Download Sampling Distribution and Confidence Interval 1

Sampling Distribution and Confidence Interval
Sampling Distr ibution
T he ore tical Probability Dis tribution of the Sa mple Statistic.
Sa mpling D istribution of Mea ns
If a random sample is taken from a normally distributed population that has a mean µ
and a standard deviation σ , the sampling distribution of the sample means, x , is normal
with
µ
µxx == µ
µ
σ
σ
σ
σxx ==
nn
W aiting tim es at a c ertain type of c linic are norm ally dis tribution wit h µµ = 8 m in. & σσ = 2
m in. If y ou s elec t random s am ples of 25 c as es , what is the s am pling dis tribution of t he
m ean?
If you select random samples of 25 cases from a normal distribution (or population) with
µ = 8 min. & σ = 2 min.
The sampling distribution of the mean is normal distribution with mean = 8 min., and
standard deviation = 2/5 = .4 min.
Example : W aiting tim es at a c ertain type of c lin ic are norm ally dis tribut ion with µµ = 8
m in. & σσ = 2 m in. If y ou s elec t random s am ples of 25 c as es , what is the probability th at
the sa mple me an of a rand om s am ple of 25 c as es is between 7.8 & 8.2 m inutes ?
xx −− µ
µ 77..88 −−88
==
== −−..50
50
σ
σ nn 22 25
25
x − µ 8. 2 − 8
zz == x − µ == 8. 2 − 8 == ..50
50
σ
σ nn 22 25
25
zz ==
The probability that the sample mean would be between 7.8 & 8.2 minutes is .383.
Central Limit Theorem
If a relative large random sample is taken from a population that has a mean µ and a
standard deviation σ , regardless of the distribution of the population, the distribution
of the sample means is approximately normal with
µ
µxx == µ
µ
σ
σ
σ
σxx ==
nn
Example: Consider the distribution of serum cholesterol levels for all 20- to 74-year-old
males living in United States has a mean of 211 mg/100 ml, and the standard deviation
of 46 mg/100 ml. If a random sample of 100 individuals from the population, what is the
probability that the average serum cholesterol level of these 100 individuals is higher
than 225?
Solution:
The sampling distribution of the mean is normal with mean = 211 and
standard error = 4.6 (= 46/10).
P( X > 225) = P(Z > [225 - 211]/4.6) = P(Z > 3.04) = 0.001
1
Sampling Distribution and Confidence Interval
A special notation:
z α = the z score that the proportion of the standard normal distribution to the right of it is
α.
z .025 = 1.96
z .010 = ?
Confide nce Inte rva l Mea n (σσ Know n)
σ
σ
σ
xx −− zzαα // 22 ⋅⋅ σ ≤≤ µ
µ ≤≤ xx ++ zzαα // 22 ⋅⋅
nn
nn
or
σ
xx ±± zzαα // 22 ⋅⋅ σ
nn
As s um ptions
• Population Sta ndard Devi ation Is Known ( It m ay c om e from prior s tudy.)
• Population Is Norm ally Dis tributed
• If Not Norm al, Can Be Approxim ated b y Norm al Dis tribution (n ≥ 30)
Example : The m ean of a ran dom s am ple of n = 25 is X = 50. Set up a 95% c onfid enc e
interval es tim ate for µ if σ = 10.
W e c an be 95% c onfident t hat the p opulat ion m ean is in (46. 08, 53.9 2).
W e c an be 95% c onfident t hat the m axim um s am pling error us ing this interval es tim ate
for es tim ating m ean is within 3.92.
σ
σ
σ
xx −− zzαα // 22 ⋅⋅ σ ≤≤ µ
µ ≤≤ xx ++ zzαα // 22 ⋅⋅
nn
nn
10
10 ≤ µ ≤ 50 + 1. 96 ⋅ 10
10
50
50 −−11..96
96⋅⋅
≤ µ ≤ 50 + 1. 96 ⋅
25
25
25
25
46
46..08
08 ≤≤ µ
µ ≤≤ 53
53..92
92
or
50 ± 3.92
Example : Yo u’re a Q/C ins pec tor for Coke. The σσ for 2-liter bottles is .05 liters . A
random s am ple of 100 bottles s howedx = 1.99 liters . W hat is the 90% c onfidenc e
interval es tim ate of the true me an am ount in 2-l iter bottl es ?
