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InferencesOn Two Samples
Overview
• We continue with confidence intervals and hypothesis testing for more advanced models
• Models comparing two means– When the two means are dependent– When the two means are independent
• Models comparing two proportions
Inference about Two Means:Dependent/paired Samples
Learning Objectives
• Distinguish between independent and dependent sampling
• Test hypotheses made regarding matched-pairs data
• Construct and interpret confidence intervals about the population mean difference of matched-pairs data
Two populations
• So far, we have covered a variety of models dealing with one population– The mean parameter for one population– The proportion parameter for one population
• However, there are many real-world applications that need techniques to compare two populations
Examples
• Examples of situations with two populations– We want to test whether a certain treatment
helps or not … the measurements are the “before” measurement and the “after” measurement
– We want to test the effectiveness of Drug A versus Drug B … we give 40 patients Drug A and 40 patients Drug B … the measurements are the Drug A and Drug B responses
Dependent Sample
• In certain cases, the two samples are very closely tied to each other
• A dependent sample is one when each individual in the first sample is directly matched to one individual in the second
• Examples– Before and after measurements (a specific person’s
before and the same person’s after)– Experiments on identical twins (twins matched with
each other
Independent Sample
• On the other extreme, the two samples can be completely independent of each other
• An independent sample is when individuals selected for one sample have no relationship to the individuals selected for the other
• Examples– Fifty samples from one factory compared to fifty
samples from another– Two hundred patients divided at random into two
groups of one hundred
Paired Samples
• The dependent samples are often called matched-pairs
• Matched-pairs is an appropriate term because each observation in sample 1 is matched to exactly one in sample 2– The person before the person after– One twin the other twin– An experiment done on a person’s left eye the
same experiment done on that person’s right eye
Test hypotheses made regarding matched-pairs sample
Analysis of Paired Samples
• The method to analyze matched-pairs is to combine the pair into one measurement– “Before” and “After” measurements – subtract the
before from the after to get a single “change” measurement
– “Twin 1” and “Twin 2” measurements – subtract the 1 from the 2 to get a single “difference between twins” measurement
– “Left eye” and “Right eye” measurements – subtract the left from the right to get a single “difference between eyes” measurement
Compute Difference d
• Specifically, for the before and after example,– d1 = person 1’s after – person 1’s before– d2 = person 2’s after – person 1’s before– d3 = person 3’s after – person 1’s before
• This creates a new random variable d• We would like to reformulate our problem
into a problem involving d (just one variable)
Test for the True Difference μd
• How do our hypotheses translate?– The two means are equal -> the mean difference is
zero -> μd = 0
– The two means are unequal -> the mean difference is non-zero -> μd ≠ 0
• Thus our hypothesis test is– H0: μd = 0
– H1: μd ≠ 0
– The standard deviation σd is unknown
• We know how to do this!
Test for the True Difference
• To solve– H0: μd = 0– H1: μd ≠ 0– The standard deviation σd is unknown
• This is exactly the test of one population mean with the standard deviation being unknown
• This is exactly the subject covered in Unit 8
Assumptions
• In order for this test statistic to be used, the data must meet certain conditions– The sample is obtained using simple random
sampling– The sample data are matched pairs– The differences are normally distributed, or
the sample size (the number of pairs, n) is at least 30
• These are the usual conditions we need to make our Student’s t calculations
Example
• An example … whether our treatment helps or not … helps meaning a higher measurement
• The “Before” and “After” results
Before After Difference
7.2 8.6 1.4
6.6 7.7 1.1
6.5 6.2 – 0.3
5.5 5.9 0.4
5.9 7.7 1.8
Example (continued)
• Hypotheses– H0: μd = 0 … no difference
– H1: μd > 0 … helps
– (We’re only interested in if our treatment makes things better or not)
– α = 0.01
• Calculations– n = 5 (i.e. 5 pairs)– = .88 (mean of the paired-difference)
– sd = .83
d
Example (continued)
• Calculations– n = 5– d = 0.88
– sd = 0.83
• The test statistic is
• This has a Student’s t-distribution with 4 degrees of freedom
3625830
08800
./.
