Transcript Hypothesis Testing - Weber State University

Chapter 9
Hypothesis Tests for
One Population Mean
Slide 1 - 60
Definition 9.1
Slide 2 - 60
Example 9.4
A company that produces snack foods uses a machine to
package 454 g bags of pretzels. We assume that the net
weights are normally distributed and that the population
standard deviation of all such weights is 7.8 g. A simple
random sample of 25 bags of pretzels has the net
weights, in grams, displayed in Table 9.1. Do the data
provide sufficient evidence to conclude that the packaging
machine is not working properly? We use the following
steps to answer the question.
Table 9.1
Slide 3 - 60
Example 9.4
a. State the null and alternative hypotheses for the
hypothesis test.
b. Discuss the logic of this hypothesis test.
c. Identify the distribution of the variable xx , that is, the
sampling distribution of the sample mean for samples
of size 25.
d. Obtain a precise criterionx for deciding whether to
reject the null hypothesis in favor of the alternative
hypothesis.
e. Apply the criterion in part (d) to the sample data and
state the conclusion.
Slide 4 - 60
Solution Example 9.4
a. The null and alternative hypotheses, as stated in
Example 9.1, are
H0 : μ = 454 g (the packaging machine is working
properly)
Ha : m ¹ 454g (the packaging machine is not working
properly).
b. If the null hypothesis is true, then the mean weight,
x , of the sample of 25 bags of pretzels should
approximately equal 454 g. However, if the sample
mean weight differs “too much” from 454 g, we would
be inclined to reject the null hypothesis and conclude
that the alternative hypothesis is true. As we show in
part (d), we can use our knowledge of the sampling
distribution of the sample mean to decide how much
difference is “too much.”
Slide 5 - 60
Solution Example 9.4
c. n = 25, σ = 7.8, weights are normally distributed,
･ μ x = μ (which we don’t know),
･ σ x = s n = 7.8 25 = 1.56 and
･ x is normally distributed.
In other words, for samples of size 25, the variable is
normally distributed with mean μ and standard
deviation 1.56 g.
d. The “95.44” part of the 68.26–95.44–99.74 rule states
that, for a normally distributed variable, 95.44% of all
possible observations lie within two standard
deviations to either side of the mean. Applying this
part of the rule to the variable and referring to part (c),
we see that 95.44% of all samples of 25 bags of
pretzels have mean weights within 2•1.56 = 3.12 g of
μ.
Slide 6 - 60
Solution Example 9.4
d. Equivalently, only 4.56% of all samples of 25 bags of
pretzels have mean weights that are not within 3.12 g
of μ, as illustrated in Fig. 9.1.
Figure 9.1
Slide 7 - 60
Solution Example 9.4
d. If the mean weight, x , of the 25 bags of pretzels
sampled is more than two standard deviations (3.12 g)
from 454 g, reject the null hypothesis and conclude
that the alternative hypothesis is true. Otherwise, do
not reject the null hypothesis.
Figure 9.2
Slide 8 - 60
Solution Example 9.4
e. The mean weight, x , of the sample of 25 bags of
pretzels whose weights are given in Table 9.1 is 450g.
So, z =( x − 454) / 1.56 = (450 − 454) /1.56 = −2.56.
That is, the
sample mean of
450 g is 2.56
standard
deviations below
the null-hypothesis
population mean
of 454 g, as shown
in Fig. 9.3.
Figure 9.3
Slide 9 - 60
Solution Example 9.4
e. Because the mean weight of the 25 bags of pretzels
sampled is more than two standard deviations from
454 g, we reject the null hypothesis (μ = 454 g) and
conclude that the alternative hypothesis ( m ¹ 454g) is
true.
The data provide sufficient evidence to conclude that
the packaging machine is not working properly.
Slide 10 - 60
The set of values for the
test statistic that leads us
to reject the null
hypothesis is called the
rejection region. In this
case, the rejection region
consists of all z-scores
that lie either to the left of
−2 or to the right of 2; that
part of the horizontal axis
under the shaded areas in
Fig. 9.5.
Figure 9.5
Slide 11 - 60
Definition 9.2
Slide 12 - 60
For a two-tailed test, the null hypothesis is rejected
when the test statistic is either too small or too large. The
rejection region consists of two parts: one on the left and
one on the right, (Fig. 9.6(a)).
For a left-tailed test, the null hypothesis is rejected
only when the test statistic is too small. The rejection
region consists of only one part, on the left, (Fig.9.6(b)).
For a right-tailed test, the null hypothesis is rejected
only when the test statistic is too large. The rejection
region consists of only one part, on the right, (Fig.9.6(c)).
