Transcript Two Sample Problems

Chapter 19: Two-Sample Problems
STAT 1450
19.0 Two-Sample Problems
Connecting Chapter 18 to our
Current Knowledge of Statistics
Population
Parameter
Point Estimate
Confidence
Interval
μ (σ known)
𝑥
𝑥 ± 𝑧∗
𝜎
𝑛
𝑧=
𝑥 − 𝜇0
𝜎 𝑛
μ (σ unknown)
s
𝑥 ± 𝑡∗
𝑠
𝑛
𝑡=
𝑥 − 𝜇0
𝑠 𝑛
Test Statistic
▸ Remember that these formulas are only valid when appropriate simple
conditions apply!
19.0 Two-Sample Problems
Connecting Chapter 18 to our
Current Knowledge of Statistics
▸ Matched pairs were covered at the end of Chapter 18.
A common situation requiring matched pairs is when before-and-after
measurements are taken on individual subjects.
▸ Example: Prices for a random sample of tickets to a 2008 Katy Perry
concert were compared with the ticket prices (for the same seats) to
her 2013 concert..
 The data could be consolidated into 1 column of differences in ticket prices.
 A test of significance, or, a confidence interval would then occur for
“1 sample of data.”
19.1 The Two-Sample Problem
The Two-Sample Problems
▸ Two-sample problems require us to compare:
 the response to two treatments
- or  the characteristics of two populations.
▸ We have a separate sample from each treatment or population.
19.1 The Two-Sample Problem
The Two-Sample Problem
▸ Example: Suppose a random samples of ticket prices for concerts by
the Rolling Stones was obtained. For comparison purposes another
random sample of Coldplay ticket prices was obtained. Note these are
not necessarily the same seats or even the same venues.
▸ Question: Are these samples more likely to be independent or
dependent?
a)
Independent
b)
Dependent
c)
Not sure
19.1 The Two-Sample Problem
The Two-Sample Problem
▸ Example: Suppose a random samples of ticket prices for concerts by
the Rolling Stones was obtained. For comparison purposes another
random sample of Coldplay ticket prices was obtained. Note these are
not necessarily the same seats or even the same venues.
▸ Question: Are these samples more likely to be independent or
dependent?
a)
Independent
b)
Dependent
c)
Not sure
19.1 The Two-Sample Problem
Two-Sample Problems
▸ The end of Chapter 18 described inference procedures for the mean
difference in two measurements on one group of subjects (e.g., pulse
rates for 12 students before-and-after listening to music).
▸ Given our answer from above, and the likelihood that each sample has
different sample sizes, variances, etc… Chapter 19 focuses on the
difference in means for 2 different groups.
Population
Parameter
Point Estimate
𝜇1 − 𝜇2
𝑥1 − 𝑥2
Confidence
Interval
Test Statistic
19.2 Comparing Two Population Means
Sampling Distribution of Two Sample Means
▸ Recall that for a single sample mean 𝑥
 The standard deviation of a statistic is estimated from data the
result is called the standard error of the statistic.
 The standard error of 𝑥 is 𝑠
𝑛
.
Inference in the two-sample problem will
require the standard error of the difference
of two sample means 𝒙𝟏 − 𝒙𝟐 .
x12
19.2 Comparing Two Population Means
Sampling Distribution of Two Sample Means
▸ The following table stems from the above comment on standard error
and statistical theory.
Variable
Parameter Point Estimate
Population
Standard Deviation
Standard Error
x1
m1
𝑥1
s1
𝑠1
𝑛1
x2
m2
𝑥2
s2
𝑠2
𝑛2
𝑥1 − 𝑥2
𝜎12 𝜎22
+
𝑛1 𝑛2
𝑠12 𝑠22
+
𝑛1 𝑛2
Diff =
x1 - x2
m1 - m2
19.2 Comparing Two Population Means
Example: SSHA Scores
▸ The Survey of Study Habits and Attitudes (SSHA) is a psychological
test designed to measure various academic behaviors (motivation,
study habits, attitudes, etc…) of college students. Scores on the SSHA
range from 0 to 200. The data for random samples 17 women
(**the outlier from the original data set was removed**) and 20 men
yielded the following summary statistics.
▸ Is there a difference in SSHA performance based upon gender?