σ
σ
σ
xx −− ZZαα // 22 ⋅⋅ σ ≤≤ µ
µ ≤≤ xx ++ ZZαα // 22 ⋅⋅
nn
nn
..05
05 ≤ µ ≤ 1.99 + 1.645 ⋅ ..05
05
11..99
99 −−11..645
645⋅⋅
≤ µ ≤ 1.99 + 1.645 ⋅
100
100
100
100
11..982
≤
µ
≤
1
.
998
982 ≤ µ ≤ 1.998
or
1.99 ± .008
2
Sampling Distribution and Confidence Interval
Confidence Interval Mean (σσ Unknown)
1. As s um ptions
• Population Sta ndard Devi ation Is Unknown
• Population Mus t Be Normall y Distribute d
2. Us e Student’s t Dis tribution
3. Confidenc e Interv al Es tim ate
s
ss
xx −− ttαα // 22,,nn−−11 ⋅⋅ s ≤≤ µ
µ≤≤ xx ++ ttαα // 22,,nn−−11 ⋅⋅
nn
nn
or
σ
xx ±± zzαα // 22 ⋅⋅ σ
nn
Example: A ran dom s am ple of n = 25 has x = 50 & s = 8. Set up a 95% c onfidenc e
interval es tim ate for µ.
ss
ss
≤≤ µ
µ ≤≤ xx ++ ttαα // 22,,nn−−11 ⋅⋅
nn
nn
88
88
50
≤≤ µ
50 −− 22..0639
0639 ⋅⋅
µ ≤≤ 50
50 ++ 22..0639
0639⋅⋅
25
25
25
25
46
.
69
≤
µ
≤
53
.
30
46.69 ≤ µ ≤ 53. 30
or 50 ± 3.30
xx −−ttαα // 22,,nn−−11 ⋅⋅
Example : You’re a t im e s tudy analys t in m anufac turing. You’ve rec orded the fo llowing
tas k tim es (m in.): 3.6, 4.2, 4.0, 3.5, 3.8, 3.1.
W hat is the 90% c onfidenc e interval es tim ate of the popu lation me an tas k tim e?
 x = 3.7
s .d. = 0.38987
n = 6, df = n - 1 = 6 - 1 = 5
s / √ n = 0.38987 / √ 6 = 0.1592
t. 05, 5 = 2.015
( 3.7 - (2.015)(0.15 92) , 3.7 + (2.015)(0.1592) )
( 3.379, 4.0 21 )
or 3.7 ± .321
3
Sampling Distribution and Confidence Interval
One-side d confide nce inte rv al
Z C. I. : Lower i nter va l :
t C. I.:
− ∞ ≤ µ ≤ x + zα ⋅
σ
n
σ
≤ µ≤ ∞
n
Upper i nter va l :
x − zα ⋅
Lower i nter va l :
− ∞ ≤ µ ≤ x + tα ⋅
Upper i nter va l:
x − tα ⋅
s
n
s
≤ µ≤ ∞
n
Example : The m ean of a ran dom s am ple of n = 25 is x = 50. Set up a upper 95%
c onfidenc e interval es tim ate for µ if σ = 10.
1 − α = .95, α = .05, zα = 1.64.
σ
, − ∞)
n
10
( 50 − 1.64 ⋅
, − ∞)
25
( 46.72 , − ∞ )
( x − zα ⋅
M argin of Error:
C.I. : x ± zα ⋅
2
σ
n
Margin of Error = B = zα ⋅
2
n=
σ
n
zα2 ⋅ σ2
2
B2
Example : W hat s am ple s iz e is needed to be 90% c onfid ent of b eing c orrec t within ± 5?
A pi lot s tudy s ugges ted that t he s tandard de viati on is 45.
22
22
zz..0505 2σσ 22 (11..645
645) (45
45)
nn ==
==
== 219
219..22 ≅≅ 220
220
BB 22
(55) 22
2
Example : Yo u plan to s urvey em ploy ees to find th eir aver age aut o ins uranc e. Yo u want
to be 95% c onfide nt that t he s am ple me an is within ± $50 .
A pi lot s tudy s howed that σσ was about $400. W hat sa mple siz e do you us e?
nn ==
22 22
2
2
zz..025
σ
(11..96
96) 2 (400
400) 2
025 σ
=
== 245
=
245..86
86 ≅≅ 246
246
22
BB22
(50
50)
4