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115
Sampling Distribution Statistical Inference Suppose we want to know the average height of an Indian or the average life length of a bulb manufactured by a company, etc. obviously we cannot burn out every bulb and find the mean life length. One chooses at random, say n bulbs, find their lifelengths
nXXX ....., 21 and take the mean life length n
X....XXX n21 +++
= as an ‘approximation’
to the actual (unknown) mean life length. Thus we make a statement about the “population” (of all life lengths) by looking at a sample of it. This is the basis behind statistical inference. The whole theory of statistical inference tells us how close we are to the true (unknown) characteristic of the population. Random Sample of size n In the above example, let X be the lifelength of a bulb manufactured by the company. Thus X is a rv which can assume values > 0. It will have a certain distribution and a certain mean µ etc. When we make n independent observations, we get n values
nxxx ...., 21 . clearly if we again take n observations, we would get nyyy ...., 21 . Thus we
may say Definition Let X be a random variable. A random sample of size n from x is a finite ordered sequence { }n21 X....,X,X of n independent rv3 such that each Xi has the same
distributions that of X. Sampling from a finite population Suppose there is an universe having a finite number of elements only (like the number of Indians, the number of females in USA who are blondes etc.). A sample of size n from the above is a subset of n elements such that each subset of n elements has the same prob of being selected.
116
Statistics Whenever we sample, we use a characteristic of the sample to make a statement about the population. For example suppose the true mean height of an Indian is µ (cms). To make a
statement about µ , we randomly select n Indians, Find their heights { }n21 X....,X,X and
then their mean namely
n
X.....XXX n21 +++
=
We use then X as an estimate of the unknown parameter µ . Remember µ is a
parameter, a constant that is unchanged. But the sample mean X is a r.v. It may assume different values depending on the sample of n Indians chosen. Definition : Let X be a r.v. Let { }n21 X.....X,X be a sample of size n from X. A statistic
is a function of the sample { }n21 X,....,X,X .
Some Important Statistics
1. The sample mean n
X.....XXX n21 +++
=
2. The sample Variance ( )2
i
n
1i
2 XX1n
1S −
−= �
=
3. The minimum of the sample { }n21 X,....,X,XminK =
4. The maximum of the sample { }.X,......X,XmaxM n21=
5. The Range of the sample KMR −= Definition
If n1 X,.....X is a random sample of size n and if ∧X is a statistic, then we remember
∧X is
also a r.v. Its distribution is referred to as the sampling distribution of ∧X .
117
The Sampling Distribution of the Sample Mean X .
Suppose X is a r.v. with mean µ and variance n212 X.....X,XLet.σ be a random sample
of size n from X. Let n
X........XXX n21 +++= be the sample mean. Then
(a) ( ) .XE µ=
(b) ( ) .n
XV2σ=
(c) If n1 X....X is a random sample from a finite population with N elements, then
Var ( ) .1NnN
nX
2
−−σ=
(d) If X is normal, X is also normal
(e) Whatever be the distribution of X, if n is “large”
n
Xσ
µ−has approximately the
standard normal distribution. (This result is known as the central limit theorem.) Explanation
(a) tells us that we can “expect” the sample mean X to be an approximation to the population mean µ .
(b) tells us that the “nearness” of X to µ is small when the sample size n is
large.
(c) says that if X has a normal distribution.
n
Xσ
µ− has a standard normal
distribution.
(d) says that whatever be the distribution of X, discrete or continuous,
n
Xσ
µ−
has approximately standard normal distribution if n is large.
118
Example 1 (See exercise 6.14, page 207) The mean of a random sample of size n = 25 is used to estimate the mean of an infinite population with standard deviation .4.2=σ What can we assert about the prob that the error will be less than 1.2 if we use
(a) Chebyshev’s theorem (b) The central limit theorem?