19.2 Comparing Two Population Means
Example: SSHA Scores
▸ Summary statistics for the two groups are below:
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women**
139.588
20.363
17
Men
122.5
32.132
20
 There is a difference in these two groups.
The women’s average was 17.5 points > than the men’s average.
19.2 Comparing Two Population Means
Example: SSHA Scores
▸ Summary statistics for the two groups are below:
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women**
139.588
20.363
17
Men
122.5
32.132
20
 There is a difference in these two groups.
The women’s average was 17.5 points > than the men’s average.
 Yet, the standard deviations are larger than this sample difference,
and the sample sizes are about the same.
19.2 Comparing Two Population Means
Example: SSHA Scores
▸ Summary statistics for the two groups are below:
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women**
139.588
20.363
17
Men
122.5
32.132
20
 There is a difference in these two groups.
The women’s average was 17.5 points > than the men’s average.
 Yet, the standard deviations are larger than this sample difference,
and the sample sizes are about the same.
 Is this difference significant enough to conclude
that 𝜇women is larger than 𝜇men?
19.2 Comparing Two Population Means
Example: SSHA Scores
▸ Summary statistics for the two groups are below:
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women**
139.588
20.363
17
Men
122.5
32.132
20
 There is a difference in these two groups.
The women’s average was 17.5 points > than the men’s average.
 Yet, the standard deviations are larger than this sample difference,
and the sample sizes are about the same.
 Is this difference significant enough to conclude
that 𝜇women is larger than 𝜇men?
Let’s learn more!
19.3 Two-Sample t Procedures
The Two-sample t Procedures: Derived
▸ Now that we have a point estimate and a formula for the standard error,
we can conduct statistical inference for the difference in two population
means.
Chapter
Parameter of Interest
Point
Estimate
Standard
Error
18
m
(σ unknown; 1-sample)
𝑥
𝑠
𝑛
19
μ 1 - μ2
(σ1, σ2 unknown;
2-samples)
Confidence Interval
𝑥 ± 𝑡∗
𝑠
𝑛
pt. estimate ± t*(standard error)
𝑥1 − 𝑥2
𝑠12 𝑠22
+
𝑛1 𝑛2
19.3 Two-Sample t Procedures
The Two-sample t Procedures: Derived
▸ Now that we have a point estimate and a formula for the standard error,
we can conduct statistical inference for the difference in two population
means.
Chapter
Parameter of Interest
Point
Estimate
Standard
Error
18
m
(σ unknown; 1-sample)
𝑥
𝑠
𝑛
19
μ 1 - μ2
(σ1, σ2 unknown;
2-samples)
Confidence Interval
𝑥 ± 𝑡∗
𝑠
𝑛
pt. estimate ± t*(standard error)
𝑥1 − 𝑥2
𝑠12 𝑠22
+
𝑛1 𝑛2
(𝑥1 − 𝑥2 ) ± t*
𝑠12
𝑛1
+
𝑠22
𝑛2
19.3 Two-Sample t Procedures
The Two-sample t Procedures: Derived
Chapter
Parameter of
Interest
Point
Estimate
Standard
Error
Test Statistic
18
μ
(σ unknown;
1-sample)
𝑥
𝑠
𝑛
𝑥 − 𝜇0
𝑡=
𝑠/ 𝑛
m 1 - μ2
19
(σ1, σ2
unknown;
2-samples)
𝑥1 − 𝑥2
𝑠12 𝑠22
+
𝑛1 𝑛2
pt. estimate – m0
standard error
Note: H0 for our purposes will be that m1=m2;
which is equivalent to there being a mean difference of ‘0.’
19.3 Two-Sample t Procedures
The Two-sample t Procedures: Derived
Chapter
Parameter of
Interest
Point
Estimate
Standard
Error
Test Statistic
18
μ
(σ unknown;
1-sample)
𝑥
𝑠
𝑛
𝑥 − 𝜇0
𝑡=
𝑠/ 𝑛
m 1 - μ2
19
(σ1, σ2
unknown;
2-samples)
𝑥1 − 𝑥2
𝑠12 𝑠22
+
𝑛1 𝑛2
pt. estimate – m0
standard error
𝑡=
(𝑥1 − 𝑥2 ) − 0
𝑠12 𝑠22
𝑛1 + 𝑛2
Note: H0 for our purposes will be that m1=m2;
which is equivalent to their being a mean difference of ‘0.’
