Download Inference about Mean (σ Unknown)

InferenceaboutMean
(σUnknown)
Whenσisknown,thesamplingdistribu>onforasample
meanisnormalifcondi>onsaresa>sfied.
Formanyyears,itwasthoughtthatwhenσwas
unknown,thiswass>llthecase.However,becauseof
theincreasedvariabilityintroducedbynotknowingσ,
thesamplingdistribu>onforasamplemeanwith
unknownσisnotnormal.
ThiswasdiscoveredbyW.S.Gosset,anIrishBrewery
qualitycontrolinspectorin1908.Healsodiscovered
thatwhenσisunknown,wecans>lldoinferenceusinga
samplingdistribu>onmodelhediscovered,thetdistribu>on.
Sec>on9.1,Page184
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t-Distribu>on
• Thet-Distribu>onisunimodalandsymmetric
• IthasmoreareainthattailsandisflaSerinthe
middlethanthenormaldistribu>on
• Itisafamilyofdistribu>ons,withadifferentcurve
foreachμ,σ,anddf(degreesoffreedom)df=n-1,
wherenisthesamplesize.
Cri$calValuesfor95%ConfidenceInterval
Knownσ:1.960fromNormalDistribu>on
Unknownσwith3df:3.18fromt-Distribu>on
Unknownσwith10df:2.23fromt-Distribu>on
Unknownσwith1000df:1.962fromt-Distribu>on
Sec>on9.1,Page184
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t-Distribu>on
WhentoUseIt
Is the population parameter of interest a
mean?
Yes
Is the value of the population
standard deviation, σ, an
unknown value?
Yes
Use the t-Distribution
Sec>on9.1,Page184
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ConfidenceInterval
(Unknownσ)
Confidence Interval = Sample Mean ± Margin of Error
Margin of Error = Critical Value x Standard Error.
Confidence Interval =
s
x ± t(df ,α ) ×
n
where s is the standard deviation of the
sample and n is the sample size, and df is
the degrees of freedom, n-1.
€
€
Conditions: The population must be normal
or the sample is large (n ≥ 30)
Sec>on9.1,Page186
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ConfidenceInterval(Unknownσ)
Illustra>veProblem-TI83Add-inPrograms
C.I.=SampleMean±Cri>calValue*StandardError
PRGM–CRITVAL–ENTER
2:T-DIST
CONFLEVEL=.95
df=19
Answer:2.0930
PRGM–STDERROR-ENTER
4:1-MEAN
n=20
Sx=1.76
Answer:.3935
C.I.=6.87±2.0930*.3935=6.87±.8236
=(6.87-.8236,6.87+.8236)=(6.05,7.69)
Sec>on9.1,Page186
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ConfidenceInterval(Unknownσ)
Illustra>veProblem-TI83BlackBoxProgram
STAT–TESTS–8:Tinterval
Inpt:Stats
6.87
x:
Sx:1.76
n:20
C-Level:.95
Calculate
Answer:(6.05,7.69)
Sec>on9.1,Page186
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Problems
a. 
b. 
c. 
d. 
e. 
Findthe98%confidenceinterval.
Findthecri>calvalue
Findthemarginoferror.
Findthestandarderror.
Whatassump>onmustwemakeaboutthethe
popula>ontohaveat-samplingdistribu>on.
f.  Whataretheproperwordstodescribethe
confidenceinterval?
g.  Ifyouwantedtohaveamarginoferrorofone
minuteandthe98%confidenceintervalforthis
data,howlargemustthesamplebe?
Problems,Page205
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Problems
a. 
b. 
c. 
d. 
e. 
Findthe93%confidenceinterval.
Findthecri>calvalue
Findthemarginoferror.
Findthestandarderror.
Whatassump>onmustwemakeaboutthepopula>on
tohaveat-samplingdistribu>on.
f.  Whataretheproperwordstodescribetheconfidence
interval?
g.  Ifyouwantedtohaveamarginoferrorof25lbs.anda
confidencelevelof93%forthisdata,howmany
studentswouldyouhavetomonitor?
Sec>on9.1,Page205
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HypothesesTest(Unknownσ)
Illustra>veProblem-TI-83Add-In
The EPA wanted to show that the mean carbon
monoxide level is higher than 4.9 parts per million.
Does a random sample of 22 readings (sample mean
= 5.1, s=1.17) present sufficient evidence to support
the EPA’s claim? Use α = .05. Assume the readings
have approximately a normal distribution.
Ho: µ = 4.9 (no higher than)
Ha: µ > 4.9 (higher than)
t-Distribution
df = 21
P-value = .2158
µx = 4.9
x = 5.1
P − Value = P(x > 5.1 given µx = 4.9)
€
€
€
PRGM – TDIST
– ENTER
LOWER BOUND = 5.1
UPPER BOUND = 2ND EE99
MEAN = 4.9
SE(x ) = 1.17 / 22
€
df = 21
Answer: .2158 (Evidence is not sufficient to
support the EPA’s claim.)
