chapter 19: two-sample problems stat 1450. connecting chapter 18 to our current knowledge of...

Chapter 19: Two-Sample Problems

STAT 1450

Connecting Chapter 18 to our Current Knowledge of Statistics

▸ Remember that these formulas are only valid when appropriate simple

conditions apply!

19.0 Two-Sample Problems

PopulationParameter

Point EstimateConfidence

IntervalTest Statistic

μ (σ known)

μ (σ unknown) s

Connecting Chapter 19 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.”


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

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.

PopulationParameter

Point EstimateConfidence

IntervalTest Statistic

19.1 The Two-Sample Problem

Sampling Distribution of Two Sample Means

▸ Recall that for a single sample mean The standard deviation of a statistic is estimated from data the

result is called the standard error of the statistic.

The standard error of is .

▸ Inference in the two-sample problem will require the standard error of the difference of two sample means .

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 PopulationStandard Deviation

Standard Error

x1m1 s1

x2m2 s2

Diff = x1 - x2

m1 - m2

1x2x1x2x

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?



▸ Summary statistics for the two groups are below:

There is a difference in these two groups. The women’s average was 17 points > than the men’s average.

Group Sample Mean

Sample Standard Deviation

Sample Size

Women** 139.588 20.363 17

Men 122.5 32.132 20





Yet, the standard deviations are larger than this sample difference, and the sample sizes are about the same.

Group Sample Mean


Sample Size

Women** 139.588 20.363 17

Men 122.5 32.132 20






Is this difference significant enough to conclude that women is different than men?

Group Sample Mean


Sample Size

Women** 139.588 20.363 17

Men 122.5 32.132 20






Is this difference significant enough to conclude that women is different than men? Let’s

learn more!

Group Sample Mean


Sample Size

Women** 139.588 20.363 17

Men 122.5 32.132 20


The Two-sample t Procedures: Derived

▸ Now that we have a point estimate and a formula for the standard error, we can determine the confidence interval for the difference in two population means.

Chapter Parameter of Interest Point Estimate

Standard Error

Confidence Interval

18 m (σ unknown; 1-sample)

19

μ1 - μ2 (σ1, σ2 unknown; 2-

samples)

pt. estimate ± t*(standard error)

19.3 Two-Sample t Procedures


▸ Now that we have a point estimate and a formula for the standard error, we can determine the confidence interval for the difference in two population means.

Chapter Parameter of Interest Point Estimate

Standard Error

Confidence Interval

18 m (σ unknown; 1-sample)

19

μ1 - μ2 (σ1, σ2 unknown; 2-

samples)

pt. estimate ± t*(standard error)

() ± t*




Chapter Parameter of Interest

Point Estimate

Standard Error

Test Statistic

18 μ (σ unknown;

1-sample)

19

m1 - μ2

(σ1, σ2 unknown;

2-samples)

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.’



Chapter Parameter of Interest

Point Estimate

Standard Error

Test Statistic

18 μ (σ unknown;

1-sample)

19

m1 - μ2

(σ1, σ2 unknown;

2-samples)

pt. estimate – m0

standard error

Note: H0 for our purposes will be that m1=m2; which is equivalent to their being a mean difference of ‘0.’

The Two-sample t Procedures

▸ Now we can complete the table from earlier:

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).

PopulationParameter

Point Estimate Confidence Interval Test Statistic


The Two-sample t Procedures

▸ Now we can complete the table from earlier:

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).

PopulationParameter

Point Estimate Confidence Interval Test Statistic

() ± t*


The Two-sample t Procedures: Confidence Intervals▸ Draw an SRS of size n1 from a large Normal population with unknown mean ,

and draw an independent SRS of size n2 from another large Normal population with unknown mean . A level C confidence interval for - is given by

() ± t*

▸ 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).


The Two-sample t Procedures: Significance Tests

▸ To test the hypothesis H0: μ1 - μ2 , calculate the two-sample t statistic

▸ Find p-values from the t distribution with df = smaller of (n1-1) and (n2-1).


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.


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.


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.



▸ The summary statistics for the SSHA scores for random samples of

men and women are below. There was neither significant skewness,

nor, strong outliers, in either data set. Use this information to construct

a 90% confidence interval for the mean difference.


Group Sample Mean


Sample Size

Women 139.588 20.363 17

Men 122.5 32.132 20

Example: 90% CI for SSHA Scores

1. Components 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(Independence)

Large enough sample? Yes. nW + nM =37 < 40

but outlier has been removed. No skewness.

Steps for Success- Constructing Confidence Intervals

for m1 - m2 .1. 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 Steps for Success-

Constructing Confidence Intervals for m1 - m2 .

1. Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?).

2. Identify the key components of the confidence interval (means, s.ds., n1 , n2



2. Components. = 139.588, sw = 20.363, nw = 17 = 122.5, sm = 32.132, nm = 20



(SRS?, N:n?, t-distribution?). 2. Identify the key components of the

confidence interval (means, s.ds., n1 , n2







confidence interval (means, s.ds., n1 , n2 3. Select t*.




3. Select t*. df =min{(nw -1), (nm -1)}=16 t*(90%, 16) = 1.746




confidence interval (means, s.ds., n1 , n2 3. Select t*.








confidence interval (means, s.ds., n1 , n2 3. Select t*.4. Construct the confidence interval.





4. Interval.




confidence interval (means, s.ds., n1 , n2 3. Select t*.4. Construct the confidence interval.





4. Interval.




confidence interval (means, s.ds., n1 , n2 3. Select t*.4. Construct the confidence interval.5. *Interpret* the interval.





4. Interval.

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.



(SRS?, N:n?, t-distribution?). 2. Identify the 3 key components of the

confidence interval (means, s.ds., n1 , n2 3. Select t*.4. Construct the confidence interval.5. *Interpret* the interval.



▸ 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.


Group Sample Mean


Sample Size

Women 139.588 20.363 17

Men 122.5 32.132 20


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.




Plan:

a.) Identify the parameter. mDiff =mWomen - mMen.

b) List all given information from the data collected.




Plan:


b) List all given information from the data collected. = 139.588, sw = 20.363, nw =

17

= 122.5, sm = 32.132, nm = 20

c) State the null (H0) and alternative (HA) hypotheses.




Plan:



17

= 122.5, sm = 32.132, nm = 20

c) State the null (H0) and alternative (HA) hypotheses. H0: mDiff 0Ha : mDiff 0




Plan:



17

= 122.5, sm = 32.132, nm = 20

c) State the null (H0) and alternative (HA) hypotheses. H0: mDiff 0Ha : mDiff 0

d) Specify the level of significance. a =.10

e) Determine the type of test. Left-tailed Right-tailed Two-

Tailed



Plan:

f) Sketch the region(s) of “extremely unlikely” test statistics.



Solve:

a) Check the conditions for the test you plan to use.

Two Simple Random Samples?

Large enough population: sample ratios?

Large enough samples?



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.



Solve:

b) Calculate the test statistic

c) Determine (or approximate) the P-Value.

t =



Solve:

b) Calculate the test statistic

c) Determine (or approximate) the P-Value. 1.96 DF = 17 - 1

1.746 < 1.96 < 2.12

.05 < P-value < .10

P-value

t =



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.




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.


Technology Tips – Constructing Confidence Intervals (2 population means, 1 , 2 unknown)

▸ TI-83/84 STAT TESTS 2SampTInt… Enter.

Select Stats. Enter , s, and n for the first sample, then for the second.

Enter the Confidence Level. Select Pooled: No. Select Calculate.

(Note: Select Data when and n are not provided. Then enter the list where the data are stored.)


Technology Tips – Constructing Confidence Intervals (2 population means, 1 , 2 unknown)

▸ JMP Enter the quantitative data into one of the columns. In the next column, enter an abridged description of the categorical variable

associated with each row of quantitative data. (Note: Pay attention to the spelling and capitalization of the abridged descriptions.)

Analyze Fit Y by X. “Click-and-Drag” (the quantitative variable) into the ‘Y, Response’ box. “Click-and-

Drag” (the categorical variable) into the ‘X, Factor’ box. Click on OK. Click on the red upside-down triangle next to the title “Oneway Analysis of …”

Proceed to ‘Means and Std Dev.’

Click on the red upside-down triangle next to the title “Oneway Analysis of …”

Proceed to ‘t Test.’


Technology Tips – Significance Tests (2 population means, 1 , 2 unknown)

▸ TI-83/84

STAT TESTS 2SampTTest Enter.

Select Stats. Enter , s, and n for the first sample, then for the second.

(Note: Select Data when and n are not provided. Then enter the list where

the data are stored.)

Select the inequality that corresponds to the alternative hypothesis.

Select Pooled: No.

Select Calculate.


Technology Tips – Significance Tests (2 population means, 1 , 2 unknown)▸ JMP

Enter the quantitative data into one of the columns.

In the next column, enter an abridged description of the categorical variable associated with

each row of quantitative data.

▸ (Note: Pay attention to the spelling and capitalization of the abridged descriptions.)

Analyze Fit Y by X.

“Click-and-Drag” (the quantitative variable) into the ‘Y, Response’ box.

