m6 Confidence Intervals for One Normal Population Mean
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Transcript of m6 Confidence Intervals for One Normal Population Mean
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Selang keyakinan bagi satu min populasi, bilatidak diketahui
Minggu 4
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Andaikan suatu pembolehubah x bagisatu populasi adalah bertaburan normaldengan min, maka, bagi sampel bersaiz
n, versi student bagi
Mempunyai taburan-t dengan (n-1)darjah kebebasan (df)
x
t =x -
s
n
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If the distribution of a population isessentially normal, then the distribution
of
is essentially a Student tDistribution for allsamples of size n.
is used to find critical values denoted by
t = x - s
n
t / 2
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The Degrees of Freedom ( f)corresponds to the number of
sample values that can vary after
certain restrictions have imposed
on all the other data values
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If ten numbers have a mean of 80, then thefirst nine numbers can be freely assigned butthe last number is determined by the first ninenumbers.
So, the df = 10 1 = 9.In general, df = N - 1
Any
#Specific#
Any
#Any
#Any
#Any
#Any
#Any
#Any
#Any
#
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Student t
distributionwith n = 3
0
Student t
distributionwith n = 12
Standardnormaldistribution
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Based on an Unknown and a Small Simple RandomSample from a Normally Distributed Population
1. t / 2has n - 1 degrees of freedom2. s is the sample standard deviation
nE = ts
2
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x - E <
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Degreesoffreedom
12
34567891011
121314151617181920
212223242526272829
Large (z)
63.6579.925
5.8414.6044.0323.7073.5003.3553.2503.1693.106
3.0543.0122.9772.9472.9212.8982.8782.8612.845
2.8312.8192.8072.7972.7872.7792.7712.7632.756
2.575
.005(one tail).01(two tails)
31.8216.965
4.5413.7473.3653.1432.9982.8962.8212.7642.718
2.6812.6502.6252.6022.5842.5672.5522.5402.528
2.5182.5082.5002.4922.4852.4792.4732.4672.462
2.327
12.7064.303
3.1822.7762.5712.4472.3652.3062.2622.2282.201
2.1792.1602.1452.1322.1202.1102.1012.0932.086
2.0802.0742.0692.0642.0602.0562.0522.0482.045
1.960
6.3142.920
2.3532.1322.0151.9431.8951.8601.8331.8121.796
1.7821.7711.7611.7531.7461.7401.7341.7291.725
1.7211.7171.7141.7111.7081.7061.7031.7011.699
1.645
3.0781.886
1.6381.5331.4761.4401.4151.3971.3831.3721.363
1.3561.3501.3451.3411.3371.3331.3301.3281.325
1.3231.3211.3201.3181.3161.3151.3141.3131.311
1.282
1.000.816
.765
.741
.727
.718
.711
.706
.703
.700
.697
.696
.694
.692
.691
.690
.689
.688
.688
.687
.686
.686
.685
.685
.684
.684
.684
.683
.683
.675
.01(one tail).02(two tails)
.025(one tail).05(two tails)
.05(one tail).10(two tails)
.10(one tail).20(two tails)
.25(one tail).50(two tails)
The t Distribution
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1. The Student tdistribution is different for differentsample sizes (eg for n= 3 and n= 12).
2. The Student tdistribution has the same general
symmetric bell shape as the normal distributionbut it reflects the greater variability (with widerdistributions) that is expected with smallsamples.
3. The Student t distribution has a mean of t= 0(just as the standard normal distribution has amean of z= 0).
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4. The standard deviation of the Student tdistributionvaries with the sample size and is greater than 1(the standard normal distribution has = 1).
5. As the sample size ngets larger, the Studenttdistribution gets closer to the normal distribution.For values of n> 30, the differences are so smallthat we can use the critical zvalues instead ofdeveloping a much larger table of critical tvalues.
Student tDistribution: Important Properties
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1. Jumlah luas di bawah lengkungt adalah 1
2. Lengkung-t tersebar ketakterhinggaan dalamkedua-dua arah, menghampiri, tetapi tidak
menyentuh paksi mengufuk.3. Lengkung-t simetri pada 0.4. Apabila darjah kebebasan (df) bertambah,
lengkung t semakin menghampiri bentuk
lengkung normal piawai.
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Using the Normal and tDistribution
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Bagi lengkung-t dengan darjah kebebasan13, tentukan t yakni cari nilai t yang
mempunyai luas 0.05 ke kanannya.
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Andaian:
1. Populasi normal atau sampel yang
banyak 2. tidak diketahui.
