Transcript Critical Values for 95% Confidence Interval

Inference about Mean
(σ Unknown)
When σ is known, the sampling distribution for a sample
mean is normal if conditions are satisfied.
For many years, it was thought that when σ was
unknown, this was still the case. However, because of
the increased variability introduced by not knowing σ,
the sampling distribution for a sample mean with
unknown σ is not normal.
This was discovered by W. S. Gosset, an Irish Brewery
quality control inspector in 1908. He also discovered
that when σ is unknown, we can still do inference using a
sampling distribution model he discovered, the tdistribution.
Section 9.1, Page 184
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t-Distribution
•The t-Distribution is unimodal and symmetric
•It has more area in that tails and is flatter in the
middle than the normal distribution
•It is a family of distributions, with a different curve
for each μ, σ, and df (degrees of freedom) df=n-1,
where n is the sample size.
Critical Values for 95% Confidence Interval
Known σ: 1.960 from Normal Distribution
Unknown σ with 3 df: 3.18 from t-Distribution
Unknown σ with 10 df: 2.23 from t-Distribution
Unknown σ with 1000 df: 1.962 from t-Distribution
Section 9.1, Page 184
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t-Distribution
When to Use It
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
Section 9.1, Page 184
3
Confidence Interval
(Unknown σ)
Confidence Interval =
Sample Mean ± Margin of Error
x
± Critical Value × Standard Error ofx
s
x  t(df , ) 
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)
Section 9.1, Page 186
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Confidence Interval (Unknown σ)
Illustrative Problem- TI 83 Add-in Programs
C.I. =Sample Mean ± Critical Value * Standard Error
PRGM – CRITVAL – ENTER
2: T-DIST
CONF LEVEL = .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)
Section 9.1, Page 186
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Confidence Interval (Unknown σ)
Illustrative Problem- TI 83 Black Box Program
STAT – TESTS – 8:Tinterval
Inpt: Stats
x :6.87
Sx: 1.76
n: 20
C-Level: .95
Calculate
Answer: (6.05, 7.69)
Section 9.1, Page 186
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Problems
a.
b.
c.
d.
e.
Find the 98% confidence interval.
Find the critical value
Find the margin of error.
Find the standard error.
What assumption must we make about the the
population to have a t-sampling distribution.
f. What are the proper words to describe the
confidence interval?
g. If you wanted to have a margin of error of one
minute and the 98% confidence interval for this
data, how large must the sample be?
Problems, Page 205
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Problems
a.
b.
c.
d.
e.
Find the 93% confidence interval.
Find the critical value
Find the margin of error.
Find the standard error.
What assumption must we make about the population
to have a t-sampling distribution.
f. What are the proper words to describe the confidence
interval?
g. If you wanted to have a margin of error of 25 lbs. and a
confidence level of 93% for this data, how many
students would you have to monitor?
Section 9.1, Page 205
8
Hypotheses Test (Unknown σ)
Illustrative Problem- TI-83 Add-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
x  4.9
x  5.1
P Value  P(x  5.1 given x  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.)

Section 9.1, Page 187
9
Hypotheses Test (Unknown σ)
Illustrative Problem- TI-83 Black Box
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
0 :4.9
x : 5.1
Sx :1.17
n  22
 : 0
Calculate
Answer : p  value  .2158
There is not sufficient evidence to support
the EPA claim.

Section 9.1, Page 187
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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, Page 209
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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, Page 205
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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, Page 206
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Inference about Proportions
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.
Section 9.2, Page 192
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Sampling Distribution for a
Proportion
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(1-p’) = #
failures in sample must be > 5.
3.The sample consists of less than 10% of the
population.
Section 9.2, Page 192
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Confidence Interval for a
Proportion, 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( ) 
p'q'
n
where q’ = proportion of failures = 1-p’ and n is the
sample size.
Section 9.2, Page 193
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Confidence Interval for Proportion
Illustrative Example TI-83 Add-in Programs
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)
Section 9.2, Page 193
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Confidence Interval for Proportion
Illustrative Example TI-83 Black Box Program
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.
Section 9.2, Page 193
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Problems
Problems, Page 206
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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, Page 206
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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, Page 206
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
Confidence Interval for Proportion
Required Sample Size
ME = Critical Value × Standard Error
pq
 z( ) 
n
ME  z( ) 
p(1 p)
n
Solving for n :
z( ) 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.
Section 9.2, Page 195
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Required Sample Size
Illustrative Problem TI-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
Section 9.2, Page 195
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Problems
Problems, Page 208
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Hypotheses Test - One Proportion
Illustrative Problem – TI-83 Add-In
Ho: p = .61
Ha: p > .61
Sampling Distribution for p’
p-value
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 ;  p  .61(1 .61) /350)
Answer: p-value = .0093 (Reject Ho, We proved that
more than 61% sleep more that 7 or more hours)
'

Section 9.2, Page 195
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Hypotheses Test - One Proportion
Illustrative Problem – TI-83 Black Box
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)
Section 9.2, Page 196
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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, Page 207
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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, Page 207
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