Solution
(a) We know the sample mean X is a rv with ( ) ( )n
XVarandXE2σ=µ=
Chebyshev’s theorem tell us that for any r.v. T,
( ) ( )( )2k
11TVark|TET|P −≥−
Taking ,XT = and noting ( ) ( ) ,XETE µ==
( ) ( ) ( ),
254.2
nXvarTvar
22
=σ== we find
.k1
154.2
.kXP 2−≥��
���
� <µ−
Desired ( )?2.1XP <µ−
25
kgives2.154.2
.k ==
Thus we can assert using Chebyshev’s theorem that
( ) 84.025211
12.1XP425
==−≥<µ−
119
(b) Central limit theorem says 54.2
n
XX µ−=µ−σ
is approximately standard normal.
Thus ( )2.1XP <µ−
( ) 9876.019938.0215.2F2
125
F225
ZP
2.1XP
54.2
n
=−×=−×=
−��
���
�=��
���
� <≈
���
�
�
���
�
�<
µ−=
σ
Example 2 (See exercise 6.15 on page 207) A random sample of size 100 is taken from an infinite population having mean 76=µ
and variance .2562 =σ What is the prob that X will be between 75 and 78? Solution
We use central limit theorem namely n
Xσ
µ− is approximately standard normal.
Required ( )78X75P <<
��
�
�
��
�
� −<µ−<−=σ
1016
n1016
7678X7675P
8284.017340.08944.0
185
45
85
45
45
85
1620
1610
=−+=
−��
���
�+��
���
�=��
���
�−−��
���
�=
��
���
� <<−=��
���
� <<−≈
FFFF
ZPZP
120
Example 3 (See Exercise 6.17 on page 217) If the distribution of weights of all men travelling by air between Dallas and El Paso has a mean of 163 pounds and a s.d .of 18 pounds, what is the prob. That the combined gross weight of 36 men travelling on a plane between these two cities is more than 6000 pounds? Solution Let X be the weight of a man traveling by air between D and E. It is given that X is a rv with mean ( ) 163XE =µ= lbs and sd .lbs18=σ
Let 3621 X.....X,X be the weights of 36 men traveling on a plane between these two cities.
Thus we can regard { }3621 X.....,X,X as a random sample of size 36 from X.
Required ( )6000X.....XXP 3621 >+++
��
���
� >≈
��
�
�
��
�
� −>µ−=
��
���
� >=
σ
1822
ZP
163XP
366000
XP
618
61000
n
by central limit theorem
( )
1112.08888.01
22.111822
1
=−=
−=��
���
� ≤−= FZP
121
The sampling distribution of the sample mean X (when σ is unknown). Theorem
Let X be a rv having normal distribution with mean ( ) µ=XE . Let X be the sample
mean and S2 the sample variance of a random sample of size n form that of X.
Then the rv. n
S
Xt
µ−= has (student’s) t-distribution with n-1 degrees of freedom.
Remark
(1) The shape of the density curve of t-distribution (with parameter ν -greek nu) is like that of standard normal distribution and is symmetrical about the y-axis.
αν ,t is that
unique number such that
( ))parameterthe(
ttP ,v
→να=> α
By symmetry αναν ,1, 1 tt −=−
The values of αν ,t for various αυ and are tabulated in Table 4.
For ν large, ααν Zt ≈, .
Example 4 (See exercise 6.20 on page 213)
A random sample of size 25 from a normal population has the mean 5.47=x and the s.d. s = 8.4. Does this information tend to support or refute the claim that the mean of the population is ?1.42=µ
122
Solution:
ns
xt
µ−= has a t-distribution with parameter 1−= nν
Here 25,4.8,1.42 === nsµ
797.2tt 005.0,24005.0,1n ==α−
Thus ( ) 005.0797.2 =>tP
Or 005.0797.2X
Pn
s=
��
�
�
��
�
�>µ−
Or 005.054.8
797.21.42XP =��
���
� ×+>
Or ( ) 005.078.46XP =>
This means when 21.4=µ only in about 0.5 percent of the cases we may get an
78.46X > . Thus we will have to refute the claim 1.42=µ (in favour of )1.42>µ
Example 5 (See exercise 6.21 on page 213) The following are the times between six calls for an ambulance (in a certain city) and the patients arrival at the hospital : 27, 15,20, 32, 18 and 26 minutes. Use these figures to judge the reasonableness of the ambulance service’s claim that it takes on the average 20 minutes between the call for an ambulance and the patients arrival at the hospital. Solution Let X = time (in minutes) between the call for an ambulance and the patient’s arrival at the hospital. We assume X has a normal distribution. (When nothing is given, we assume normality). We want to judge the reasonableness of the claim that ( ) 20XE =µ= minutes.