.
n/s
dt d
Example (continued)
• Use the Student’s t-distribution with 4 degrees of freedom
• The right-tailed α = 0.01 critical value is 3.75 (i.e. t0.01;4 d.f. = 3.75)
• 2.36 is less than 3.75 (the classical method)• Thus we do not reject the null hypothesis• There is insufficient evidence to conclude that our
method significantly improves the situation• We could also have used the P-Value method. P value
is 0.039 (note: tcdf(2.36, E99, 4) = 0.039)
Example (continued)
• Matched-pairs tests have the same various versions of hypothesis tests– Two-tailed tests– Left-tailed tests (the alternatively hypothesis
that the first mean is less than the second)– Right-tailed tests (the alternatively hypothesis
that the first mean is greater than the second)• Each can be solved using the Student’s t
Classical and P-value Approaches
• Each of the types of tests can be solved using either the classical or the P-value approach
Summary of the Method
• A summary of the method– For each matched pair, subtract the first observation
from the second– This results in one data item per subject with the data
items independent of each other– Test that the mean of these differences is equal to 0
• Conclusions
– Do not reject that μd = 0
– Reject that μd = 0 ... Reject that the two populations have the same mean
Construct and interpret confidence intervals about the population mean
difference of matched-pairs data
Confidence Interval for the Paired Difference
• We’ve turned the matched-pairs problem in one for a single variable’s mean / unknown standard deviation– We just did hypothesis tests– We can use the techniques taught in Unit 7
(again, single variable’s mean / unknown standard deviation) to construct confidence intervals
• The idea – the processes (but maybe not the specific calculations) are very similar for all the different models
Confidence Interval for the Paired Difference
• Confidence intervals are of the form
Point estimate ± margin of error• This is precisely an application of our results for a
population mean / unknown standard deviation– The point estimate
d
and the margin of error
for a two-tailed test
ns
t d/ 2
Confidence Interval for the Paired Difference
• Thus a (1 – α) • 100% confidence interval for the difference of two means, in the matched-pair case, is
where tα/2 is the critical value of the Student’st-distribution with n – 1 degrees of freedom
n
std d
/ 2
Example
Salt-free diets are often prescribed for people with high blood pressure. The following data was obtained from an experiment designed to estimate the reduction in diastolic blood pressure as a result of following a salt-free diet for two weeks. Assume diastolic readings to be normally distributed.
Find a 99% confidence interval for the mean reduction
Before 93 106 87 92 102 95 88 110
After 92 102 89 92 101 96 88 105
Difference 1 4 -2 0 1 -1 0 5
Example (continued)
1. Population Parameter of InterestThe mean reduction (difference) in diastolic blood pressure
2. The Confidence Interval Criteria
a. Assumptions: Both sample populations are assumed normal
b. Test statistic: t with df = 8 1 = 7
c. Confidence level: 1 = 0.99
39.2 and ,0.1 ,8 dsdn
3. Sample evidence
Sample information:
Example
4. The Confidence Interval
a. Confidence coefficients:Two-tailed situation, /2 = 0.005t(df, /2) = t(7, 0.005) = 3.50
b. Maximum error:
c. Confidence limits:
5. The Results1.957 to 3.957 is the 99% confidence interval estimate for the amount of reduction of diastolic blood pressure, d..
95728
392503 .)
