STAT 1450

19.0 Two-Sample Problems
Connecting Chapter 18 to our
Current Knowledge of Statistics
Population
Parameter
μ (σ known)
μ (σ unknown)
Point Estimate s
𝑥
Confidence
Interval
𝑥 ± 𝑧 ∗ 𝜎 𝑛
Test Statistic 𝑧 =
𝑥 − 𝜇
0 𝜎 𝑛
𝑥 ± 𝑡 ∗ 𝑠 𝑛 𝑡 =
𝑥 − 𝜇
0 𝑠 𝑛
▸ 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 𝜇
1
− 𝜇
2
Point Estimate
𝑥
1
− 𝑥
2
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 𝒙
𝟏 𝒙
𝟐
.

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 x
1 x
2
Diff = x
1
- x
2 m
1 m
2 m
1
m
2
𝑥
1
𝑥
2
𝑥
1
− 𝑥
2 s
1 s
2 𝜎
1
2 𝑛
1
+ 𝜎
2
2 𝑛
2 𝑠
1 𝑛
1 𝑠
2 𝑛
2 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2

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
Women**
Sample
Mean
139.588
Sample Standard
Deviation
20.363
Sample
Size
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
Women**
Sample
Mean
139.588
Sample Standard
Deviation
20.363
Sample
Size
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
Women**
Sample
Mean
139.588
Sample Standard
Deviation
20.363
Sample
Size
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 𝜇 women is larger than 𝜇 men
?

19.2 Comparing Two Population Means
Example: SSHA Scores
▸ Summary statistics for the two groups are below:
Group
Women**
Sample
Mean
139.588
Sample Standard
Deviation
20.363
Sample
Size
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 𝜇 women is larger than 𝜇 men
? 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
18
19
Parameter of Interest m
(
σ unknown; 1-sample)
(
σ
1
μ
1
,
σ
2
μ
2 unknown;
2-samples)
Point
Estimate
𝑥
𝑥
1
− 𝑥
2
Standard
Error 𝑠 𝑛 𝑠 2
1 𝑛
1
+ 𝑠 2
2 𝑛
2
Confidence Interval
𝑥 ± 𝑡 ∗ 𝑠 𝑛 pt. estimate ± t*(standard error)

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
18
19
Parameter of Interest m
(
σ unknown; 1-sample)
(
σ
1
μ
1
,
σ
2
μ
2 unknown;
2-samples)
Point
Estimate
𝑥
𝑥
1
− 𝑥
2
Standard
Error 𝑠 𝑛 𝑠 2
1 𝑛
1
+ 𝑠 2
2 𝑛
2
Confidence Interval
𝑥 ± 𝑡 ∗ 𝑠 𝑛 pt. estimate ± t*(standard error)
( 𝑥
1
− 𝑥
2
) ± t * 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2

19.3 Two-Sample t Procedures
The Two-sample t Procedures: Derived
Chapter Parameter of
Interest
18
Point
Estimate
μ
(
σ unknown;
1-sample)
𝑥
Standard
Error 𝑠 𝑛
Test Statistic 𝑡 =
𝑥 − 𝜇
0 𝑠/ 𝑛
19 m
1
μ
2
( σ
1
, σ
2 unknown;
2-samples)
𝑥
1
− 𝑥
2 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2 pt. estimate – m
0 standard error
0
m
1
m
2

19.3 Two-Sample t Procedures
The Two-sample t Procedures: Derived
Chapter Parameter of
Interest
18
Point
Estimate
μ
(
σ unknown;
1-sample)
𝑥
Standard
Error 𝑠 𝑛
Test Statistic 𝑡 =
𝑥 − 𝜇
0 𝑠/ 𝑛 m
1
μ
2 pt. estimate – m
0 standard error
19
( σ
1
, σ
2 unknown;
2-samples)
𝑥
1
− 𝑥
2 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2 𝑡 =
( 𝑥
1 𝑥
2
) − 0 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2
0
m
1
m
2

19.3 Two-Sample t Procedures
The Two-sample t Procedures
▸ Now we can complete the table from earlier:
Population
Parameter
Point Estimate Confidence Interval Test Statistic 𝜇
1
− 𝜇
2
𝑥
1
− 𝑥
2 t* is the critical value for confidence level C for the t distribution with df = smaller of (n
1
-1) and (n
2
-1).
Find P -values from the t distribution with df = smaller of (n
1
-1) and (n
2
-1).

19.3 Two-Sample t Procedures
The Two-sample t Procedures
▸ Now we can complete the table from earlier:
Population
Parameter
Point Estimate Confidence Interval 𝜇
1
− 𝜇
2
𝑥
1
− 𝑥
2
( 𝑥
1
− 𝑥
2
) ± t * 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2
Test Statistic t* is the critical value for confidence level C for the t distribution with df = smaller of (n
1
-1) and (n
2
-1).
Find P-values from the t distribution with df = smaller of (n
1
-1) and (n
2
-1).