For this we recorded the times for 6 calls. So we have a random sample of size 6 from X with
123
( )
.236
138
6/261832201527XThus.26X,18X,32X,20X,15X,27X 654321
==
+++++=======
( ) ( ) ( ) ( ) ( ) ( )[ ]
[ ]5
2049258196416
51
23262318233223202315232716
1 2222222
=+++++=
−+−+−+−+−+−−
=S
Hence 5
204=S
We calculate
150.16/
2023xt
5204
ns
=−=µ−=
Now 05.0015.2,5,1 ===− ααα forttn
= 10.0476.1 =αfor
Since our observed 10.5150.1 tt <=
We can say that it is reasonable to assume that the average time is 20=µ minutes
Example 6 A process for making certain bearings is under control if the diameters of the bearings have a mean of 0.5000 cm. What can we say about this process if a sample of 10 of these bearings has a mean diameter of 0.5060 cm and sd 0.0040 cm?
( ),504.0506.0
01.0504.0492.0
01.025.35.0
25.3.int10
004.
>=
=<<
=��
�
�
��
�
�<−<−
XSince
xPor
XPH
the process is not under control.
124
Sampling Distribution of S2 (The sample variance) Theorem If S2 is the sample variance of a random sample of size n taken from the normal
population with (population) variance ,2σ then
( ) ( )2
i
n
1i22
22 XX
1S1n −
σ=
σ−=Χ �
=
is a random variable having chi-square distribution with parameter .1−= nν Remark
Since S2 > 0, the rv has +ve density only to right of the origin. 2,ανΧ is that unique
number such that ( ) α=Χ>Χ αν2
,2P and is tabulated for some ss and να in table 5.
Example 7 (See exercise 6.24 on page 213) A random sample of 10 observations is taken from a normal population having the
variance 5.422 =σ . Find approximately the prob of obtaining a sample standard deviation S between 3.14 and 8.94 Solution Required ( )94.814.3 << SP
( ( ) ( ) )
( ) ( ) ( )
( )925.16088.2
94.85.42
9114.3
5.429
94.814.3
2
222
2
222
<Χ<=
��
���
� ×<−<×=
<<=
P
Sn
P
Sp
σ
(From Table 5, 088.299.0,,919.1605 29
29 =Χ=Χ )
( ) ( )( ))(94.005.099.0
919.16088.2 22
approx
approxPP
=−=>Χ−>Χ=
125
Example 8 (See exercise 6.23 on page 213)
The claim that the variance of a normal population is 3.212 =σ is rejected if the variance of a random sample of size 15 exceeds 39.74. What is the prob that the claim
will be rejected even though ?3.212 =σ Solution The prob that the claim is rejected
( )
( ) ( )
( )12.21,5tablefromAs025.0
12.21P74.393.21
14S
1nP
74.29SP
2025.0,14
222
2
=Χ=
>Χ=��
���
� ×>σ−=
>=
Theorem
If 22
21 , SS are the variances of two independent random samples of sizes 21 ,nn
respectively taken from two normal populations having the same variance, then
22
21
SS
F =
is a rv having the (Snedecor’s) F distribution with parameters 11 2211 −=−= nandn νν
Remark
1. 11 −n is called the numerator degrees of freedom and 12 −n is called the
denominator degrees of freedom.
2. If F is a rv having ( )21 ,νν degrees of freedom, then ανν ,, 21F is that unique number
such that
126
( ) αανν => ,21FFP and is tabulated for 05.0=α in table 6(a) and for 01.0=α in table
6(b).