Figure 9.6
Slide 13 - 60
Definition 9.3
Slide 14 - 60
Definition 9.4
Slide 15 - 60
Key Fact 9.1
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Key Fact 9.2
Slide 17 - 60
Key Fact 9.3
Slide 18 - 60
Procedure 9.1
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Procedure 9.1 (cont.)
Slide 20 - 60
Key Fact 9.4
Slide 21 - 60
Definition 9.6
Slide 38 - 60
If the null hypothesis is true, this test statistic has the
standard normal distribution, and its probabilities equal
areas under the standard normal curve. If we let z0
denote the observed value of the test statistic z, we
obtain the P-value as follows:
Two-tailed test: The P-value is the probability of
observing a value of the test statistic z that is at least as
large in magnitude as the value actually observed, which
is the area under the standard normal curve that lies
outside the interval from − |z0 | to |z0 |, as in Fig. 9.21(a).
Figure 9.21
Slide 39 - 60
Left-tailed test: The P-value is the probability of
observing a value of the test statistic z that is as small as
or smaller than the value actually observed, which is the
area under the standard normal curve that lies to the left
of z0, as in Fig. 9.21(b).
Right-tailed test: The P-value is the probability of
observing a value of the test statistic z that is as large as
or larger than the value actually observed, which is the
area under the standard normal curve that lies to the
right of z0, as in Fig. 9.21(c).
Figure 9.21
Slide 40 - 60
The P-value can be interpreted as the observed
significance level of a hypothesis test. Suppose that the
value of the test statistic for a right-tailed z-test turns out
to be 1.88. Then the P-value of the hypothesis test is 0.03
(actually 0.0301), the shaded area in Fig. 9.24. The null
hypothesis would be
rejected for a test at the
0.05 significance level
but would not be
rejected for a test at the
0.01 significance level.
In fact, the P-value is
precisely the smallest
significance level at
which the null hypothesis
would be rejected.
Figure 9.24
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Procedure 9.2
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Procedure 9.2 (cont.)
Slide 43 - 60
Table 9.11
Slide 44 - 60
Table 9.12
Slide 45 - 60
Example 9.16
Use Table IV to estimate the P-value of each one-mean ttest.
a. Left-tailed test, n = 12, and t = −1.938
b. Two-tailed test, n = 25, and t = −0.895
Solution a. Because the test is left-tailed, the P-value is
the area under the t-curve with df = 12−1 = 11 that lies to
the left of −1.938, as shown in Fig. 9.27(a).
Figure 9.27
Slide 46 - 60
Solution Example 9.16
A t-curve is symmetric about 0, so the area to the left of
−1.938 equals the area to the right of 1.938, which we
can estimate by using Table IV. In the df = 11 row of Table
IV, the two t-values that straddle 1.938 are
t0.05 = 1.796 and t0.025 = 2.201. Therefore the area under
the t-curve that lies to the right of 1.938 is between 0.025
and 0.05, as shown in Fig. 9.27(b). Consequently, the
area under the t-curve that lies to the left of −1.938 is also
between 0.025 and 0.05, so 0.025 < P < 0.05. Hence we
can reject H0 at any significance level of 0.05 or larger,
and we cannot reject H0 at any significance level of 0.025
or smaller. For significance levels between 0.025 and
0.05, Table IV is not sufficiently detailed to help us to
decide whether to reject H0.
Slide 47 - 60
Solution Example 9.16
b. Because the test is two tailed, the P-value is the area
under the t-curve with df = 25 − 1 = 24 that lies either to
the left of −0.895 or to the right of 0.895, as shown in Fig.
9.28(a).
Figure 9.28
Slide 48 - 60
Solution Example 9.16
Because a t-curve is symmetric about 0, the areas to the
left of −0.895 and to the right of 0.895 are equal. In the
df = 24 row of Table IV, 0.895 is smaller than any other
t-value, the smallest being t0.10 = 1.318. The area under
the t-curve that lies to the right of 0.895, therefore, is
greater than 0.10, as shown in Fig. 9.28(b).
Consequently, the area under the t-curve that lies either
to the left of −0.895 or to the right of 0.895 is greater than
0.20, so P > 0.20. Hence we cannot reject H0 at any
significance level of 0.20 or smaller. For significance
levels larger than 0.20, Table IV is not sufficiently detailed
to help us to decide whether to reject H0.
Slide 49 - 60
Procedure 9.3
Slide 50 - 60
Procedure 9.3 (cont.)
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Flowchart for
choosing the
correct hypothesis
testing procedure
for a population
mean.
Figure 9.34
Slide 59 - 60
Homework
When able, do the following by hand and check your answer
with minitab.
9.54, 55, 57, 59, 62
9.117, 120
9.141, 142, 143
Slide 60 - 60