19.3 Two-Sample t Procedures
The Two-sample t Procedures
▸ Now we can complete the table from earlier:
Population
Parameter
Point Estimate
𝜇1 − 𝜇2
𝑥1 − 𝑥2
Confidence Interval
Test Statistic
t* is the critical value for confidence level C for the
t distribution with df = smaller of (n1-1) and (n2-1).
Find P-values from the t distribution with df = smaller of (n1-1) and (n2-1).
19.3 Two-Sample t Procedures
The Two-sample t Procedures
▸ Now we can complete the table from earlier:
Population
Parameter
𝜇1 − 𝜇2
Point Estimate
𝑥1 − 𝑥2
Confidence Interval
(𝑥1 − 𝑥2 ) ± t*
𝑠12
𝑛1
+
Test Statistic
𝑠22
𝑛2
t* is the critical value for confidence level C for the
t distribution with df = smaller of (n1-1) and (n2-1).
Find P-values from the t distribution with df = smaller of (n1-1) and (n2-1).
19.3 Two-Sample t Procedures
The Two-sample t Procedures
▸ Now we can complete the table from earlier:
Population
Parameter
𝜇1 − 𝜇2
Point Estimate
𝑥1 − 𝑥2
Confidence Interval
(𝑥1 − 𝑥2 ) ± t*
𝑠12
𝑛1
+
𝑠22
𝑛2
Test Statistic
𝑡=
(𝑥1 − 𝑥2 ) − 0
𝑠12 𝑠22
+
𝑛1 𝑛2
t* is the critical value for confidence level C for the
t distribution with df = smaller of (n1-1) and (n2-1).
Find P-values from the t distribution with df = smaller of (n1-1) and (n2-1).
19.3 Two-Sample t Procedures
The Two-sample t Procedures: Confidence
Intervals
▸ Draw an SRS of size n1 from a large Normal population with unknown mean
𝜇1 , and draw an independent SRS of size n2 from another large Normal
population with unknown mean 𝜇2 . A level C confidence interval for 𝜇2 -𝜇1 is
given by
(𝑥1 − 𝑥2 ) ± t*
𝑠12
𝑛1
𝑠2
+ 𝑛2
2
▸ Here t* is the critical value for confidence level C for the t distribution with
degrees of freedom from either Option 1(computer generated) or
Option 2 (the smaller of n1 – 1 and n2 – 1).
19.3 Two-Sample t Procedures
The Two-sample t Procedures: Significance
Tests
▸ To test the hypothesis H0: μ1 - μ2 , calculate the two-sample t statistic
𝑡=
(𝑥1 − 𝑥2 )
𝑠12 𝑠22
𝑛1 + 𝑛2
▸ Find p-values from the t distribution with df = smaller of (n1-1) and (n2-1).
19.0 Two-Sample Problems
Conditions for Inference Comparing TwoSample Means and Robustness of t Procedures
▸ The general structure of our necessary conditions is an extension of
the one-sample cases.
 Simple Random Samples:
 Do we have 2 simple random samples?
 Population : Sample Ratio:
 The samples must be independent and from two large populations of
interest.
19.0 Two-Sample Problems
Conditions for Inference Comparing TwoSample Means and Robustness of t Procedures
 Large enough sample:
Both populations will be assumed to be from a Normal distribution and
 when the sum of the sample sizes is less than 15, t procedures can
be used if the data close to Normal (roughly symmetric, single peak, no
outliers)? If there is clear skewness or outliers then, do not use t.
 when the sum of the sample sizes is between 15 and 40, t procedures
can be used except in the presences of outliers or strong skewness.
 when the sum of the sample sizes is at least 40, the t procedures can
be used even for clearly skewed distributions.
19.0 Two-Sample Problems
Conditions for Inference Comparing TwoSample Means and Robustness of t Procedures
▸ Note: In practice it is enough that the two distributions have similar
shape with no strong outliers. The two-sample t procedures are even
more robust against non-Normality than the one-sample procedures.
▸ Now that we have a point estimate and a formula for the standard error,
we can conduct statistical inference for the difference in two population
means.