We also note the fact : ανν
ανν−
=1,,
,,
21
22
1F
F
Thus 36.077.211
05.0,10,2095.0,20,10 ===
FF
Example 9
(a) 38.062.211
05.0,12,1595.0,15,12 ===
FF
(b) 135.040.711
01.0,6,2099.0,20,6 ===
FF
Example 10 (See Exercise on page 213) If independent random samples of size 821 == nn come from two normal populations
having the same variance, what is the prob that either sample variance will be at least seven times as large as the other? Solution
Let 22
21 , SS be the sample variances of the two samples.
Reqd ( )21
22
22
21 S7SORS7SP >>
( )72
77 21
22
22
21
>=
���
����
�>>=
FP
SS
orSS
P
where F is a rv having F distribution with (7,7) degrees of freedom = 2 x 0.01 = 0.02 (from table 6(b)).
127
Example 11 (see exercise 6.38 on page 215) If two independent random samples of size 169 21 == nandn are taken from a normal
population, what is the prob that the variance of the first sample will be at least four times as large as the variance of the second sample?
Hint : Reqd prob ( )22
21 4SSP >=
( )
( )4Fas01.0
4FP4SS
P
01.0,15,8
22
21
==
>=���
����
�>=
Example 12 (See Exercise 6.29 on page 214) The F distribution with (4,4) degrees of freedom is given by
( ) ( )�
≤>+=
−
00016 4
F
FFFFf
If random samples of size 5 are taken from two normal populations having the same variance, find the prob that the ratio of the larger to the smaller sample variance will exceed 3? Solution
Let 22
21 S,S be the sample variance of the two random samples.
Reqd ( )21
22
22
21 33 SSorSSP >>
( )3232 22
21 >=��
�
����
�>= FP
SS
P
where F is a rv having (4,4) degrees of freedom
128
( ) ( ) ( )
( ) ( )
165
192125
1921
321
12
F131
F121
12
dFF1
1F1
112dF
F1F6
2
32
3434
3
=×=��
��
� −=
��
��
�
++
+−=
��
��
�
+−
+=
+= ��
∞∞
Inferences Concerning Means We shall discuss how we can make statement about the mean of a population from the knowledge about the mean of a random sample. That is we ‘estimate’ the mean of a population based on a random sample. Point Estimation Here we use a statistic to estimate the parameter of a distribution representing a population. For example if we can assume that the lifelength of a transistor is a r.v. having exponential distribution with (unknown) parameter ββ , can be estimated by
some statistic, say X the mean of a random sample. Or we may say the sample mean is an estimate of the parameter β .
Definition
Let θ be a parameter associated with the distribution of a r.v. A statistic ∧θ (based on a
random sample of size n) is said to be an unbiased estimate ( ≡ estimator) of θ if
θθ =��
���
� ∧E . That is,
∧θ will be on the average close to θ .
Example
Let X be a rv; µ the mean of X. If X is the sample mean then we know ( ) µ=XE . Thus
we may say the sample mean X is an unbiased estimate of µ (Note X is a rv, a
statistic, n
X.....XXX n21 +++= a function of the random sample
129
( ) n21n21 ....,If.X.....,X,X ωωω are any n non-ve numbers 1≤ such that
,1...... n21 =ω++ω+ω then we can easily see that nn2211 x.....xx ω++ω+ω is also an
unbiased estimate of µ . (Prove this). X is got as a special case by taking
.1
....21 nn ==== ωωω Thus we have a large number of unbiased estimates for µ .
Hence the question arises : If ∧∧
21 ,θθ are both unbiased estimates of θ , which one do we
prefer? The answer is given by the following definition. Definition
Let ∧∧
21 ,θθ be both unbiased estimates of the parameter θ . We say ∧θ is more efficient than
.VarVarif 212 ��
���
�θ≤��
���
�θθ∧∧∧
Remark That is the above definition says prefer that unbiased estimate which is “more closer” to
θ . Remember the variance is a measure of the “closeness’ of X∧θ to θ .
