Means - People

Report
Hypothesis Testing Using a
Single Sample
© 2010 Pearson Prentice Hall. All rights reserved
Part I: Means
A P-value is the probability of observing a
sample statistic as extreme or more
extreme than the one observed under the
assumption that the null hypothesis is true.
10-3
Step 1: A claim is made regarding the population
mean. The claim is used to determine the
null and alternative hypotheses. Again,
the hypothesis can be structured in one of
three ways:
10-4
To test hypotheses regarding the population
mean assuming the population standard
deviation is unknown, we use the tdistribution rather than the Z-distribution.
When we replace  with s,
x
s
n
follows Student’s t-distribution with n-1
degrees of freedom.
10-5

Properties of the t-Distribution
1. The t-distribution is different for different
degrees of freedom.
10-6
Properties of the t-Distribution
1. The t-distribution is different for different
degrees of freedom.
2. The t-distribution is centered at 0 and is
symmetric about 0.
10-7
Properties of the t-Distribution
1. The t-distribution is different for different
degrees of freedom.
2. The t-distribution is centered at 0 and is
symmetric about 0.
3. The area under the curve is 1. Because of the
symmetry, the area under the curve to the
right of 0 equals the area under the curve to
the left of 0 equals 1/2.
10-8
Properties of the t-Distribution
4. As t increases (or decreases) without bound,
the graph approaches, but never equals, 0.
10-9
Properties of the t-Distribution
4. As t increases (or decreases) without bound,
the graph approaches, but never equals, 0.
5. The area in the tails of the t-distribution is a
little greater than the area in the tails of the
standard normal distribution because using s
as an estimate of  introduces more variability
to the t-statistic.
10-10
Properties of the t-Distribution
6. As the sample size n increases, the density
curve of t gets closer to the standard normal
density curve. This result occurs because as
the sample size increases, the values of s get
closer to the values of  by the Law of Large
Numbers.
10-11
Testing Hypotheses Regarding a Population
Mean with σ Unknown
To test hypotheses regarding the population
mean with  unknown, we use the following
steps, provided that:
1. The sample is obtained using simple random
sampling.
2. The sample has no outliers, and the
population from which the sample is drawn
is normally distributed or the sample size is
large (n≥30).
10-12
Setup: Clearly indicate/describe the parameter
of interest to the study.
Step 1: Determine the null and alternative
hypotheses. The hypotheses can be
structured in one of three ways:
10-13
Step 2: Select a level of significance, , based
on the seriousness of making a
Type I error.
10-14
Step 3: Compute the test statistic
t0 
x  0
s
n
which follows Student’s t-distribution
with n-1 degrees of freedom.

10-15
P-Value Approach
Step 4: Use Table VI to estimate the P-value
using n-1 degrees of freedom.
10-16
P-Value Approach
Two-Tailed
10-17
P-Value Approach
Left-Tailed
10-18
P-Value Approach
Right-Tailed
10-19
P-Value Approach
Step 5: If the P-value < , reject the null
hypothesis.
If the P-value ≥ α, fail to reject the null
hypothesis
10-20
Step 6: State the conclusion in the
context of the problem.
10-21
Parallel Example 1: Testing a Hypothesis about a
Population Mean, Large Sample
Assume the resting metabolic rate (RMR) of healthy males
in complete silence is 5710 kJ/day. Researchers measured
the RMR of 45 healthy males who were listening to calm
classical music and found their mean RMR to be 5708.07
with a standard deviation of 992.05.
At the =0.05 level of significance, is there evidence to
conclude that the mean RMR of males listening to calm
classical music is different than 5710 kJ/day?
10-22
Solution
We assume that the RMR of healthy males is 5710
kJ/day. This is a two-tailed test since we are interested
in determining whether the RMR differs from 5710
kJ/day.
Since the sample size is large, we follow the steps for
testing hypotheses about a population mean for large
samples.
10-23
Solution
Step 1: H0: =5710
versus
H1: ≠5710
Step 2: The level of significance is =0.05.
Step 3: The sample mean is x = 5708.07 and the
sample standard deviation is s=992.05. The test statistic
is
t0 
10-24
 .07  5710
5708
992 .05
45
  0.013
Solution: P-Value Approach
Step 4: Since this is a two-tailed test, the P-value is the
area under the t-distribution with n-1=45-1=44
degrees of freedom to the left of -t0.025= -0.013
and to the right of t0.025=0.013. That is, P-value
= P(t < -0.013) + P(t > 0.013) = 2 P(t > 0.013).
0.50 < P-value.
Step 5: Since the P-value is greater than the level of
significance (0.05<0.5), we fail to reject the
null hypothesis.
10-25
Solution
Step 6: There is insufficient evidence at the =0.05
level of significance to conclude that the mean
RMR of males listening to calm classical music
differs from 5710 kJ/day.
10-26
Parallel Example 2: Testing a Hypothesis about a
Population Mean, Small Sample
According to the United States Mint, quarters weigh 5.67
grams. A researcher is interested in determining whether the
“state” quarters have a weight that is different from 5.67
grams. He randomly selects 18 “state” quarters, weighs
them and obtains the following data.
5.70
5.67
5.73
5.61
5.70
5.67
5.65
5.62
5.73
5.65
5.79
5.73
5.77
5.71
5.70
5.76
5.73
5.72
At the =0.05 level of significance, is there evidence to
conclude that state quarters have a weight different than
5.67 grams?
10-27
Solution
We assume that the weight of the state quarters is 5.67
grams. This is a two-tailed test since we are interested
in determining whether the weight differs from 5.67
grams.
Since the sample size is small, we must verify that the
data come from a population that is normally distributed
with no outliers before proceeding to Steps 1-6.
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Assumption of normality appears reasonable.
10-29
No outliers.
10-30
Solution
Step 1: H0: =5.67
H1: ≠5.67
versus
Step 2: The level of significance is =0.05.
Step 3: From the data, the sample mean is calculated to
be 5.7022 and the sample standard deviation is
s=0.0497. The test statistic is
t0 
10-31
5 .7022  5 .67
.0497
18
 2 .75
Solution: P-Value Approach
Step 4: Since this is a two-tailed test, the P-value is the
area under the t-distribution with n-1=18-1=17
degrees of freedom to the left of -t0.025= -2.75
and to the right of t0.025=2.75. That is, P-value
= P(t < -2.75) + P(t > 2.75) = 2 P(t > 2.75).
0.01 < P-value < 0.02.
Step 5: Since the P-value is less than the level of
significance (0.02<0.05), we reject the null
hypothesis.
10-32
Solution
Step 6: There is sufficient evidence at the =0.05 level
of significance to conclude that the mean
weight of the state quarters differs from
5.67 grams.
10-33

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