19.3 Two-Sample t Procedures
The Two-sample t Procedures
▸ Now we can complete the table from earlier:
Population
Parameter
Point Estimate Confidence Interval Test Statistic 𝜇
1
− 𝜇
2
𝑥
1
− 𝑥
2
( 𝑥
1
− 𝑥
2
) ± t * 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
2 𝑡 =
( 𝑥
1
− 𝑥
2
) − 0 𝑠 2
1 𝑛
1
+ 𝑠 2
2 𝑛
2 t* is the critical value for confidence level C for the t distribution with df = smaller of (n
1
-1) and (n
2
-1).
Find P-values from the t distribution with df = smaller of (n
1
-1) and (n
2
-1).

19.3 Two-Sample t Procedures
The Two-sample t Procedures: Confidence
Intervals
▸ Draw an SRS of size n
1 𝜇
1 from a large Normal population with unknown mean
, and draw an independent SRS of size n
2 from another large Normal population with unknown mean 𝜇
2
. A level C confidence interval for 𝜇
2
𝜇
1 is given by
( 𝑥
1
− 𝑥
2
) ± t * 𝑠
2
1 𝑛
1
+ 𝑠
2
2 𝑛
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 n
1
– 1 and n
2
– 1).

19.3 Two-Sample t Procedures
The Two-sample t Procedures: Significance
Tests
▸ To test the hypothesis H
0
: μ
1
μ
2
, calculate the two-sample t statistic 𝑡 =
( 𝑥
1
− 𝑥
2
) 𝑠
2
1 𝑛
1
+ 𝑠 𝑛
2
2
2
▸ Find p-values from the t distribution with df = smaller of ( n
1
-1 ) and ( n
2
-1 ).

19.0 Two-Sample Problems
Conditions for Inference Comparing Two-
Sample 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 Two-
Sample 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 Two-
Sample 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.
µ
Women
µ
Men
= 0 b.
µ
Women
µ
Men
≠ 0
(There is no difference.)
(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
Women
Men
Sample
Mean
139.588
122.5
Sample Standard
Deviation
20.363
32.132
Sample
Size
17
20

18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
Steps for Success-
1. Components

Constructing Confidence Intervals for m
1
- m
2
.
1.
Confirm that the 3 key conditions are satisfied
Do we have two simple random samples?
(SRS?, N:n?, t-distribution?).
Yes. It was stated.
 Large enough population: sample ratio?
Yes. N
W
N
M
> 20*17 = 340
> 20*20 = 400
 Large enough sample?
Yes. n
W
+ n
M
=37 < 40 but outlier has been removed.
No skewness.

18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙 𝒘 𝒙 𝒎
= 139.588, s w
= 122.5, s m
= 20.363, n w
= 32.132, n m
= 17
= 20
Steps for Success-
Constructing Confidence Intervals for m
1
- m
2
.
1.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
2.
Identify the 3 key components of the
, n
2
).
confidence interval (means, s.ds., n
1
3.
Select t* .
4. Construct the confidence interval .
5.
* Interpret * the interval.

18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙 𝒘 𝒙 𝒎
= 139.588, s w
= 122.5, s m
= 20.363, n w
= 32.132, n m
= 17
= 20
3. Select t*.
df =min{(n w
-1), (n m
-1)}=16 t*(90%, 16) = 1.746
Steps for Success-
Constructing Confidence Intervals for m
1
- m
2
.
1.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
2.
Identify the 3 key components of the
, n
2
).
confidence interval (means, s.ds., n
1
3.
Select t* .
4. Construct the confidence interval .
5.
* Interpret * the interval.

