting to establish.
Summary of Forms for Null and Alternative
Hypotheses about a Population Mean

The equality part of the hypotheses always appears
in the null hypothesis.
 In general, a hypothesis test about the value of a
population mean  must take one of the following
three forms (where 0 is the hypothesized value of
the population mean).
H 0 :   0
H a :   0
One-tailed
(Right)
H 0 :   0
H a :   0
H 0 :   0
H a :   0
One-tailed Two-tailed
(Left)
Using the Test Statistic
 The test statistic z has a standard normal probability
distribution.
 We can use the standard normal probability
distribution table to find the z-value with an area
of a in the lower (or upper) tail of the distribution.
 The value of the test statistic that established the
boundary of the rejection region is called the
critical value for the test.

The rejection rule is:
• Lower tail: Reject H0 if z < z.
• Upper tail: Reject H0 if z > z.
Using the p-Value
 The p-value is the probability of obtaining a sample
result that is at least as unlikely as what is observed.
 Reject H0 if the p-value <  .
Steps of Hypothesis Testing
1. Determine the null and alternative hypotheses.
2. Specify the level of significance .
3. Select the test statistic that will be used to test the
hypothesis.
Using the Test Statistic
4. Use to determine the critical value for the test
statistic and state the rejection rule for H0.
5. Collect the sample data and compute the value
of the test statistic.
6. Use the value of the test statistic and the rejection
rule to determine whether to reject H0.
Steps of Hypothesis Testing
Using the p-Value
4. Collect the sample data and compute the value of
the test statistic.
5. Use the value of the test statistic to compute the
p-value.
6. Reject H0 if p-value < .
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)

Hypotheses
H0:   
or
Ha: 

Test Statistic
 Known
z

x  0
/ n
H0:   
Ha: 
 Unknown
z
x  0
s/ n
Rejection Rule
Left tail: Reject H0 if z < z.
Right tail: Reject H0 if z > z.
One-Tailed Test about a Population Mean:
Large-Sample Case (n > 30)
Let  = .05
Sampling
distribution
of z  x   0
/ n
Reject H0
Do Not Reject H0

z
0
z = 1.645
One-Tailed Test about a Population Mean:
Large-Sample Case (n > 30)
Let  = .10
Sampling
distribution
of z  x   0
/ n
Reject H0
1
Do Not Reject H0
z
z = 1.28
0
Confidence Level
Critical Value (Z-score)
0.90
1.645
0.91
1.70
0.92
1.75
0.93
1.81
0.94
1.88
0.95
1.96
0.96
2.05
0.97
2.17
0.98
2.33
0.99
2.575
Example 1:
The response times for a random
sample of 40 medical emergencies
were tabulated. The sample mean
is 13.25 minutes and the sample
standard deviation is 3.2
minutes.
The director of medical services
wants to perform a hypothesis test, with a
.05 level of significance, to determine whether or not the
service goal of 12 minutes or less is being achieved.
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
1. Determine the hypotheses.
H0:  1
Ha: 1
2. Specify the level of significance.
3. Select the test statistic.
z
 = .05
x  0
s/ n
( is not known)
4. State the rejection rule.
Reject H0 if z > 1.96
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
5. Compute the value of the test statistic.
x   13.25  12
z

 2.47
s / n 3.2 / 40
6. Determine whether to reject H0.
Because 2.47 > 1.96, we reject H0.
We are 95% confident that Metro EMS is
not meeting the response goal of 12 minutes.
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the p-Value
4. Compute the value of the test statistic.
x   13.25  12
z

 2.47
s / n 3.2 / 40
5. Compute the p–value.
For z = 2.47, cumulative probability = .9932.
p–value = 1  .9932 = .0068
6. Determine whether to reject H0.
Because p–value = .0068 <  = .05, we reject H0.
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
• Using the p-value
 = .05
p-value

z
0
z =
1.645
z=
2.47
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)

Hypotheses
H 0 :   0
H a :   0

Test Statistic
 Known
x  0
z
/ n

 Unknown
z
x  0
s/ n
Rejection Rule
Lower tail: Reject H0 if z < z.
Upper tail: Reject H0 if z > z.
Example 2:
• Two-Tailed Tests about a Population Mean:
Large n
The production line for Glow toothpaste
is designed to fill tubes with a mean weight
of 6 oz. Periodically, a sample of 30 tubes
will be selected in order to check the
filling process.
Quality assurance procedures call for
the continuation of the filling process if the
sample results are consistent with the assumption that
the mean filling weight for the population of toothpaste
tubes is 6 oz.; otherwise the process will be adjusted.
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
1. Determine the hypotheses.
H0:  
Ha:   6
(two-tailed test)
2. Specify the level of significance.
3. Select the test statistic.
4. State the rejection rule.
 = .05
x  0
s/ n
( is not known)
z
Reject H0 if |z| > 1.96
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
Sampling
distribution
x  0
of z 
/ n
Reject H0
Reject H0
Do Not Reject H0


