Sample Answers to Chapter 5 - Computer Sciencemlittman/courses/ml03/ans_ch...90% confidence interval...
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Sample Answers to Chapter 5 5.2 (1) Standard Deviation
( )( )(1 ( ))
ss s
error herror h error h
nσ −
≈
( / )*(1 / )r n r nn−
=
((100 83) /100)*(1 (100 83) /100)100
− − −=
= 0.03756 (2) Confidence Interval
100 83( ) 0.17100
srerror hn
µ −= = = =
When confidence interval = 95%, the constant 1.96Nz = .
Therefore, the confidence interval is ( ) ( )( ,N s N serror h error hz z )µ σ µ σ− + , which is (0.0964,
0.2436). 5.3
10( ) 0.153865
srerror hn
µ = = = =
( )( )(1 ( ))
ss s
error herror h error h
nσ −
≈
( / )*(1 / )r n r nn−
=
(10 / 65)*(1 10 / 65)65−
=
0.04475=
Two-sided bounds: When confidence interval = 90%, the constant 1.64Nz = .
90% confidence interval = ( ) ( )( ,N s N serror h error hz z )µ σ µ σ− + , which is (0.08041,
0.22719). One-sided bounds: Because 100(1-α)% confidence interval with lower bound L and upper bound U implies a 100(1-α/2)% confidence interval with upper bound U and no lower bound. Therefore, 95% one-sided confidence interval and 90% two-sided interval share the same upper bound U.
For 95% one-sided confidence interval, U = 0.22719.
Moreover, 90% one-sided interval and 80% two-sided interval share the same upper bound U. For 80% two-sided confidence interval, the constant . Thus the
upper bound U of 90% one-sided interval is
1.28Nz =
( )N serror hzµ σ+ = 0.21108.
5.4 Given 95% confidence interval, 1.96Nz = ; The range between 0.2 and 0.6 implies that
0.2 0.6 0.42
µ += = .
Moreover, ( ) ( )1.96*N s serror h error hU zµ σ µ σ= + = + (1)
( ) ( )1.96*N s serror h error hL zµ σ µ σ= − = − (2)
(1)-(2) represents the width of the two-sided 95% confidence interval, which is required to be smaller than 0.1,
→2 ( )*1.96* 0.1serror hσ ≤
→( )(1 ( ))*1.96* 0.1s serror h error h
n−
≤2
→(1 )*1.96* 0.1n
µ µ−≤2
→0.4*(1 0.4)*1.96* 0.1
n−
≤2
→ n 368.7936≥Thus the minimum number of examples should be 369.