€
Sec>on9.1,Page187
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HypothesesTest(Unknownσ)
Illustra>veProblem-TI-83BlackBox
The EPA wanted to show that the mean carbon
monoxide level is higher than 4.9 parts per million.
Does a random sample of 22 readings (sample mean =
5.1, s=1.17) present sufficient evidence to support the
EPA’s claim? Use α = .05. Assume the readings have
approximately a normal distribution.
Ho: µ = 4.9 (no higher than)
Ha: µ > 4.9 (higher than)
STAT − TESTS − TTest
Inpt :Stats
µ0 :4.9
x : 5.1
Sx :1.17
n = 22
µ :> µ0
Calculate
Answer : p − value = .2158
There is not sufficient evidence to support
the EPA claim.
€
Sec>on9.1,Page187
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Problems
In a study of computer use, 95 randomly selected
Internet Canadian users were asked how much
time they spend using the Internet in a typical
week. The mean of the sample was 12.7 hours
and the standard deviation is 4.1 hours. Is this
convincing evidence that Canadians use the
internet more than the USA average of 12.3
hours a week?
a. 
b. 
c. 
d. 
State the correct hypothesis
Find the p-value.
State your conclusion.
What is the name of the probability model
used for the sampling distribution
e.  What is the mean of the sampling
distribution?
f.  What is the value of the standard error?
g.  If your conclusion is in error, what type of
error is it?
Problems,Page209
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Problems
The recommended number of hours of sleep per
night is 8 hours, but everybody “knows” that the
average college student sleeps less than 7 hours.
The number of hours slept last night by 10 randomly
selected college students is listed here.
5.2
6.8 5.5 7.8 5.8 7.1 8.1 6.9 5.7 7.2
Is there convincing evidence that the students sleep
less than 7 hours? Assume the population is
normal.
a. 
b. 
c. 
d. 
State the correct hypothesis.
Find the p-value.
State your conclusion.
What is the name of the probability model used
for the sampling distribution
e.  What is the mean of the sampling distribution?
f.  What is the value of the standard error?
g.  If your conclusion is in error, what type of error is
it?
Problems,Page205
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Problems
a. 
b. 
c. 
d. 
State the appropriate hypotheses.
Find the p-value.
State your conclusion.
What is the name of the probability model used
for the sampling distribution
e.  What is the mean of the sampling distribution?
f.  What is the value of the standard error?
g.  If your conclusion is in error, what type of error is
it?
Problems,Page206
13
InferenceaboutPropor>ons
A binomial experiment is a experiment that has only
two outcomes. Binomial experiments relate to
categorical variables. Consider the variable,
Supports Obama. The variable has two categories,
yes and no. We will consider the yes category as a
“success” and the no category a “failure”.
Suppose we have a population of 1000 voters, and
550 support Obama. We define proportion of
success for the population, p = 550/1000 = .55.
The proportion of failures is q = 1-p = .45.
Suppose we take a sample to estimate p, the true
proportion of voters that support Obama. We take
a random sample of 100 voters and 53 support
Obama. Our sample proportion is p’ = 53/100=.53.
Our sample q’ = 1-p’ = .47.
Sec>on9.2,Page192
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Problems
Problems,Page206
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SamplingDistribu>onfora
Propor>on
In practice, the following conditions will insure the
sampling distribution for a proportion is normal.
1.  The sample size is greater than 20
2.  The product np and nq are both greater than 5.
Where we do not know p, we substitute p’, np’ =
# success in sample must be > 5 and nq’ =n(1p’) = # failures in sample must be > 5.
3.  The sample consists of less than 10% of the
population.
Sec>on9.2,Page192
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ConfidenceIntervalfora
Propor>on,p
Given a sample proportion p’, that has a normal
sampling distribution, the confidence interval for the
true population proportion, p, is:
sample proportion ± margin of error
sample proportion ± critical value × standard error of p’
p'±z(α ) ×
p'q'
n
where q’ = proportion of failures = 1-p’ and n is the
sample size.
€
Sec>on9.2,Page193
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ConfidenceIntervalforPropor>on
Illustra>veExampleTI-83Add-inPrograms
In a discussion about the cars that fellow students
drive, several statements were made about types,
ages, makes colors, and so on. Dana decided he
wanted to estimate the proportion of convertibles
students drive, so he randomly identified 200 cars in
the student parking lot and found 17 to be
convertibles. Find the 90% confidence Interval for p,
the true proportion of convertibles.
Check conditions for normal sampling condition:
n = 200 > 20
# successes = 17 > 5
# failures = 200 – 17 = 183 > 5
Conditions are satisfied.
