1

Chapter 24

Comparing Means

2

Inference for Who? Students at I.S.U. What? Time (minutes). When? Fall 2000. Where? Lied Recreation Athletic Center. How? Measure time from when student arrives on 2nd floor until she/he leaves.

Why? Part of a Stat 101 data collection project.

21 μμ −

3

Inference for Do males and females at I.S.U. spend the same amount of time, on average, at the Lied Recreation Athletic Center?

21 μμ −

4

1. Female 2. MalePopulations

Samples

random selection

random

selectionIn

fere

nce

5

Time (minutes)

1. Females 2. Males

63, 32, 86, 53, 49 52, 75, 74, 68, 9373, 39, 56, 45, 67 77, 41, 87, 72, 5349, 51, 65, 54, 56 84, 65, 66, 69, 62

6

Time (minutes)

Sex=F Sex=M

Mean 55.87 Mean 69.20

Std Dev 13.527 Std Dev 13.790

Std Err Mean 3.4927 Std Err Mean 3.5606

N 15 N 15

7

Comment This sample of I.S.U. females spends, on average, 13.33 minutes less time at the Lied Recreation Athletic Center than this sample of I.S.U. males.

8

Conditions & Assumptions Randomization Condition 10% Condition Nearly Normal Condition Independent Groups Assumption

How were the data collected?

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Conditions & Assumptions Randomization Condition

Random sample of males. Random sample of females.

Independence Assumption Two separate random samples.

10% Condition

10

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Normal Quantile Plot

1

2

3

4

5

Count

30 40 50 60 70 80 90 100

Time (min)

Females

11

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Normal Quantile Plot

1

2

3

4

5

Count

30 40 50 60 70 80 90 100

Time (min)

Males

12

Nearly Normal Condition The female sample data could have come from a population with a normal model.

The male sample data could have come from a population with a normal model.

13

Confidence Interval for 21 μμ −

y1 −y2( ) ±t*SE y1 −y2( )

SE y1 −y2( ) =s1

2

n1

+s2

2

n2

SE y1 −y2( ) = SE y1( )⎡⎣ ⎤⎦2+ SE y2( )⎡⎣ ⎤⎦

2

14

SE y1 −y2( ) =s1

2

n1

+s2

2

n2

=13.527( )2

15+

13.792( )2

15= 24.88 =4.988

15

SE y1 −y2( ) = SE y1( )⎡⎣ ⎤⎦2+ SE y2( )⎡⎣ ⎤⎦

2

= 3.4927[ ]2 + 3.5606[ ]

2

= 24.88 =4.988

16

Finding t*

Use Table T. Confidence Level in last row. df = a really nasty formula (so the value will be given to you). df = 28 for our example.

17

Table T

Confidence Levels 80% 90% 95% 98% 99%

df 1 2 3 4 28M

2.048

18

Confidence Interval for 21 μμ −

( ) ( )( ) ( )

11.3 to55.23

22.1033.13

988.4048.22.6987.55

SE 21

*

21

−−

±−

±−

−±− yytyy

19

Interpretation We are 95% confident that I.S.U. females spend, on average, from 3.11 to 23.55 minutes less time at the Lied Recreation Athletic Center than I.S.U. males do.

20

Inference for Do males and females at I.S.U. spend the same amount of time, on average, at the Lied Recreation Athletic Center?

Could the difference between the population mean times be zero?

21 μμ −

21

Test of Hypothesis for Step 1: Set up the null and alternative hypotheses.

21 μμ −

0:or H :H

0:or H :H

21A21A

210210

≠−≠=−=

μμμμμμμμ

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Test of Hypothesis for Step 2: Check Conditions.

Randomization Condition•Two Independent Random Samples

10% Condition Nearly Normal Condition

21 μμ −

23

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Normal Quantile Plot

1

2

3

4

5

Count

30 40 50 60 70 80 90 100

Time (min)

Females

24

.01

.05

.10

.25

.50

.75

.90

.95

.99

-3

-2

-1

0

1

2

3

Normal Quantile Plot

1

2

3

4

5

Count

30 40 50 60 70 80 90 100

Time (min)

Males

25

Nearly Normal Condition The female sample data could have come from a population with a normal model.

The male sample data could have come from a population with a normal model.

26

Test of Hypothesis for Step 3: Compute the value of the test statistic and find the P-value.

21 μμ −

( )( )

( )2

2

2

1

2

121

21

21

SE

SE

0

n

s

n

syy

yy

yyt

+=−

−

−−=

27

Time (minutes)

Sex=F Sex=M

Mean 55.87 Mean 69.20

Std Dev 13.527 Std Dev 13.790

Std Err Mean 3.4927 Std Err Mean 3.5606

N 15 N 15

28

SE y1 −y2( ) =s1

2

n1

+s2

2

n2

=13.527( )2

15+

13.792( )2

15= 24.88 =4.988

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Test of Hypothesis for Step 3: Compute the value of the test statistic and find the P-value.

21 μμ −

( )( )

( )672.2

988.4

20.6987.55

SE

0

21

21 −=−

=−

−−=

yy

yyt

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Table T

df 1 2 3 4 28

M

Two tail probability 0.20 0.10 0.05 0.02 P-value 0.01

1.313 1.701 2.048 2.467 2.672 2.763

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Test of Hypothesis for Step 4: Use the P-value to make a decision. Because the P-value is small (it is between 0.01 and 0.02), we should reject the null hypothesis.

21 μμ −

32

Test of Hypothesis for Step 5: State a conclusion within the context of the problem. The difference in mean times is not zero. Therefore, on average, females and males at I.S.U. spend different amounts of time at the Lied Recreation Athletic Center.

21 μμ −

33

Comment This conclusion agrees with the results of the confidence interval.

Zero is not contained in the 95% confidence interval (–23.55 mins to –3.11 mins), therefore the difference in population mean times is not zero.

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Alternatives

)(Pr prob tailTwo ,:H

)(Pr prob tailOne ,:H

) (Pr prob tailOne ,:H

:H

21A

21A

21A

210

t

t

t

>≠

>><<

=

μμμμμμμμ

35

JMP Data in two columns.

Response variable: •Numeric – Continuous

Explanatory variable:•Character – Nominal

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JMP Starter Basic – Two-Sample t-Test

Y, Response: Time X, Grouping: Sex

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Time

30

40

50

60

70

80

90

100

F M

Sex

FM

Level 15 15

Number 55.8667 69.2000

Mean 13.5270 13.7903

Std Dev 3.4927 3.5606

Std Err Mean 48.376 61.563

Lower 95% 63.358 76.837

Upper 95%

Means and Std Deviations

F-MAssuming unequal variances

DifferenceStd Err DifUpper CL DifLower CL DifConfidence

-13.333 4.988 -3.116 -23.550

0.95

t RatioDFProb > |t|Prob > tProb < t

-2.6732627.98961

0.0124 0.9938 0.0062

-15 -10 -5 0 5 10 15

t Test

Oneway Analysis of Time By Sex