Two Population MeansTwo Population Means

Hypothesis Testing and Hypothesis Testing and Confidence IntervalsConfidence Intervals

With Known Standard DeviationsWith Known Standard Deviations

SITUATION: 2 PopulationsSITUATION: 2 Populations

Population 1

Mean = 1

St’d Dev. = 1

Population 2

Mean = 2

St’d Dev. = 2

Salaries in Chicago Salaries in St. Louis

Women’s Test Scores Men’s Test Scores

Lakers Attendance Clippers Attendance

Anaheim Sales Irvine Sales

KEY ASSUMPTIONSKEY ASSUMPTIONSSampling is done from two populations.

– Population 1 has mean µ1 and variance σ12.

– Population 2 has mean µ2 and variance σ22.

– A sample of size n1 will be taken from population 1.

– A sample of size n2 will be taken from population 2.

– Sampling is random and both samples are drawn independently.

– Either the sample sizes will be large or the populations are assumed to be normally distribution.

1

21

1

111 n

σ variance,

n

σ deviation standard ,μ mean :X variableRandom

2

22

2

222 n

σ variance,

n

σ deviation standard ,μ mean :X variableRandom

The ProblemThe Problem1 and 2 are unknown

1 and 2 may or may not be known

(In this module we assume they are known.)

OBJECTIVESOBJECTIVES• Test whether 1 > 2 (by a certain amount)

– or whether 1 2

• Determine a confidence interval for the difference in the means: 1 - 2

21 XX

Key Concepts About the Key Concepts About the Random Variable .Random Variable .

• is the difference in two sample means.1. Its meanmean is the difference of the two individual means:

2. If the variables are independent (which we assumed), the variancevariance (not the standard deviation) of the random variable of the differences = the sum (not the difference) of the two variances:

3. Thus its standard deviationstandard deviation is:

4. Its distributiondistribution is:• Normal if σ1 and σ2 are known • t if σ1 and σ2 are unknown

21 XX

2

22

1

21

n

σ

n

σ

21 μμ

2

22

1

21

n

σ

n

σ

Random Variable XDistribution normal normal normal

Mean 1 2

Standard Deviation

21 XX X

n

σ

2

22

1

21

n

σ

n

σ

21 XX ,X X, Variable Random

Hypothesis Test Statistics forHypothesis Test Statistics forDifference in Means, Known Difference in Means, Known σσ’s’s

• We will be performing hypothesis tests with null hypotheses, H0, of the form:

• From the general form of a test statistic, the required test statistic will be:

2

22

1

21

21

nσ

nσ

v)xx(

ErrorStandard

Value) zed(Hypothesi - imate)(Point Estz

HH00: : µµ11 - µ - µ22 = v = v

Confidence Intervals for Confidence Intervals for µµ11 - µ - µ22

Known Known σσ’s’s• Recall the general form of a confidence interval is:

Thus when the σ’s are known this becomes:

(Point Estimate) ± zα/2(Appropriate Standard Error)

2

22

1

21

α/221 n

σ

n

σzxx

EXAMPLEEXAMPLEHypothesis Test: Hypothesis Test: 11, , 2 2 KnownKnown

Test whether starting salaries for secretaries in Chicago are at least $5 more per week than those in St. Louis.

GIVEN:Salaries assumed to be normal

Standard Deviations known: Chicago $10; St. Louis $15

Sample ResultsSampled 100 secretaries in Chicago; 75 secretaries in St. Louis

Sample averages: Chicago - $550, St. Louis -$540

Hypothesis TestHypothesis Test

H0: 1 - 2 = 5

HA: 1 - 2 > 5

Use = .05

Reject H0 (Accept HA) if z > z.05 = 1.645

Calculating zCalculating zRemember

Deviation Standard eAppropriat

Value) zed(HypothesiEstimate)(Point z

2

22

1

21

21

nσ

nσ

5)x-x(z

2.5

75

15

100

10

5540)-(550z

22

ConclusionConclusion

• Since 2.5 > 1.645– It can be concluded that the average starting salary for

secretaries in Chicago is at least $5 per week greater than the average starting salary in St. Louis.

• The p-value:– The area above z= 2.5 on the normal curve = 1 - .9938

= .0062– Since .0062 is low (compared to α), it can be concluded

that the average starting salary for secretaries in Chicago is at least $5 per week greater than the average starting salary in St. Louis.

EXAMPLEEXAMPLEConfidence Interval: Confidence Interval: 11, , 2 2 KnownKnown

• Construct a 95% confidence for the difference in average between weekly starting salaries for secretaries in Chicago and St. Louis.

2

22

1

21

.02521 n

σ

n

σz )x-x(

75

15

100

101.96 540)-(550

22

$10 ± $3.92

$6.08 ↔ $13.92

Excel ApproachExcel Approach

• Suppose, as shown on the next slide the data for Chicago is given in column A (A2:A101) and the data for St. Louis is given in column B (B2:B76).

• The analysis can be done using an entry from the Data Analysis Menu:

z-test: Two Sample for Meansz-test: Two Sample for Means

Select

z-Test: Two Sample For Means

Go to Data

Then Data Analysis

Select

z-Test: Two Sample For Means

For 1-tail tests, input columns so that the test is

a “>” test.Enter

Hypothesized Difference

Enter Variances

Not

Standard Deviations

Check

Labels

Enter

Beginning Cell

For Output

p-value for “>” test

p-value for “” test

=(E4-F4)-NORMSINV(0.975)*SQRT(E5/E6+F5/F6)

Highlight formula in cell E15—press F4.

Drag to cell E16 and change “-” to “+”.

Estimating Sample SizesEstimating Sample Sizes• Usual Assumptions:

– Same sample size from each pop.: n1 = n2 = n

– Standard deviations, 1 2 known

• Calculate n from the “±” part of the confidence interval for known 1and 2

En

σσ1.96

n

σ

n

σ1.96

22

21

22

21

ExampleExample• How many workers would have to be

surveyed in Chicago and St. Louis to estimate the true average difference in starting weekly salary to within $3?

surveyd! be would total workers278

cityeach in 139138.72(11.78)n

78.113

3251.96 n

3n

15101.96

n

σ

n

σ1.96

2

2222

21

ReviewReview• Mean and standard deviation for X1 -X2

• Assumptions for tests and confidence intervals

• z-tests for differences in means when 1 and 2 are known: – By Formula– By Excel data analysis tool

• Confidence intervals for differences in means when 1 and 2 are known:

– By Formula– By Excel data analysis tool

• Estimating Sample Sizes