.)(.(E
957.3 to957.1
957.20.1 to957.20.1
to
EdEd
Summary
• Two sets of data are dependent, or matched-pairs, when each observation in one is matched directly with one observation in the other
• In this case, the differences of observation values should be used
• The hypothesis test and confidence interval for the difference is a “mean with unknown standard deviation” problem, one which we already know how to solve
Inference about Two Means:Independent Samples
Learning Objectives
• Test hypotheses regarding the difference of two independent means
• Construct and interpret confidence intervals regarding the difference of two independent means
Independent Samples
• Two samples are independent if the values in one have no relation to the values in the other
• Examples of not independent– Data from male students versus data from
business majors (an overlap in populations)– The mean amount of rain, per day, reported in
two weather stations in neighboring towns (likely to rain in both places)
Independent Samples
• A typical example of an independent samples test is to test whether a new drug, Drug N, lowers cholesterol levels more than the current drug, Drug C
• A group of 100 patients could be chosen– The group could be divided into two groups of
50 using a random method– If we use a random method (such as a simple
random sample of 50 out of the 100 patients), then the two groups would be independent
Test of Two Independent Samples
• The test of two independent samples is very similar, in process, to the test of a single population mean
• The only major difference is that a different test statistic is used
• We will discuss the new test statistic through an analogy with the hypothesis test of one mean
Test hypotheses regarding the difference of two independent
means
Test Statistic for a Single Mean
• For the test of one mean, we have the variables– The hypothesized mean (μ)– The sample size (n)– The sample mean (x)– The sample standard deviation (s)
• We expect that x would be close to μ
Test statistic for the Difference of Two Means
• In the test of two means, we have two values for each variable – one for each of the two samples
– The two hypothesized means μ1 and μ2
– The two sample sizes n1 and n2
– The two sample means x1 and x2
– The two sample standard deviations s1 and s2
• We expect that x1 – x2 would be close to μ1 – μ2
Standard Error of the Test Statistic for a Single Mean
• For the test of one mean, to measure the deviation from the null hypothesis, it is logical to take
x – μ
which has a standard deviation/standard error of approximately
ns2
Standard Error of the Test Statistic for the Difference of Two Means
• For the test of two means, to measure the deviation from the null hypothesis, it is logical to take
(x1 – x2) – (μ1 – μ2)
which has a standard deviation/standard error of approximately
2
22
1
21
ns
ns
t -Test Statistic for a Single Mean
• For the test of one mean, under certain appropriate conditions, the difference
x – μ
is Student’s t with mean 0, and the test statistic
has Student’s t-distribution with n – 1 degrees of freedom
ns
xt
2
t - Test Statistic for the Difference of Two Means
• Thus for the test of two means, under certain appropriate conditions, the difference
(x1 – x2) – (μ1 – μ2)
is approximately Student’s t with mean 0, and the test statistic
has an approximate Student’s t-distribution
2
22
1
21
2121
ns
ns
)()xx(t
Distribution of the t-statistic
• This is Welch’s approximation, that
has approximately a Student’s t-distribution• The degrees of freedom is the smaller of
n1 – 1 and n2 – 1
2
22
1
21
2121
ns
ns
)()xx(t
Note: Some computer or calculator calculates the degrees of freedom for this t test statistic with a somewhat complicated formula. But, we’ll use the smallerof n1 – 1 and n2 – 1 as the degrees of freedom.