“Click-and-Drag” (the categorical variable) into the ‘X, Factor’ box. Click on OK.


Proceed to ‘Means and Std Dev.’


Proceed to ‘t Test.’


SSHA Scores (via Technology)

▸ Use technology to compute a 98% confidence interval for the mean

difference in SSHA scores between women and men.

▸ Use technology to conduct the test of significance for the mean

difference in SSHA scores at a = .02.


98% confidence interval for the mean difference in SSHA scores between women & men.

▸ TI-83/84

STAT >> TESTS >> 2SampTInt… >> Enter

Note: Select Data when and are not provided. Then

enter the list where the data are stored.

(for this example)

Inpt >> STATS

1:139.588 >> sx1: 20.363 >> n1: 17 2:122.5 >> sx1: 32.13 >> n1: 20

C-Level : 98

Pooled : No

Calculate (ENTER)


98% confidence interval for the mean difference in SSHA scores between women & men.

▸ TI-83/84

STAT >> TESTS >> 2SampTInt… >> Enter

Note: Select Data when and are not provided. Then

enter the list where the data are stored.

(for this example)

Inpt >> STATS

1:139.588 >> sx1: 20.363 >> n1: 17 2:122.5 >> sx2: 32.13 >> n2: 20

C-Level : 98

Pooled : No

Calculate (ENTER) ( -4.241, 38.417)


Test of Significance for the mean difference in SSHA scores between women & men.▸ TI-83/84. STAT TESTS 2SampleTest Enter.

Select Stats. Enter , s and n for each sample. Select Calculate. (Note: Select Data when and n are not provided. Then enter the list where the data

are stored.).

(for this example) Inpt >> STATS

1:139.588 >> sx1: 20.363 >> n1: 17 2:122.5 >> sx2: 32.13 >> n2: 20

1m : ≠ 2 m >> Pooled: No Calculate (ENTER) t= 1.96

p= .0586


Test of Significance for the mean difference in SSHA scores between women & men.▸ TI-83/84. STAT TESTS 2SampleTest Enter.

Select Stats. Enter , s and n for each sample. Select Calculate. (Note: Select Data when and n are not provided. Then enter the list where the data

are stored.).

(for this example) Inpt >> STATS

1:139.588 >> sx1: 20.363 >> n1: 17 2:122.5 >> sx1: 32.13 >> n2: 20

1m : ≠ 2 m >> Pooled: No Calculate (ENTER) t= 1.96

p= .0586 .0586 > .02

df=32.56 => p-value > a . Fail to Reject H0,

There is not enough evidence (at a = .02) that there is a significant difference in SSHA scores between men and

women. (Note: we did have enough evidence at a = .10).



▸ Technology output for Two Sample Means:

JMP output for the Two-Sample Means’ test of significance and and confidence interval: Means and Std Deviations

Level Number Mean Std Dev Std Err Mean Lower 95% Upper 95% Men 20 122.500 32.1321 7.1850 107.46 137.54 Women 17 139.588 20.3625 4.9386 129.12 150.06

t Test of SSHA by Gender (women – men)

60

80

100

120

140

160

180

200

SSHA

m w

Gender

-30 -20 -10 0 10 20 30

Difference 17.088 t Ratio 1.959977

Std Err Dif 8.719 DF 32.56308

Upper CL Dif 34.835 Prob > |t| 0.0586

Lower CL Dif -0.659 Prob > t 0.0293*

Confidence 0.95 Prob < t 0.9707


Closing Caveats and Comments

▸ The two-sample t statistic has an approximate (but accurate) t

distribution. The approximate distribution of the two-sample t has an

elaborate degrees of freedom computation (p.480). Computers use

this formula in determining degrees of freedom.

▸ We will use Option 2 (p.470). This has df= smaller of (n1-1) and (n2-1).

▸ Because of the above fact, output in JMP (or other software packages)

might have different df and p-values from manual analyses.



▸ We will not use “pooled” two-sample procedures. This assumes that

the population variance is known and is equal for both variables. Use

of our “Option 1” for two-sample t procedures yields more accurate

results than the “pooled t.” The only caveat is when the sample sizes

are equal; then our results and the “pooled t” would be equal.

▸ Do not use two-sample t procedures for inference regarding standard

deviations. The F-test is more appropriate in those cases.



▸ Practitioners prefer having equal sample sizes for the two groups when

possible.

▸ Exercises for this chapter will all assume that the SRS is from a

Normal distribution.


Five-Minute Summary

▸ List at least 3 concepts that had the most impact on your knowledge of

two-sample problems.

_______________ ______________

____________

chapter 19: two-sample problems stat 1450. connecting chapter 18 to our current knowledge of...

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