19.3 Two-Sample t Procedures
Poll: SSHA Scores
▸ Suppose we have a goal of measuring the mean difference in SSHA
between women and men. Which seems more plausible?
a.
µWomen-µMen = 0
(There is no difference.)
b.
µWomen - µMen ≠ 0
(There is some difference.)
19.3 Two-Sample t Procedures
Example: SSHA Scores
▸ The summary statistics for the SSHA scores for random samples of
men and women are below. Use this information to construct a 90%
confidence interval for the mean difference.
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women
139.588
20.363
17
Men
122.5
32.132
20
18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
1. Components
1.
Do we have two simple random samples?
Yes. It was stated.
Large enough population: sample ratio?
Yes.
NW > 20*17 = 340
NM > 20*20 = 400
Large enough sample?
Yes.
nW + nM =37 < 40
but outlier has been removed.
No skewness.
Steps for SuccessConstructing Confidence Intervals
for m1 - m2 .
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5, sm = 32.132, nm = 20
Steps for SuccessConstructing Confidence Intervals
for m1 - m2 .
1.
2.
3.
4.
5.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
Identify the 3 key components of the
confidence interval (means, s.ds., n1 , n2 ).
Select t*.
Construct the confidence interval.
*Interpret* the interval.
18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5, sm = 32.132, nm = 20
3. Select t*.
df =min{(nw -1), (nm -1)}=16
t*(90%, 16) = 1.746
Steps for SuccessConstructing Confidence Intervals
for m1 - m2 .
1.
2.
3.
4.
5.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
Identify the 3 key components of the
confidence interval (means, s.ds., n1 , n2 ).
Select t*.
Construct the confidence interval.
*Interpret* the interval.
18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5, sm = 32.132, nm = 20
Steps for SuccessConstructing Confidence Intervals
for m1 - m2 .
1.
2.
3.
4.
5.
3. Select t*.
df =min{(nw -1), (nm -1)}=16
t*(90%, 16) = 1.746
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
Identify the 3 key components of the
confidence interval (means, s.ds., n1 , n2 ).
Select t*.
Construct the confidence interval.
*Interpret* the interval.
4. Interval.
139.588 − 122.5 ± 1.746
20.3632
17
+
32.1322
20
17.088 ± 15.222 = 1.866 to 32.31
18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5, sm = 32.132, nm = 20
Steps for SuccessConstructing Confidence Intervals
for m1 - m2 .
1.
2.
3.
4.
5.
3. Select t*.
df =min{(nw -1), (nm -1)}=16
t*(90%, 16) = 1.746
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
Identify the 3 key components of the
confidence interval (means, s.ds., n1 , n2 ).
Select t*.
Construct the confidence interval.
*Interpret* the interval.
4. Interval.
139.588 − 122.5 ± 1.746
20.3632
17
+
32.1322
20
17.088 ± 15.222 = 1.866 to 32.31
5. Interpret.
We are 90% confident that the mean women’s SSHA score is
between 1.866 and 32.31 points higher than men’s.
19.3 Two-Sample t Procedures
Example: SSHA Scores
▸ Let’s continue with this example by now conducting a test of
significance for the mean difference in SSHA by gender at a=0.10.
Does our decision align with the results from the earlier poll?
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women
139.588
20.363
17
Men
122.5
32.132
20
19.3 Two-Sample t Procedures
Example: SSHA Scores
State: Is there a difference in the mean SSHA scores between men and women?
(i.e., mDiff ≠ 0, mWomen − mMen ≠ 0, mWomen ≠ mMen )
Plan:
a.) Identify the parameter.
19.3 Two-Sample t Procedures
Example: SSHA Scores
State: Is there a difference in the mean SSHA scores between men and women?
(i.e., mDiff ≠ 0, mWomen − mMen ≠ 0, mWomen ≠ mMen )
Plan:
a.) Identify the parameter.
mDiff =mWomen - mMen.
b) List all given information from the data collected.
19.3 Two-Sample t Procedures
Example: SSHA Scores
State: Is there a difference in the mean SSHA scores between men and women?