A Special Case
• For the particular case where be believe that the two population means are equal, or μ1 = μ2, and the two sample sizes are equal, or n1 = n2, then the test statistic becomes
with n – 1 degrees of freedom, where n = n1 = n2
n
ss
)xx(t
22
21
21
General Test Procedure
• Now for the overall structure of the test– Set up the hypotheses– Select the level of significance α– Compute the test statistic– Compare the test statistic with the appropriate
critical values– Reach a do not reject or reject the null
hypothesis conclusion
Assumptions
• In order for this method to be used, the data must meet certain conditions– Both samples are obtained using simple
random sampling– The samples are independent– The populations are normally distributed, or
the sample sizes are large (both n1 and n2 are at least 30)
• These are the usual conditions we need to make our Student’s t calculations
State Hypotheses & level of significance
• State our two-tailed, left-tailed, or right-tailed hypotheses
• State our level of significance α, often 0.10, 0.05, or 0.01
Compute the Test Statistic
• Compute the test statistic
and the degrees of freedom, the smaller ofn1 – 1 and n2 – 1
• Compute the critical values (for the two-tailed, left-tailed, or right-tailed test
2
22
1
21
2121
ns
ns
)()xx(t
Make a Statistical Decision
• Each of the types of tests can be solved using either the classical or the P-value approach
• Based on either of these methods, do not reject or reject the null hypothesis
Example
• We have two independent samples– The first sample of n = 40 items has a sample mean of
7.8 and a sample standard deviation of 3.3– The second sample of n = 50 items has a sample
mean of 11.6 and a sample standard deviation of 2.6– We believe that the mean of the second population is
exactly 4.0 larger than the mean of the first population– We use a level of significance α = .05
• We test versus 4211
:H4210
:H
Example (continued)
• The test statistic is
• This has a Student’s t-distribution with 39 degrees of freedom
• The two-tailed critical value is -2.02, so we do not reject the null hypothesis (notice: invT(.025,39) = -2.02 or use a t-table)
• Or, compute the p-value which is 0.093 greater than 0.05 level of significance. (Notice that: 2*tcdf(-E99,-1.72,39) = 0.093)
• We do not have sufficient evidence to state that the deviation from 4.0 is significant
721
50
62
40
33
049128722
2
2
2
1
2
1
2121 ...
).()..(
n
s
n
s
)()xx(t
Construct and interpret confidence intervals regarding the difference of
two independent means
Confidence Interval of
• Confidence intervals are of the form
Point estimate ± margin of error• We can compare our confidence interval with the
test statistic from our hypothesis test
– The point estimate is x1 – x2
– We use the denominator of the test statistic as the standard error
– We use critical values from the Student’s t
Confidence Interval of
• Thus (1- confidence interval is
Point estimate ± margin of error
2
22
1
21
221 ns
ns
t)xx( /
Standard errorPoint estimate
where t has the degrees of freedom that is the smaller of n1-1 and n2-1 .
Example
A recent study reported the longest average workweeks for non-supervisory employees in private industry to be chef and construction
Industry n Average Hours/Week Standard Deviation
Chef 18 48.2 6.7
Construction 12 44.1 2.3
Find a 95% confidence interval for the difference in mean length of workweek between chef and construction. Assume normality for the sampled populations and that the samples were selected randomly.
Example
1. Parameter of interestThe difference between the mean hours/week for chefs and the mean hours/week for construction workers, 1 - 2
2. The Confidence Interval Criteria
a. Assumptions: Both populations are assumed normal and the samples were random and independently selected
b. Test statistic: t with df = 11;the smaller of n1 1 = 18 1 = 17 or n2 1 = 12 1 = 11
c. Confidence level: 1 = 0.953. The Sample Evidence
Sample information given in the tablePoint estimate for 1 - 2:
1414424821
...xx
Example4. The Confidence Interval
a. Confidence coefficients: t0.025, 11d.f.= 2.20
b. Margin of error:
c. Confidence limits:
4.1 – 3.77 = 0.33 to 4.1 + 3.77 = 7.875. The Results
0.33 to 7.87 is a 95% confidence interval for the difference in mean hours/week for chefs and construction workers. ( It also means that there is a significant difference between the mean hours/week for chefs and the mean hours/week for construction workers at 0.05 level of significance, since the interval does not contain zero.)
77312
32
18
76202
22
.)..