-1.96
0
1.96
z
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
5. Compute the value of the test statistic.
x  0
6.1  6
z

 2.74
s / n .2 / 30
6. Determine whether to reject H0.
Because 2.74 > 1.96, we reject H0.
We are 95% confident that the mean filling
weight of the toothpaste tubes is not 6 oz.
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
• Using the p-Value
Suppose we define the p-value for a two-tailed test
as double the area found in the tail of the distribution.
With z = 2.74, the cumulative standard normal
probability table shows there is a 1.0 - .9969 = .0031
probability of a z–score greater than 2.74 in the upper
tail of the distribution.
Considering the same probability of a z-score less
than –2.74 in the lower tail of the distribution, we have
p-value = 2(.0031) = .0062.
The p-value .0062 is less than
 = .05, so H0
is rejected.
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the p-Value
1/2
p-value
= .0031
1/2
p-value
= .0031
 
 


z
z = -2.74
-z/2 = -1.96
0
z/2 = 1.96
z = 2.74
Tests about a Population Mean:
Small-Sample Case (n < 30)
• Test Statistic
 Known
x  0
z
/ n
 Unknown
t
x  0
s/ n
This test statistic has a t distribution
with n - 1 degrees of freedom (df).
Tests about a Population Mean:
Small-Sample Case (n < 30)

Rejection Rule
H0:   
Reject H0 if t > t
H0:   
Reject H0 if t < -t
H0: 
Reject H0 if |t| > t
p -Values and the t Distribution

The format of the t distribution table provided in most
statistics textbooks does not have sufficient detail
to determine the exact p-value for a hypothesis test.
 However, we can still use the t distribution table to
identify a range for the p-value.
 An advantage of computer software packages is that
the computer output will provide the p-value for the
t distribution.
Example: Highway Patrol
• One-Tailed Test about a Population Mean:
Small n
A State Highway Patrol periodically samples
vehicle speeds at various locations
on a particular roadway.
The sample of vehicle speeds
is used to test the hypothesis
H0:  < 65
The locations where H0 is rejected are deemed
the best locations for radar traps.
Example 3: Highway Patrol
One-Tailed Test about a Population Mean:
Small n
At Location F, a sample of 16 vehicles shows a
mean speed of 68.2 mph with a
standard deviation of 3.8 mph.
Use a = .10 to test the hypothesis.

One-Tailed Test about a Population Mean:
Small-Sample Case (n < 30)
 Using the Test Statistic
1. Determine the hypotheses.
H0:  < 65
Ha:  > 65
2. Specify the level of significance.
3. Select the test statistic.
t
 = .10
x  0
s/ n
( is not known)
4. State the rejection rule.
Reject H0 if t > 1.753
(d.f. = 16-1 = 15)
One-Tailed Test about a Population Mean:
Small-Sample Case (n < 30)
Reject H0
Do Not Reject H0
0
1
1.753
(Critical value)
t
One-Tailed Test about a Population Mean:
Small-Sample Case (n < 30)
 Using the Test Statistic
5. Compute the value of the test statistic.
t
x   0 68.2  65

 3.37
s / n 3.8 / 16
6. Determine whether to reject H0.
Because 3.37 > 1.753, we reject H0.
We are at least 90% confident that the mean speed of
vehicles at Location F is greater than 65 mph.
Location F is a good candidate for a radar trap.
Activity 1:
1. An environmentalist collects a liter of water from 45 different locations along
the banks of a stream. He measures the amount of dissolved oxygen in each
specimen. The mean oxygen level is 4.62 mg, with the overall standard deviation of
0.92. A water purifying company claims that the mean level of oxygen in the water
is 5 mg. Conduct a hypothesis test with a =0.01 to determine whether the mean
oxygen level is less than 5 mg.
2. A bus company advertised a mean time of 160 minutes for a trip between two
cities. A consumer group had reason to believe that the mean time was more than
163 minutes. A sample of 40 trips showed a mean x̄= 153 minutes and a
standard deviation s=7.5 minutes. At the .05 level of significance, test the
consumer group’s belief.
3. Last year the government made a claim that the average income of the
American people was $34,050. However, a sample of 50 people taken recently
showed an average income of $34,076 with a population standard deviation of
$324. Is the government’s estimate too low? Conduct a significance test to see if
the true mean is more than the reported average. Use an a =0.01.
4. Your company sells exercise clothing and equipment on the Internet. To
design the clothing, you collect data on the physical characteristics of your
different types of customers. The company believed that the mean weight was
more than 60. We take a sample of 24 male runners and find their mean
weight to be 61.79 kilograms. Assume that the population standard deviation
is σ = 4.5. Calculate a 95% level of confidence
Tests about a Population Mean:
Small-Sample Case (n < 30)

Rejection Rule
H0:   
Reject H0 if t > t
H0:   
Reject H0 if t < -t
H0: 
Reject H0 if |t| > t
Tests about a Population Mean:
Large-Sample Case (n < 30)
The rejection rule is:
• Lower tail: Reject H0 if z < z.
• Upper tail: Reject H0 if z > z.