Maximum Error in estimating Xbyµ
Let X be the sample mean of a random sample of size n from a population with
(unknown) mean µ . Suppose we use X to estimate µ . X - µ is called the error in
estimating µ by X . Can we find an upperbound on this error? We know if X is normal
(or if n is large) then by Cantral Limit Theorem.
n
Xσ
µ− is a r.v. having (approximately) the standard normal distribution. And we can say
α−=��
�
�
��
�
�<µ−<− αα σ
1ZX
ZP22
n
130
Thus we can say with prob ( )α−1 that the max absolute error µ−X in estimating µ by
X is atmost n
Zσ
α2
. (Here obviously we assume, σ the population s.d. is known. And
2αZ is that unique no. such that ( )
2ZZP
2
α=> α .
We also say that we can say with ( )α−1100 percent confidence that the max. abs error is
atmost n
Zσ
α2
. The book denotes, this by E.
Estimation of n Thus to find the size n of the sample so that we may say with ( )α−1100 percent
confidence, the max. abs. error is a given quantity E, we solve for n, the equation
.2
En
Z =σα
or n
2
2
��
�
��
�
�=
E
Z σα
Example 1 What is the maximum error one can expect to make with prob 0.90 when using the mean
of a random sample of size n = 64 to estimate the mean of a population with ?56.22 =σ Solution
Substituting 645.1ZZand6.1,64n 05.02
===σ= α (Note 90.01 =−α implies 05.02
=α)
in the formula for the maximum error n
ZEσ
α2
= we get
3290.02.0645.186.1
445.164
6.1645.1 =×=×=×=E
Thus the maximum error one can expect to make with prob 0.90 is 0.3290.
131
Example 2 If we want to determine the average mechanical aptitude of a large group of workers, how large a random sample will we need to be able to assert with prob 0.95 that the sample mean will not differ from the population mean by more than 3.0. points? Assume that it is known from past experience that .200=σ Solution
Here 95.01 =−α so that 025.02
=α, hence 96.1025.0
2
== ZZ α
Thus we want n so that we can assert with prob 0.95 that the max error E = 3.0
74.1703
2096.1 22
2 =��
��
� ×=��
�
��
�
�=∴
E
Zn
σα
Since n must be an integer, we take it as 171. Small Samples If the population is normal and we take a random sample of size n (n small) from it, we note
ns
Xt
µ−= ( X sample mean, S = Sample s.d)
is a rv having t-distribution with (n-1) degrees of freedom.
Thus we can assert with prob α−1 that 22
,1,1 αα −−≤nn
twherett is that unique no such that
( )22
,1
αα => −n
ttP . Thus if we use X to estimate µ , we can assert with prob ( )α−1 that
the max error will be
n
StE
n2
,1 α−=
(Note : If n is large, then t is approx standard normal. Thus for n large, the above
formula will become n
SZE
2α= )
132
Example 3 20 fuses were subjected to a 20% overload, and the times it took them to blow had a
mean x = 10.63 minutes and a s.d. S = 2.48 minutes. If we use x = 10.63 minutes as a point estimate of the true average it takes for such fuses to blow with a 20% overload, what can we assert with 95% confidence about the maximum error? Solution
Here n = 20 (fuses) x = 10.63, S = 2.478
95.010095
1 ==−α so that 025.02
=α
Hence 093.2025.0,19,1
2
==− ttn α
Hence we can assert with 95% confidence (ie with prob 0.95) that the max error will be
16.120
48.2093.2
2,1
=×== − n
StE
n α
Interval Estimation
If X is the mean of a random sample of size n from a population with known sd σ , then we know by central limit theorem,
n
XZ
σ
µ−=
is (approximately) standard normal. So we can say with prob ( )α−1 that
22
ZX
Zn
αα <µ−<−σ
.
which can be rewritten as
22
Zn
XZn
X αα
σ+<µ<σ−
133
Thus we can assert with Prob ( ) ( )( )confidencewithie %1001.1 ×−≡− αα that µ lies in
the interval .Zn
X,Zn
X22��
���
� σ+−σ− αα
We refer to the above interval as a ( ) %1001 α− confidence interval for µ . The end
points 2
Zn
X α
σ± are known as ( ) %1001 α− . confidence limits for µ .
Example 4 Suppose the mean of a random sample of size 25 from a normal population (with 2=σ )
is x = 78.3. Obtain a 99% confidence interval for µ , the population mean.
