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Section 11.1 – Inferences from Matched Pairs
A sampling method is independent when the individuals selected for one sample do not dictate which
individuals are to be in the second sample.
A sampling method is dependent (matched pairs) when the individuals selected to be in one sample
are used to determine the individuals in the second sample.
Requirements:
1. Sample data consist of matched pairs
2. Samples are simple random samples
3. The number of matched pairs is greater than 30 or differences are normally distributed.
d = individual difference between two values in a matched pair
d = mean value of the differences d for the paired sample data (average of “x – y” values)
sd = standard deviation of the differences d for the paired sample data
n = number of pairs of data
μd = mean value of the differences d for the population of paired data
Hypotheses:
Ho :
H1 :
μd = 0
μd ≠ 0 or
μd > 0 or
μd < 0
CRITICAL VALUES come from the “t-chart” (Table VI)
Degrees of Freedom = n –1
(you will have to put the appropriate sign on the critical value, +, –, or ±)




d 

TEST STATISTIC t 
and P-VALUES from the calculator

 sd  

 

 n 

T-Test (#2)
Enter “before” data in L1.
Enter “after” data in L2.
Make L3 = L1 – L2
When you run the T-test, choose “data” and “L3”
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Example: In low-speed crash test of five BMW cars, the repair costs were computed for a factory
authorized repair center and an independent repair center. Is there sufficient evidence at
α = 0.01 to support the claim that the independent center has lower repair costs?
The results were:
Authorized repair center
Independent repair center
I.
$797
$523
$571
$488
$904
$875
$1147 $418
$911 $297
Ho:
H1:
II.
α =
Critical Value(s) =
III.
T.S. = t =
IV.
(picture)
V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
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When you calculate a confidence interval in this chapter (a confidence interval of the difference
between 2 parameters), there are 3 possible outcomes:
 , 
means that the difference between the two values is negative.
There is a significant difference between p1 and p2.
 ,  
 ,  
means that the difference between the two values is positive.
There is a significant difference between p1 and p2.
means that there is a possibility that the difference is 0.
There is NOT a significant between p1 and p2.
*If the confidence interval contains 0, we say there is NO significant difference between
the values.
*If the confidence interval does not contain 0, we say there IS a significant difference
between the values.
Confidence Interval Estimate for the population mean difference μd :
d  E
, dE

where E = t
sd
2
n
(T-Interval (#8) on calc)
and Degrees of freedom = n – 1
Example: In an experiment conducted online at the University of Mississippi, participants are asked to
react to a stimulus. In one experiment, the participant must press a key upon seeing a blue
screen. Their reaction time is measured (in seconds). The same person is asked to press a
key upon seeing a red screen, again with reaction time measured. The results for 6
randomly sampled participants are as follows. Construct a 95% confidence interval about
the population mean difference in response time. Is there a significant difference in the
response times?
Blue
0.582
0.481
0.841
0.267
0.685
0.450
Red
0.408
0.407
0.512
0.402
0.456
0.533
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Section 11.2 – Inferences About Two Means – Independent Samples
Requirements:
1. The two samples are independent
2. Both samples are simple random samples.
3. Both sample sizes are greater than 30 or both samples come from
populations that are normally distributed.
Hypotheses:
Ho :
H1 :
μ1 = μ2
μ1 ≠ μ2 or μ1 > μ2 or μ1 < μ2
CRITICAL VALUES come from the “t-chart” (Table VI)
Degrees of Freedom = the smaller of n1 – 1 and n2 – 1
“NOT POOLED”
(you will have to put the appropriate sign on the critical value, +, –, or ±)





( x1  x2 )  ( 1   2 ) 
TEST STATISTIC  t 
 and P-VALUES from the calculator.
2
2
s
s


1
 2


n
n
1
2


2-SampT-Test (#4)
Example: For 1,657 men (ages 65-74), the mean weight is 164 lbs and the standard deviation is 27 lbs.
For 804 men (ages 25-34), the mean weight is 176 lbs and the standard deviation is 35 lbs.
Test the claim that the older men come from a population with a mean weight that is less
than the mean weight for men in the younger age bracket. Use α = 0.01.
I.
Ho:
H1:
II.
α =
Critical Value(s) =
4
III.
T.S. = t =
IV.
(picture)
V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
Example: A researcher wishes to determine whether the salaries of professional nurses employed by
private hospitals are higher than those of nurses employed by government-owned hospitals.
At α = 0.01, can the researcher conclude that the private hospitals pay more than the
government hospitals?
Private
Government
x1 = $26,800
x 2 = $25,400
s1 = $600
s2 = $450
n1 = 50
n2 = 68
I.
Ho:
H1:
II.
α =
Critical Value(s) =
III.
T.S. = t =
IV.
(picture)
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V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
Confidence Interval Estimate for μ1 – μ2:
 x1  x2   E
,
x1  x2   E 
(2-SampT-Int (#0) on the calculator, NOT POOLED)
where E = t
2
s12 s 22

n1 n 2
Example: A randomized trial tested the effectiveness of diets on adults. Among 40 subjects using the
Weight Watchers diet, the mean weight loss after one year was 3 lb with a standard
deviation of 4.9 lb. Among 40 subjects using the Atkins diet, the mean weight loss after one
year was 2.1 lb with a standard deviation of 4.8 lb. Construct a 95% confidence interval
estimate of the difference between the population means. Does there appear to be a
significant difference between the effectiveness of the two diets?
Section 11.3 – Inferences About Two Population Proportions
Requirements:
1. Proportions are from two independent simple random samples.
2. For both samples, npq ≥ 10.
n1 = size of sample 1
x1 = number of successes in sample 1
p1 = proportion for population 1
x
pˆ 1  1
(sample proportion)
n1
(complement of the sample proportion)
qˆ1  1  pˆ 1
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(similar definitions for p2, n2, x2, etc.)
Hypotheses:
Ho :
H1 :
p1 = p2
p1 ≠ p2 or p1 > p2 or p1 < p2
CRITICAL VALUES come from the little “z-chart” in the formula packet




