Transcript 7.2 Hypothesis Testing for the Mean (Large Samples Statistics Mrs. Spitz Spring 2009 Objectives/Assignment     How to find critical values in a normal distribution How to use the.

7.2 Hypothesis Testing for the
Mean (Large Samples
Statistics
Mrs. Spitz
Spring 2009
Objectives/Assignment




How to find critical values in a normal
distribution
How to use the z-test to test a mean 
How to find P-values and use them to
test a mean
Assignment: pp. 324-327 #1-36
Critical Values in a Normal
Distribution

In Chapter 5, you learned that when
the sample size is at least 30, the
sampling distribution for x (the sample
mean) is normal.
Ex.1: Finding a critical value
for a left-tailed test


Find the critical value and
rejection region for a lefttailed test with  = 0.01.
Solution: The graph at the
right shows a standard
normal curve with a shaded
area of 0.01 in the left tail.
Using Table 4, the z-score
that corresponds to an area
of 0.01 is -2.33. So, the
critical value is zo = -2.33.
The rejection region is the left
of this critical value.
Reminder: If you cannot find
the exact area in Table 4, use
the area that is closest. For
instance, in Ex. 1, the area
closest to 0.01 is 0.0099
Try it Yourself 1

a.
b.
c.
Find the critical value and
rejection region for a lefttailed test with  = 0.10.
Draw a standard normal
curve with an area of  in
the left tail.
Use Table 4 to locate the
area that is closest to .
(Standard normal table front
of book)
Find the z-score that
corresponds to this area.
2. 0.1003
3. -1.28
Ex. 2: Finding a critical value
for a Right-Tailed Test

a.
b.
Find the critical value and
rejection region for a righttailed test with  = 0.04.
Draw a standard normal
curve with an area of  in
the right tail. The area to
the left of zo is 1 - =0.96
Use Table 4 to locate the
area that is closest to 0.96 is
1.75. So the critical value of
zo = 1.75
a. O.96
3. 1.75
Ex. 3: Finding Critical Values
for a Two-Tailed Test


Find the critical values and
rejection regions for a twotailed test with  = 0.05.
Solution: The graph at the
right shows a standard
normal curve with shaded
areas ½ = 0.025 in each
tail. The area to the left of
–zo is ½ = 0.025, and the
area to the left of zo is 1 ½ = 0.975.
Using Table 4, the z-scores
that correspond to the
areas 0.025 and 0.975 are 1.96 and 1.96 respectively,
so the critical values are –zo
= -1.96 and zo =1.96

Study Tip

Notice in Ex. 3 that
the critical values
are opposites. This
is always true for
two-tailed z-tests.
The table lists the
critical values for
commonly used
levels of
significance.
WRITE THIS DOWN!!!! YOU
WILL NEED IT!!!!
The z-test for a Mean 

The z-test for the mean is used in
populations for which the sampling
distribution of sample means is normal.
To use the z-test, you need to find the
standardized value for your test
statistic, x .

With all hypothesis tests, it is helpful to
sketch the sampling distribution. Your sketch
should include any critical values, rejection
regions, and the standardized test statistic.
Ex. 4: Testing  with a Large
Sample

Employees in a large accounting firm claim
that the mean salary of the firm’s
accountants is less than that of its
competitor’s which is $45,000. A random
sample of 30 of the firm’s accountants has a
mean salary of $43,500 with a standard
deviation of $5200. At  = 0.05, test the
employees’ claim.
Solution to ex. 4

The claim is “the mean salary is less than
$45,000.” So, the null and alternative
hypotheses are written as:
Ho:  ≥ $45,000
Ha:  < $45,000 (Claim)
Solution to ex. 4
Because the test is a left-tailed test and the level of
significance is  = 0.05, the critical value is zo = 1.645 and the rejection region is z < -1.645.
Because the sample size is at least 30, the
standardized test statistic for the z-test is:
x   43,500 45,000
z

 1.58
s
5200
n
30
Solution to ex. 4

The graph shows the location
of the rejection region and the
standardized test statistic, z.
Because z is not in the
rejection region, you fail to
reject the null hypothesis. In
other words, there is not
enough evidence to at the 5%
level of significance to support
the employees’ claim that the
mean salary is less than
$45,000
A word of advice

Be sure you understand the decision
made in Example 4. Even though your
sample has a mean of $43,500, you
cannot (at a 5% level of significance)
support the claim that the mean of all
accountants’ salaries is less than
$45,000. The difference between your
test statistic and the hypothesized mean
is probably due to sampling error.
Ex. 5: Testing  with a Large
Sample