Solution
Here ( ) 99.010079
1,2,25 ==−== ασn
575.2005.02 005.0
2
==∴=∴ ZZ α
α
x = 78.3 Hence a 99% confidence interval for µ is
( )
( )33.79,27.77
0300.13.78,0300.13.78
25
2575.23.78,
25
2575.23.78
,22
=
+−=
���
����
� ×+×−=
���
����
� +−n
Zxn
Zxσσ
αα
134
σ unknown
Suppose X is the sample mean and S is the sample sd of a random sample of size n taken from a normal population with (unknown) mean µ . Then we know the r.v.
n
sX
tµ−=
has a t-distribution with (n-1) degrees of freedom. Thus we can say with prob α−1 that
22,1,1 αα −− <<−
nnttt
or 2
,1n,1nt
n
SX
t2
α−− <µ−<− α
orn
StX
n
StX
2,1n
2,1n
α−
α−
+<µ<−
Thus a ( ) %1001 α− confidence interval for µ is
���
����
�+− α
−α
− n
StX,
n
StX
2,1n
2,1n
Note :
(1) If n is large, t has approx the standard normal distribution. In which case the ( ) %1001 α− confidence interval for µ will be
���
����
�+−
n
SZx
n
SZx
22
, αα
(2) If nothing is mentioned, we assume that the sample is taken from a normal
population so that the above is valid.
135
Example 5 Material manufactured continuously before being cut and wound into large rolls must be monitored for thickness (caliper). A sample of ten measurements on paper, in mm, yielded
32.2, 32.0, 30.4, 31.0, 31.2, 31.2, 30.3, 29.6, 30.5, 30.7 Obtain a 95% confidence interval for the mean thickness. Solution Here n = 10
262.2
025.02
95.01
7880.041.30
0025.0,92
,1==∴
==−
==
−tt
or
Sx
nα
αα
Hence a 95% confidence interval for µ is
( )46.31,34.30
10
7880.0262.29.30,
10
7880.0262.29.30
=
���
����
� ×+×−
Example 6: Ten bearings made by a certain process have a mean diameter of 0.5060 cm with a sd of 0.0040 cm. Assuming that the data may be looked upon as a random sample from a normal population, construct a 99% confidence interval for the actual average diameter of bearings made by this process.
136
Solution
Here 0040.0,5060.0,10 === Sxn
( )
250.3tt
005.0Hence.99.010099
1
005.0,92
,1n==∴
=α==α−
α−
Thus a 99% confidence interval for the mean
( )5101.0,5019.0
10
0040.0250.35060.0,
10
0040.0250.35060.0
,2
,12
,1
=
���
����
� ×+×−=
���
����
�+−=
−− n
stx
n
Stx
nnαα
Example 7 In a random sample of 100 batteries the lifetimes have a mean of 148.2 hours with a s.d. of 24.9 hours. Construct a 76.60% confidence interval for the mean life of the batteries. Solution
Here 9.24,2.148,100 === Sxn
19.1
1170.02
7660.100
60.761
1170.01170.0,992
,1=≈=
===−
−ZttThus
thatso
nα
αα
Hence a 76.60% confidence interval is
( ).2.151,2.145100
9.2419.12.148,
100
9.2419.12.148
=
���
����
� ×+×−
137
Example 8 A random sample of 100 teachers in a large metropolitan area revealed a mean weekly salary of $487 with a sd of $48. With what degree of confidence can we assert that the average weekly salary of all teachers in the metropolitan area is between $472 and $502? Solution Suppose the degree of confidence is ( ) %1001 ×−α
Thus 502$2
,1=+
− n
Stx
nα
Here 100,48,487 === nSx
22,99
αα Zt ≈∴
Thus we get 5021048
4872
=+ αZ
Or 125.38.4
15
2
==αZ
9982.010009.02
=−=∴ ααor
∴ We can assert with 99.82% confidence that the true mean salaries will be between $472 and $502. Maximum Likelihood Estimates (See exercise 7.23, 7.24) Definition Let X be a rv. Let ( ) ( )xXPxf ==θ, be the point prob function if X is discrete and let
( )θ,xf be the pdf of X if X is continuous (here θ is a parameter). Let n21 X.....X,X be a
random sample of size n from X. Then the likelihood function based on the random sample is defined as
138
( ) ( ) ( ) ( ) ( ).,xf.....,xf,xf;x,....x,xLL n21n21 θθθ=θ=θ
Thus the likelihood function ( ) ( ) ( ) ( )nn xxPxxPxxPL ==== ...2211θ if X is discrete and
is the joint pdf of n1 X,...X when X is continous. The maximum likelihood estimate
(MLE)of θ is that ∧θ which maximizes ( )θL .
