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Chapter 10
Lesson 10.4
Hypotheses and Test Procedures
10.4: Hypothesis Tests for a Population
Mean
The One-Sample t-test for a
Population Mean
Null hypothesis:
Test Statistic:
H0: m = hypothesized value
x m
t 
s
n
Alternative Hypothesis:
P-value:
Ha: m > hypothesized value Area to the right of calculated t
with df = n-1
Ha: m < hypothesized value Area to the left of calculated t
with df = n-1
Ha: m ≠ hypothesized value 2(Area to the right of t) of +t
or 2(Area to the left of t) of -t
The One-Sample t-test for a
Population Mean
Assumptions:
1. Random Sample
1. The sample size n is large (n > 30) so CLT applies
or the population distribution is at least
approximately normal.
A study conducted by researchers at Pennsylvania State
University investigated whether time perception, an
indication of a person’s ability to concentrate, is impaired
during nicotine withdrawal.
After a 24-hour smoking abstinence, 20 smokers were
asked to estimate how much time had elapsed during a
45-second period. Researchers wanted to see whether
smoking abstinence had a negative impact on time
perception, causing elapsed time to be overestimated.
Suppose the resulting data on perceived elapsed time (in
seconds) were as follows:
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
What is the mean and standard
x = 59.30 s = 9.84
n = 20
deviation
of the sample?
Smoking Abstinence Continued . . .
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
x = 59.30 s = 9.84 n = 20
Where m is the true mean perceived elapsed
H0: m = 45
Since the boxplot
is approximately
time for
smokers
who have abstained from
Verify
Ha:symmetrical,
m > 45
it is plausible
that the
smoking
for 24-hours
State the
assumptions.
population
distribution
is
Assumptions:
approximately normal.
1) It is a random sample.
hypotheses.
2) Since the
sizewe
is not
Tosample
do this,
need to graph the
at least data
30, weusing
must a histogram, boxplot,
determine ifor
it normal
is plausible
probability plot
that the population
distribution is approximately
40 50 60 70
normal.
Smoking Abstinence Continued . . .
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
x = 59.30 s = 9.84 n = 20
H0: m = 45
Ha: m > 45
Where m is the true mean perceived elapsed
time for smokers whoCompute
have abstained
from
the test
smoking for 24-hours
Test statistic:
P-value ≈ 0
statistic and P-value.
59.30  45
t 
 6.50
9.84
20
a = .05
Since P-value < a, we reject H0. There is convincing
evidence that the mean perceived elapsed time is
greater than the actual elapsed time of 45 seconds.
Smoking Abstinence Continued . . .
69 65 72 73 59 55 39 52 67 57
56 50 70 47 56 45 70 64 67 53
x = 59.30 s = 9.84 n = 20
Compute
the appropriate
Where
m
is
the
true
mean
perceived elapsed
H0: m = 45
confidence
time for smokers
who have interval.
abstained from
Ha: m > 45
smoking for 24-hours
Since P-value < a, we reject H0.
Notice that the
a = .05
hypothesized value
 9.84
 
59
.
30

1
.
729
of 45 is NOT in the
20 

90%
confidence
Since
this is a one-tailed
test,
a )
(55.497
, 63.103
interval
wetail. .05 goes in
goes inand
thethat
upper
“rejected”
H0! leaving .90 in the
the lower tail,
middle.
A growing concern of employers is time spent in activities
like surfing the Internet and emailing friends during work
hours. The San Luis Obispo Tribune summarized the
findings of a large survey of workers in an article that ran
under the headline “Who Goofs Off More than 2 Hours a
Day? Most Workers, Survey Says” (August 3, 2006).
Suppose that the CEO of a large company wants to
determine whether the average amount of wasted time
during an 8-hour day for employees of her company is less
than the reported 120 minutes. Each person in a random
sample of 10 employees was contrasted and asked about
daily wasted time at work. The resulting data are the
following:
108 112 117 130 111
131 113 113 105 128
What is the mean and standard
x = 116.80 s = 9.45 n = 10
deviation of the sample?
Surfing Internet Continued . . .
108 112 117 130 111 131 113 113 105 128
x = 116.80 s = 9.45 n = 10
H0: m = 120 Where m is the true mean daily wasted
time forsome
employees
of this company
Ha:The
m < boxplot
120 reveals
skewness,
but there is no outliers. It is plausible
Assumptions:
the
that the population distribution isState
Verify
the
1) The given
sample
was
a
random
sample
of
employees
hypotheses.
approximately normal.
assumptions.
2) Since the sample size is not
at least 30, we must
determine if it is plausible
that the population
distribution is approximately
normal.
110
120
130
Surfing Internet Continued . . .
108 112 117 130 111 131 113 113 105 128
x = 116.80 s = 9.45 n = 10
H0: m = 120
Ha: m < 120
Where m is the true mean daily wasted
time for employees of this company
the
testwe
116potential
.80 Compute
120 error
What
could

 and
1.07
Test Statistic: thave
statistic
TypeP-value.
II
9made?
.45
P-value =.150
10
a = .05
Since p-value > a, we fail to reject H0. There is not
sufficient evidence to conclude that the mean daily
wasted time for employees of this company is less than
120 minutes.
Homework
• Pg.610: #10.43, 45, 48, 57 (49 if
you want extra practice)