(i.e., mDiff ≠ 0, mWomen − mMen ≠ 0, mWomen ≠ mMen )
Plan:
a.) Identify the parameter.
mDiff =mWomen - mMen.
b) List all given information from the data collected. 𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5,
c) State the null (H0) and alternative (HA) hypotheses.
sm = 32.132, nm = 20
19.3 Two-Sample t Procedures
Example: SSHA Scores
State: Is there a difference in the mean SSHA scores between men and women?
(i.e., mDiff ≠ 0, mWomen − mMen ≠ 0, mWomen ≠ mMen )
Plan:
a.) Identify the parameter.
mDiff =mWomen - mMen.
b) List all given information from the data collected. 𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5,
c) State the null (H0) and alternative (HA) hypotheses.
sm = 32.132, nm = 20
H0: mDiff = 0
Ha : mDiff ≠ 0
19.3 Two-Sample t Procedures
Example: SSHA Scores
State: Is there a difference in the mean SSHA scores between men and women?
(i.e., mDiff ≠ 0, mWomen − mMen ≠ 0, mWomen ≠ mMen )
Plan:
a.) Identify the parameter.
mDiff =mWomen - mMen.
b) List all given information from the data collected. 𝒙𝒘 = 139.588, sw = 20.363, nw = 17
𝒙𝒎 = 122.5,
c) State the null (H0) and alternative (HA) hypotheses.
sm = 32.132, nm = 20
H0: mDiff = 0
Ha : mDiff ≠ 0
d) Specify the level of significance. a =.10
e) Determine the type of test.
Left-tailed
Right-tailed
Two-Tailed
19.3 Two-Sample t Procedures
Example: SSHA Scores
Plan:
f) Sketch the region(s) of “extremely unlikely” test statistics.
19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve:
a)
Check the conditions for the test you plan to use.
 Two Simple Random Samples?
 Large enough population: sample ratios?
 Large enough samples?
19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve:
a)
Check the conditions for the test you plan to use.
 Two Simple Random Samples?
Yes.
Stated as a random sample.
 Large enough population: sample ratios?
Yes. Both populations are arbitrarily large;
much greater than, NW > 20*17 = 340; NM > 20*20 = 400
 Large enough samples?
Yes. nW + nM =37 < 40 outlier has been removed. No skewness.
19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve:
b)
c)
Calculate the test statistic
t=
𝑥𝑤 −𝑥𝑚
𝑠𝑤 2 𝑠𝑚 2
+
𝑛𝑤
𝑛𝑚
=
Determine (or approximate) the P-Value.
19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve:
b)
c)
Calculate the test statistic
t=
𝑥𝑤 −𝑥𝑚
𝑠𝑤 2 𝑠𝑚
+
𝑛𝑤
𝑛𝑚
=
2
Determine (or approximate) the P-Value.
139.588−122.5
20.3632 32.1322
+ 20
17
=
17.088
8.719
= 1.96
19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve:
b)
c)
Calculate the test statistic
t=
𝑥𝑤 −𝑥𝑚
𝑠𝑤 2 𝑠𝑚
+
𝑛𝑤
𝑛𝑚
=
2
Determine (or approximate) the P-Value.
139.588−122.5
20.3632 32.1322
+ 20
17
1.96
 1.746 < 1.96 < 2.12
 .05 < P-value < .10
P-value
=
17.088
8.719
DF = 17 - 1
= 1.96
19.3 Two-Sample t Procedures
Example: SSHA Scores
Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
19.3 Two-Sample t Procedures
Example: SSHA Scores
Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
Because the approximate P-value is smaller than 0.10,
we reject the null hypothesis.
b) Interpret the decision in the context of the original claim.
19.3 Two-Sample t Procedures
Example: SSHA Scores
Conclude:
a) Make a decision about the null hypothesis (Reject H0 or Fail to reject H0).
Because the approximate P-value is smaller than 0.10,
we reject the null hypothesis.
b) Interpret the decision in the context of the original claim.
There is enough evidence (at a=.10)
that there is a difference in the mean SSHA score
between men and women.
19.3 Two-Sample t Procedures
Example: SSHA Scores
▸ Let’s continue with this example by now conducting a test of
significance for the mean difference in SSHA by gender at a=0.10.
Does our decision align with the results from the earlier poll?
________
Group
Sample
Mean
Sample Standard
Deviation
Sample
Size
Women
139.588
20.363
17
Men
122.5
32.132
20