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Langkah 1:
Bagi aras keyakinan 1- , gunakanJadual t untuk mencari dengan df=n-1 di mana n adalah saiz sampel.
Langkah 2 : Selang keyakinan bagi adalah dari ke di mana t
/2diperolehi dari Langkah 1dan min dan s dikira dari sampel.
t/2
n
stx .
2
n
stx .
2
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A study of 12 Proto Sigma cars involved incollisions found that the repairs averaged RM
26,227 with a standard deviation of RM 15,873.
Find the 95% interval estimate of , the mean
repair cost for all Proto Sigma cars involved in
collisions. (The 12 cars distribution appears to be
bell-shaped.)
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x = 26, 227
s = 15, 873
= 0.05
/2= 0.025
Fi d t l f th t Di t ib ti
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Degreesoffreedom
12
34567891011
121314151617181920
212223242526272829
Large (z)
63.6579.925
5.8414.6044.0323.7073.5003.3553.2503.1693.106
3.0543.0122.9772.9472.9212.8982.8782.8612.845
2.8312.8192.8072.7972.7872.7792.7712.7632.756
2.575
.005(one tail).01(two tails)
31.8216.965
4.5413.7473.3653.1432.9982.8962.8212.7642.718
2.6812.6502.6252.6022.5842.5672.5522.5402.528
2.5182.5082.5002.4922.4852.4792.4732.4672.462
2.327
12.7064.303
3.1822.7762.5712.4472.3652.3062.2622.2282.201
2.1792.1602.1452.1322.1202.1102.1012.0932.086
2.0802.0742.0692.0642.0602.0562.0522.0482.045
1.960
6.3142.920
2.3532.1322.0151.9431.8951.8601.8331.8121.796
1.7821.7711.7611.7531.7461.7401.7341.7291.725
1.7211.7171.7141.7111.7081.7061.7031.7011.699
1.645
3.0781.886
1.6381.5331.4761.4401.4151.3971.3831.3721.363
1.3561.3501.3451.3411.3371.3331.3301.3281.325
1.3231.3211.3201.3181.3161.3151.3141.3131.311
1.282
1.000.816
.765
.741
.727
.718
.711
.706
.703
.700
.697
.696
.694
.692
.691
.690
.689
.688
.688
.687
.686
.686
.685
.685
.684
.684
.684
.683
.683
.675
.01(one tail).02(two tails)
.025(one tail).05(two tails)
.05(one tail).10(two tails)
.10(one tail).20(two tails)
.25(one tail).50(two tails)
Find t value from the t Distribution
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x = 26,227
s = 15,873
= 0.05
/2= 0.025
t/2= 2.201
E = t 2 s = (2.201)(15,873) = 10,085.29
n 12
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x - E <
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26,227 - 10,085.3 < < 26,227 + 10,085.3
x - E < < x + E
E = t 2 s = (2.201)(15,873) = 10,085.29
n 12
A study of 12 Proto Sigmas involved in collisions resulted in repairs averagingRM26,227 and a standard deviation of RM15,873. Find the 95% intervalestimate of , the mean repair cost for all Proto Sigmas involved in collisions.(The 12 cars distribution appears to be bell-shaped.)
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26, 227 - 10, 085.3 < < 26, 227 + 10, 085.3
x - E < < x + E
$16,141.7 < < $36,312.3
A study of 12 Proto Sigmas involved in collisions resulted in repairs averagingRM26,227 and a standard deviation of RM15,873. Find the 95% interval
estimate of , the mean repair cost for all Proto Sigmas involved in collisions.(The 12 cars distribution appears to be bell-shaped.)
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26, 227 - 10, 085.3 < < 26, 227 + 10, 085.3
x - E < < x + E
$16,141.7 < < $36,312.3
A study of 12 Proto Sigmas involved in collisions resulted in repairs averagingRM26,227 and a standard deviation of RM15,873. Find the 95% interval
estimate of , the mean repair cost for all Proto Sigmas involved in collisions.(The 12 cars distribution appears to be bell-shaped.)
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26, 227 - 10, 085.3 < < 26, 227 + 10, 085.3
x - E < < x + E
$16,141.7 < < $36,312.3We are 95% confident that this interval contains the average cost of repairinga Proto Sigma.
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Gunakan data berikut bagi mencari 95%selang keyakinan bagi min, ( n=25,
= 513.32, s = 262.23)
x
447 2 7 627 43 883313 844 253 397 214217 768 1 64 26 587833 277 8 5 653 549649 554 57 223 443