)(.(E
Summary
• Two sets of data are independent when observations in one have no affect on observations in the other
• In this case, the differences of the two means should be used in a Student’s t-test
• The overall process, other than the formula for the standard error, are the general hypothesis test and confidence intervals process
Inference aboutTwo Population Proportions
Learning Objectives
• Test hypotheses regarding two population proportions
• Construct and interpret confidence intervals for the difference between two population proportions
Test hypotheses regarding two population proportions
Inference about Two Proportions
• This progression should not be a surprise
• One mean and one proportion– Unit 7 – confidence intervals– Unit 8 – hypothesis tests
• Two means– Unit 9 - hypothesis tests and confidence
intervals
• Now for two proportions …
Examples
• We now compare two proportions, testing whether they are the same or not
• Examples– The proportion of women (population one) who have
a certain trait versus the proportion of men (population two) who have that same trait
– The proportion of white sheep (population one) who have a certain characteristic versus the proportion of black sheep (population two) who have that same characteristic
Two Population Proportions
• The test of two populations proportions is very similar, in process, to the test of one population proportion and the test of two population means
• The only major difference is that a different test statistic is used
• We will discuss the new test statistic through an analogy with the hypothesis test of one proportion
Test of One Proportion
• For the test of one proportion, we had the variables of– The hypothesized population proportion (p0)
– The sample size (n)– The number with the certain characteristic (x)– The sample proportion ( )
• We expect that should be close to p0
n/xp̂ p̂
Test of Two Proportions
• In the test of two proportions, we have two values for each variable – one for each of the two samples– The two hypothesized proportions (p1 and p2)
– The two sample sizes (n1 and n2)
– The two numbers with the certain characteristic (x1 and x2)
– The two sample proportions ( and )
• We expect that should be close to p1 – p2
111 n/xp̂ 222 n/xp̂
21 p̂p̂
Test Statistic of One Proportion
• For the test of one proportion, to measure the deviation from the null hypothesis, we took
which has a standard deviation of
n)p(p 00 1
0pp̂
Test Statistic of Two Proportions
• For the test of two proportions, to measure the deviation from the null hypothesis, it is logical to take
which has a standard deviation of
2
22
1
11 11n
)p(pn
)p(p
)pp()p̂p̂( 2121
Test Statistic for One Proportion
• For the test of one proportion, under certain appropriate conditions, the difference
is approximately normal with mean 0, and the test statistic
has an approximate standard normal distribution
n)p(p
pp̂z
00
01
0pp̂
Test Statistic for Two Proportions
• Thus for the test of two proportions, under certain appropriate conditions, the difference
is approximately normal with mean 0, and the test statistic
has an approximate standard normal distribution
2
22
1
11
212111n
)p(pn
)p(p)pp()p̂p̂(
z
)pp()p̂p̂( 2121
Test Statistic for Equal Proportions
• For the particular case where we believe that the two population proportions are equal, or p1 = p2 (i.e.
p1 – p2 = 0). Thus
and
21
2121
11
n
)p̂(p̂
n
)p̂(p̂
)pp()p̂p̂(z
cccc
212121 p̂p̂)pp()p̂p̂(
21
21
111
nn)p̂(p̂
p̂p̂
cc
Here, since two population proportions are the same under the null hypothesis, we use , an estimated common proportion for both p1 and p2, which is computed by combining two samples together to calculate an estimated common sample proportion. That is,
cp̂
21
21
nn
xxp̂c
General Test Procedure
• Now for the overall structure of the test– Set up the hypotheses– Select the level of significance α– Compute the test statistic– Compare the test statistic with the appropriate
critical values– Reach a do not reject or reject the null
hypothesis conclusion
Assumptions
• In order for this method to be used, the data must meet certain conditions– Both samples are obtained independently
using simple random sampling– Each sample size is large
• These are the usual conditions we need to make our test of proportions calculations
Hypotheses and Level of Significance
• State our two-tailed, left-tailed, or right-tailed hypotheses
• State our level of significance α, often 0.10, 0.05, or 0.01
Test Statistic and Critical Values
• Compute the test statistic
which has an approximate standard normal distribution
• Compute the critical values (for the two-tailed, left-tailed, or right-tailed test)
21
2121
111
nn)P̂(P̂
)pp()p̂p̂(z
cc
Make Statistical Decision• Each of the types of tests can be solved using
either the classical or the P-value approach
• Based on either of these two methods, do not reject the null hypothesis
Example
• We have two independent samples– 55 out of a random sample of 100 students at one
university are commuters– 80 out of a random sample of 200 students at another
university are commuters– We wish to know of these two proportions are equal– We use a level of significance α = .05
• Both samples sizes are large so our method can be used
Example (continued)
• The test statistic is
Notice that
• The critical values for a two-tailed test using the normal distribution are ± 1.96, thus we reject the null hypothesis
• Or, we calculate P-value which is 0.014 less than the 0.05 level of significance. ( Notice: 2*normalcdf(2.46,E99) = 0.014)
• We conclude that the two proportions are significantly different
462
200
1
100
14501450
400550
111
21
2121 .