18.3 One-Sample t Confidence Intervals
Example: 90% CI for SSHA Scores
2. Components.
𝒙 𝒘 𝒙 𝒎
= 139.588, s w
= 122.5, s m
= 20.363, n w
= 32.132, n m
= 17
= 20
3. Select t*.
df =min{(n w
-1), (n m
-1)}=16 t*(90%, 16) = 1.746
Steps for Success-
Constructing Confidence Intervals for m
1
- m
2
.
1.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
2.
Identify the 3 key components of the
, n
2
).
confidence interval (means, s.ds., n
1
3.
Select t* .
4. Construct the confidence interval .
5.
* Interpret * the interval.
4. Interval.
139.588 − 122.5 ± 1.746
20.363
2
17
+
32.132
2
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.
𝒙 𝒘 𝒙 𝒎
= 139.588, s w
= 122.5, s m
= 20.363, n w
= 32.132, n m
= 17
= 20
3. Select t*.
df =min{(n w
-1), (n m
-1)}=16 t*(90%, 16) = 1.746
Steps for Success-
Constructing Confidence Intervals for m
1
- m
2
.
1.
Confirm that the 3 key conditions are satisfied
(SRS?, N:n?, t-distribution?).
2.
Identify the 3 key components of the
, n
2
).
confidence interval (means, s.ds., n
1
3.
Select t* .
4. Construct the confidence interval .
5.
* Interpret * the interval.
4. Interval.
139.588 − 122.5 ± 1.746
20.363
2
17
+
32.132
2
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
Women
Men
Sample
Mean
139.588
122.5
Sample Standard
Deviation
20.363
32.132
Sample
Size
17
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., m
Diff
≠ 0, m
Women
− m
Men
≠ 0 , m
Women
≠ m
Men
)
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., m
Diff
≠ 0, m
Women
− m
Men
≠ 0 , m
Women
≠ m
Men
)
Plan: a.) Identify the parameter.
m
Diff
= m
Women
m
Men
.
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., m
Diff
≠ 0, m
Women
− m
Men
≠ 0 , m
Women
≠ m
Men
)
Plan: a.) Identify the parameter.
m
Diff
= m
Women
m
Men
.
b) List all given information from the data collected.
𝒙 𝒘
= 139.588, s w
= 20.363, n w
= 17 𝒙 𝒎
= 122.5, s m
= 32.132, n m
= 20 c) State the null (H
0
) and alternative (H
A
) hypotheses.

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., m
Diff
≠ 0, m
Women
− m
Men
≠ 0 , m
Women
≠ m
Men
)
Plan: a.) Identify the parameter.
m
Diff
= m
Women
m
Men
.
b) List all given information from the data collected.
𝒙 𝒘
= 139.588, s w
= 20.363, n w
= 17 𝒙 𝒎
= 122.5, s m
= 32.132, n m
= 20 c) State the null (H
0
) and alternative (H
A
) hypotheses.
H
0
: m
Diff
= 0 H a
: m
Diff
≠ 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., m
Diff
≠ 0, m
Women
− m
Men
≠ 0 , m
Women
≠ m
Men
)
Plan: a.) Identify the parameter.
m
Diff
= m
Women
m
Men
.
b) List all given information from the data collected.
𝒙 𝒘
= 139.588, s w
= 20.363, n w
= 17 𝒙 𝒎
= 122.5, s m
= 32.132, n m
= 20 c) State the null (H
0
) and alternative (H
A
) hypotheses.
H
0
: m
Diff
= 0 H a
: m
Diff
≠ 0 d) Specify the level of significance. a
=.10
e) Determine the type of test.
Left-tailed Right-tailed Two-Tailed

Example: SSHA Scores
Plan: f)
Sketch the region(s) of “extremely unlikely” test statistics.
19.3 Two-Sample t Procedures

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, N
W
> 20*17 = 340; N
M
> 20*20 = 400
 Large enough samples?
Yes. n
W
+ n
M
=37 < 40 outlier has been removed. No skewness.

Example: SSHA Scores
Solve: b) Calculate the test statistic 𝑥 𝑤
−𝑥 𝑚 𝑠𝑤 2 𝑛𝑤
+ 𝑠𝑚 2 𝑛𝑚
= c) Determine (or approximate) the P-Value.
19.3 Two-Sample t Procedures

19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve: b) Calculate the test statistic 𝑥 𝑤
−𝑥 𝑚 𝑠𝑤 2 𝑛𝑤
+ 𝑠𝑚 2 𝑛𝑚
=
139.588−122.5
20.3632
17
+
32.1322
20
=
17.088
8.719
= 1.96
c) Determine (or approximate) the P-Value.

19.3 Two-Sample t Procedures
Example: SSHA Scores
Solve: b) Calculate the test statistic 𝑥 𝑤
−𝑥 𝑚 𝑠𝑤 2 𝑛𝑤
+ 𝑠𝑚 2 𝑛𝑚
=
139.588−122.5
20.3632
17
+
32.1322
20
=
17.088
8.719
= 1.96
c) Determine (or approximate) the P-Value.
1.96 DF = 17 - 1
 1.746 < 1.96 < 2.12
 .05 < P -value < .10
P -value

19.3 Two-Sample t Procedures
Example: SSHA Scores
Conclude: a) Make a decision about the null hypothesis ( Reject H
0 or Fail to reject H
0
).

19.3 Two-Sample t Procedures
Example: SSHA Scores
Conclude: a) Make a decision about the null hypothesis ( Reject H
0 or Fail to reject H
0
).
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 H
0 or Fail to reject H
0
).
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
Women
Men
Sample
Mean
139.588
122.5
Sample Standard
Deviation
20.363
32.132
Sample
Size
17
20