chapter 24
DESCRIPTION
Chapter 24. Comparing Means. 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 2 nd floor until she/he leaves. Why? Part of a Stat 101 data collection project. Inference for. - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 24
Comparing Means
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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 −
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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 −
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1. Female 2. MalePopulations
Samples
random selection
random
selectionIn
fere
nce
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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
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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
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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.
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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
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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.
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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
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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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SE y1 −y2( ) = SE y1( )⎡⎣ ⎤⎦2+ SE y2( )⎡⎣ ⎤⎦
2
= 3.4927[ ]2 + 3.5606[ ]
2
= 24.88 =4.988
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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.
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Table T
Confidence Levels 80% 90% 95% 98% 99%
df 1 2 3 4 28M
2.048
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Confidence Interval for 21 μμ −
( ) ( )( ) ( )
11.3 to55.23
22.1033.13
988.4048.22.6987.55
SE 21
*
21
−−
±−
±−
−±− yytyy
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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.
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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 −
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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.
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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
+=−
−
−−=
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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
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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 −
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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 −
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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
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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