).(.
..
nn)p̂(p̂
)pp()p̂p̂(z
cc
450200100
8055.p̂
c
Confidence Interval of p1 – p2
• Thus confidence intervals are
Point estimate ± margin of error
cp̂
2
22
1
11221
11n
)p̂(p̂n
)p̂(p̂z)p̂p̂( /
Standard errorPoint estimate
Here, for calculating the standard error, we use separate estimates of the population proportions, instead of the common estimate21
p̂,p̂
Example
A consumer group compared the reliability of two similar microcomputers from two different manufacturers. The proportion requiring service within the first year after purchase was determined for samples from each of two manufacturers.
Find a 98% confidence interval for p1 p2, the difference in proportions needing service
Manufacturer Sample Size Proportion Needing Service
1 200 0.15
2 250 0.09
Example (continued)
1. Population Parameter of Interest : The difference between the proportion of microcomputers needing service for manufacturer 1 and the proportion of microcomputers needing service for manufacturer 2, that is, p1- p2
2. Point estimate:
3. Confidence coefficients:z(/2) = z(0.01) = 2.33
06009015021
...p̂p̂
0 z
98.0 01.001.0
33.2
z(0.01
Example (continued)
• Margin of error:
• Confidence limits: 0.06 – 0.0724 = -0.0124 to 0.06 + 0.0724 = 0.1324
Results 0.0124 to 0.1324 is a 98% confidence interval for the difference in
proportions
07240250
910090
200
850150332 .
).)(.().)(.(.E
Summary
• We can compare proportions from two independent samples
• We use a formula with the combined sample sizes and proportions for the standard error
• The overall process, other than the formula for the standard error, are the general hypothesis test and confidence intervals process
Inferenceson Two Samples
Summary
Summary
• The process of hypothesis testing is very similar across the testing of different parameters
• The major steps in hypothesis testing are– Formulate the appropriate null and alternative
hypotheses– Calculate the test statistic– Determine the appropriate critical value or
values– Reach the reject / do not reject conclusions
Tests for Means and Proportions
• Similarities in hypothesis test processes
Parameter Mean (one population)
Two Means(Independent)
Two Means(Dependent)
TwoProportions
H0: μ = μ0 μ1 = μ2 μ1 = μ2 p1 = p2
(2-tailed) H1: μ ≠ μ0 μ1 ≠ μ2 μ1 ≠ μ2 p1 ≠ p2
(L-tailed) H1: μ < μ0 μ1 < μ2 μ1 < μ2 p1 < p2
(R-tailed) H1: μ > μ0 μ1 > μ2 μ1 > μ2 p1 > p2
Test statistic Difference Difference Difference Difference
Critical value Normal Normal Student t Normal
Summary
• We can test whether sample data from two different samples supports a hypothesis claim about a population mean or proportion
• For two population means, there are two cases– Dependent (or matched-pair) samples– Independent samples
• All of these tests follow very similar processes, differing only in their test statistics and the distributions for their critical values
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