m6 confidence intervals for one normal population mean

8/13/2019 m6 Confidence Intervals for One Normal Population Mean

1/26

Selang keyakinan bagi satu min populasi, bilatidak diketahui

Minggu 4


2/26

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


3/26

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


4/26

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


5/26

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

#


6/26

Student t

distributionwith n = 3

0

Student t

distributionwith n = 12

Standardnormaldistribution


7/26

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


8/26

x - E <


9/26

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






The t Distribution


10/26

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).


11/26

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


12/26

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.


13/26

Using the Normal and tDistribution


14/26

Bagi lengkung-t dengan darjah kebebasan13, tentukan t yakni cari nilai t yang

mempunyai luas 0.05 ke kanannya.


15/26

Andaian:

1. Populasi normal atau sampel yang

banyak 2. tidak diketahui.


16/26

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


17/26

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.)


18/26

x = 26, 227

s = 15, 873

= 0.05

/2= 0.025

Fi d t l f th t Di t ib ti


19/26

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


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






Find t value from the t Distribution


20/26

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 <


22/26

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.)


23/26

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.)


24/26

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.)


25/26

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.


26/26

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

m6 confidence intervals for one normal population mean

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