PRGM: CRITVAL 1 – CONF LEVEL = .90
Answer: 1.6449
PRGM: STDERROR – 1:1 PROP –
p’ = 17/200; n = 200
Answer: .0197
Confidence Interval = 17/200 ± 1.6449*.0197 =
0.0850 ± 0.0324 = (.0850 - .0324, .0850 + .0324)=
(.0526, .1174)
Sec>on9.2,Page193
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ConfidenceIntervalforPropor>on
Illustra>veExampleTI-83BlackBoxProgram
In a discussion about the cars that fellow students
drive, several statements were made about types,
ages, makes colors, and so on. Dana decided he
wanted to estimate the proportion of convertibles
students drive, so he randomly identified 200 cars in
the student parking lot and found 17 to be
convertibles. Find the 90% confidence Interval for
p, the true proportion of convertibles.
STAT – TESTS – A:1 – PROPZInt
x: 17 (The number of success in p’: must be integer)
n: 200
C-Level: .90
Calculate
Answer: (.0526, .1174)
We are 90% confident that the true proportion of
convertibles is in the interval.
Sec>on9.2,Page193
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Problems
A bank randomly selected 250 checking account
customers and found that 110 of them also had
savings accounts at the same bank.
a.  Construct a 90% confidence interval for the
true proportion of checking account
customers who also have savings accounts.
b.  What is the name of the sampling distribution
used for the confidence interval?
c.  Show that the necessary conditions for a
sampling distribution are satisfied.
d.  What is the standard error of the sampling
distribution?.
Problems,Page206
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Problems
In a sample of 60 randomly selected students, only
22 favored the amount being budgeted for next
year’s intramural and interscholastic sports.
a.  Construct the 99% confidence for the proportion
of students who support the proposed budget
amount.
b.  What is the name of the sampling distribution?
c.  Show that the necessary conditions for a
sampling distribution are satisfied.
d.  What is the standard error of the sampling
distribution?.
Problems,Page206
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ConfidenceIntervalforPropor>on
RequiredSampleSize
ME = Critical Value × Standard Error
pq
= z(α ) ×
n
ME = z(α ) ×
p(1− p)
n
Solving for n :
⎛ z(α ) ⎞ 2
n =⎜
⎟ × p(1− p)
⎝ ME ⎠
€
If we have a good estimate of p, we use it. If
we have no good estimate of p, we estimate p
= .5. This will produce the largest sample for
the given conditions. If p’ turns out to be
different that p, our ME will be less than we
initially required.
Sec>on9.2,Page195
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RequiredSampleSize
Illustra>veProblemTI-83
Consider a manufacturer that purchases bolts from
a supplier who claims the bolts are approximately
5% defective. How large a sample do we need to
estimate the true proportion to be within ± .02 with
90% confidence?
PRGM – SAMPLSIZ – 1:PROPORTION
CONF LEVEL = .90
ME = .02
p EST = .05 (Problem estimate)
Answer: 322
Sec>on9.2,Page195
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Problems
Problems,Page208
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HypothesesTest-OnePropor>on
Illustra>veProblem–TI-83Add-In
Ho: p = .61
Ha: p > .61
p-value
Sampling Distribution for p’
p’
P − Value = P( p' > .6714 given p = .61)
€
p=.61
p’ = 235/350 = .6714
PRGM – NORMDIST – 1
LOWER BOUND = .6714 – UPPERBOUND = 2ND EE99
MEAN = .61 ; σ p = .61(1− .61) /350)
Answer: p-value = .0093 (Reject Ho, We proved that
more than 61% sleep more that 7 or more hours)
'
€
Sec>on9.2,Page195
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HypothesesTest-OnePropor>on
Illustra>veProblem–TI-83BlackBox
Ho: p = .61
Ha: p > .61
STAT – TESTS – 5:1-PropZTest
po: .61
x: 235 (# of successes in p’; must be integer)
n: 350
prop > po
Calculate
Answer: p-value = .0092 (Ho is rejected)
Sec>on9.2,Page196
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Problems
A politician claims that she will receive at least 60%
of the vote in an upcoming election. The results of a
properly designed random sample of 100 voters
showed that 57 of them will vote for her. Is the
sample evidence sufficient to prove that her claim is
false?
a.  Check the conditions for a normal sampling
distribution.
b.  State the hypotheses.
c.  Find the p-value.
d.  State your conclusion.
e.  If you make an error in your conclusion, what
type is it?
f.  Find the mean of the sampling distribution.
g.  Find the standard error of the sampling
distribution.
Problems,Page207
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Problems
a.  Check the conditions for a normal sampling distribution.
b.  State the hypotheses.
c. 
Find the p-value.
d.  State your conclusion
e.  If you make an error in your conclusion, what type is it?
f. 
Find the mean of the sampling distribution.
g.  Find the standard error of the sampling distribution.
Problems,Page207
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