The U.S. Department of Agriculture reports
that the mean cost of raising a child from
birth to age 2 in a rural area is $8390. You
believe that this value is incorrect, so you
select a random sampling of 900 children
(age 2) and find the mean cost is $8275 with
a standard deviation of $1540. At  = 0.05,
is there enough evidence to conclude that
the mean cost is different from $8390?
Solution to ex. 5

You want to support the claim that “the
mean cost is different from $8390.” So the
null and alternative hypotheses are:
Ho:  = $8390
Ha:   $8390 (Claim)
Solution to ex. 5
Because the test is a two-tailed test and the level of
significance is  = 0.05, the critical values are
-zo = -1.96 and are zo = 1.96 the rejection region is
z < -1.96 and z > 1.96. Because n ≥ 30, the
standardized test statistic for the z-test is:
x   8275 8390
z

 2.24
s
1540
n
900
Solution to ex. 5

The graph shows the location
of the rejection region and the
standardized test statistic, z.
Because z is in the rejection
region, you should decide to
reject the null hypothesis. In
other words, you have enough
evidence to conclude that the
mean cost of raising a child
from birth to age 2 in a rural
area is significantly different
from $8390 at the 5% level of
significance.
Using P-values for a z-Test

Another way to reach a conclusion in a
hypothesis test is to use a P-value for
the sample statistic. Many people
prefer this method when using
technology.

If a P-value is less than 0.01, the null hypothesis
will be rejected at the common levels of 0.01,
0.05, and 0.10. If the P-value is greater than
0.10, then you would fail to reject Ho for these
common levels. This conclusion will be reached
no matter whether you use the critical value
method or the P-value method.
Study Tip:

The lower the P-value, the more
evidence there is in favor of rejecting
Ho. The P-value gives you the lowest
level of significance for which the
sample statistic allows you to reject the
null hypothesis. In example 6, you
would reject Ho at any level of
significance greater than 0.0237.
Ex. 6: Interpreting a P-value




The P-value for a hypothesis test is p =
0.0237. What is your decision if the level of
significance is 1)  = 0.05 and 2)  = 0.01?
SOLUTION:
Because 0.0237 < 0.05, you should reject the
null hypothesis.
Because 0.0237 > 0.01, you should fail to
reject the null hypothesis.
Ex. 7: Using a Technology Tool
to Interpret a z-test

What decision should you make for the
following printout using a level of significance
 = 0.05?
Ex. 8: Hypothesis Testing
Using P-values

In an advertisement, a pizza shop
claims that its mean delivery time is less
than 30 minutes. A random selection of
36 delivery times has a sample mean of
28.5 minutes and a standard deviation
of 3.5 minutes. Is there enough
evidence to support the claim at  =
0.01? Use a P-value.
Solution to ex. 8

the claim is “the mean delivery time is less
than 30 minutes.” So the null and alternative
hypotheses are:
Ho:  ≥ 30 minutes
Ha:  < 30 minutes (Claim)
Solution to ex. 8
The standardized test statistic for the z-test is:
x   28.5  30
z

 2.57
s
3 .5
n
36
Solution to ex. 8

Using Table 4, the area
corresponding to z = -2.57 is
0.0051. Because this test is a
left-tailed test, the P-value is
equal to the area to the left of z =
-2.57. So, P – 0.0051. Because
the P-value is less than  = 0.01,
you should decide to reject the
null hypothesis. So, at the 1%
level of significance, you have
sufficient evidence to conclude
that the mean delivery time is
less than 30 minutes.
Ex. 9: Hypothesis Testing
Using P-values

You think that the average
franchise investment
information given in the
graph is incorrect, so you
randomly select 30
franchises and determine
the necessary investment
for each. The sample
mean is $135,000 with a
standard deviation of
$30,000. Is there enough
evidence to support your
claim at  = 0.05? Use a
P-value.
Ex. 9 - SOLUTION

The claim is “the mean is different from
$143,260.” so, the null and alternative
hypotheses are:
Ho:  = $143,260
Ha:   $143,260 (Claim)
Solution to ex. 8
The level of significance is  = 0.05. Using the
z-test, the standardized test statistic is:
x   135,000 143,260
z

 1.51
s
30,000
n
30
Solution to ex. 9
Using Table 4, the area corresponding to
z = -1.51 is 0.0655. Because the test is a two-tailed test, the
P-value is equal to twice the area to the left of z = -1.51.
So,
P  2(0.0655)  0.1310
Because the P-value is greater than , you should fail to
reject the null hypothesis. So there is not enough
evidence at the 5% level of significance to conclude that
the mean franchise investment is not $143,260.