( pˆ 1  pˆ 2 )  ( p1  p2 ) 

TEST STATISTIC  z 
 and P-VALUE come from the calculator
p
q
p
q





n
n
1
2


Note:
p1 – p2 = 0 and p is the pooled estimate of p1 and p2. p 
x1  x2
and q  1  p
n1  n2
2-PropZ-Test (#6)
Example: When conducting tests of auto parts stores, the Arizona Department of Weights and
Measures conducted 100 inspections of AutoZone stores and found that 63% of those
inspections failed. Among 37 inspections at NAPA Auto Parts stores, 81% failed. Use a
0.05 significance level to determine whether there is a significant difference between those
two rates of failures.
I.
Ho:
H1:
II.
α =
Critical Value(s) =
III.
T.S. = z =
7
IV.
(picture)
V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
Example: In a study of red/green color blindness, 500 men and 2,100 women were randomly selected
and tested. Among the men, 45 demonstrated color blindness. Among the women, 6
demonstrated color blindness. Is there sufficient evidence at α = 0.05 to support the claim
that men have a higher rate of red/green color blindness than women?
I.
Ho:
H1:
II.
α =
Critical Value(s) =
III.
T.S. = z =
IV.
(picture)
V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
Confidence Interval Estimate for p1 – p2 :
 ( pˆ1  pˆ 2 )  E ,
( pˆ1  pˆ 2 )  E 
(2-PropZ-Int (#B) on the calculator)
where E = z
2
pˆ 1 qˆ1 pˆ 2 qˆ 2

n1
n2
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Example: A study was done on the effectiveness of hospital gloves. Among 240 vinyl gloves, 63%
leaked viruses, and among 240 latex gloves, 7% leaked viruses. Construct at 99%
confidence interval estimate of the difference between the two leakage rates. Based on the
result, does there appear to be a difference between the two rates of leakage?
Example: Among 5000 items of randomly selected baggage handled by American Airlines, 22 were
lost. Among 4000 items of randomly selected baggage handled by Delta Airlines, 15 were
lost. Use the sample data to construct a 95% confidence interval estimate of the difference
between the two rates of lost baggage. Based on the result, does there appear to be a
difference between the two rates of lost luggage?
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Section 11.4 – Inference for Two Population Standard Deviations
Requirements:
1. Two independent simple random samples
2. The two populations are each normally distributed
s1 = sample standard deviation from population 1
n1 = sample size from population 1
 1 = standard deviation of population 1
(similar definitions for population 2)
Hypotheses:
Ho :  1 =  2
H1 :  1 ≠  2
or
 1 >  2 or
1 <  2
For two normally distributed populations with equal variances, the sampling distribution of the test
statistic is the F-Distribution. The F distribution is skewed right, and has no negative values.
CRITICAL VALUES come from the F-table (Table VIII)
1. Go to α for one tail, α/2 for two tails
2. Numerator degrees of freedom = n1 – 1
3. Denominator degrees of freedom = n2 – 1
F , n1 1, n2 1
Critical Value for a Right-Tailed Test
1
Critical Value for a Left-Tailed Test
Critical Values for a Two-Tailed Test
F , n2 1, n1 1
Split alpha in half first, then find
the right and left critical values

s2 
TEST STATISTIC  F  12  and P-VALUE come from the calculator
s2 

2-SampF-Test (#D)
We will not do confidence intervals for F-tests!!
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Example:
Find the Critical Value for a right-tailed test where α = 0.025, n1 = 16, and n2 = 13.
Example:
Find the Critical Value for a left-tailed test where α = 0.10, n1 = 8, and n2 = 16.
Example:
Find the Critical Value for a two-tailed test where α = 0.10, n1 = 21, and n2 = 8.
Example: An experiment was conducted to test the effects of alcohol. The errors were recorded in a
test of visual and motor skills for a treatment group of people who drank ethanol and
another group given a placebo. The results are below. Use a 0.05 significance level to test
the claim that the treatment group has scores that vary more than the scores of the placebo
group.
Treatment
Placebo
n = 22
n = 22
x = 4.2
x = 1.71
s = 2.2
s = 0.72
I.
Ho:
H1:
II.
α =
Critical Value(s) =
III.
T.S. = F =
IV.
(picture)
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V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
Example: Researchers wanted to investigate the relationship between birth weight and IQ.
Researchers recorded the birth weight and the IQ score at age 8. The results follow.
Using a 0.05 significance level, test the claim that babies with extremely low birth weights
and babies with normal birth weights have different amounts of variation in their IQ scores.
Extremely low birth weight
(less than 2.2 lb)
n = 125
x = 95.5
s = 16
I.
Normal birth weight
n = 60
x = 104.9
s = 14.1
Ho:
H1:
II.
α =
Critical Value(s) =
III.
T.S. = F =
IV.
(picture)
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V.
P-value =
VI.
Reject Ho or Fail to Reject Ho
VII.
(conclusion)
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