Example 8 Let X be a rv having Poisson distribution with parameter λ .
Thus ( ) ( ) .......2,1,0x;!x
exXP,xfx
=λ===λ λ−
Hence the likelihood function is
( )!
....!! 21
21
n
xxx
xe
xe
xeL
nλλλλ λλλ −−−=
.......2,1,0x;!x!.....x!x
ei
n21
x....xxn n21
=λ=+++λ−
To find ∧λ the value of λ which maximizes ( )λL , we use calculus.
First we take ln (log to base e� natural logarithm)
( ) ( ) ( )!x!....xlnlnx.....xnLln n1n1 −λ+++λ−=λ
Differentiating w.r.t. λ (noting nxx .....1 are not be varied)
We get ( )( )
λλλnxx
nL
L....1 1 +
+−=∂∂
nx....x
gives0 n1 ++=λ=
139
We can ‘easily’ verify 2
2
λ∂∂ L
is <0 for this λ .
Hence the MLE of λ is xn
xx n =+
=∧ ....1λ (The sample mean)
Example 9 MLE of Proportion Suppose p is the proportion of defective bolts produced by a factory. To estimate p, we proceed as follows. We take n bolts at random and calculate fD = Sample proportion of defectives.
noneschosenntheamongfounddefectivesofNo=
we show fD ist he MLE of p. We define a rv X as follows.
�
=defectiveischosenbolttheif1
defectivenotischosenbolttheif0X
Thus X has the prob distribution
x 0 1
Prob 1-p p
It is clear that the point prob function
( )( )Xofpxf ; is given by
( ) ( ) 1,0x;p1pp;xf x1x =−= −
(Note ( ) ( ) ( ) ( ) )p1xP1;xf&p10xP0;xf ===−===
Choosing n bolts at random amounts to choosing a random sample { }n21 X...,X,X from X
where Xi = 0 if the ith bolt chosen is not defective and = 1 if it is defective (I=1,2…n).
140
Hence n21 X....XX ++ (can you guess?)
= no of defective bolts among the n chosen. The likelihood function of the sample is
( ) ( ) ( ) ( )
( ) ( )
( ) ( )n1sns
ix...xxnx...x
n21
x....xsp1p
n,....1iallfor1or0xp1p
p;xf.....p,xfp;xfpL
n21nT1
+=−=
==−=
=
−
+++−+
Taking ln and differentiating (partially) wrt p, We get
( )psn
ps
pL
L −−−=
∂∂
11
for maximum, p1sn
ps
or0pL
−−==
∂∂
(i.e) n
x.....xxns
p n21 +++==
defectivesofproportionSample
nchosenntheamongdefectivesofNo
=
=
(One can easily see this p makes 0pL2
2
<∂∂
so that L is maximum for this p).
Example 10 Let X be a rv having exponential distribution with parameter β (unknown). Hence the
density of X is ( ) ( )01
; >=−
xexfxβ
ββ
141
Let { }n21 X.....,X,X be a random sample of size n. Hence the likelihood function is
( ) ( ) ( ) ( )
( )
( )01
;....;;
....
21
21
>=
=
+++−
i
xxx
n
n
xe
xfxfxfL
n
β
β
ββββ
Taking ln and differentiating (partially) w.r.t. ,β we get
( )imummaxfor0x....xnL
L1
2n1 =
β++
+β
−=β∂
∂
gives xn
xxx n =+++
=....21β
Thus the sample mean x is the MLE of β .
Example 11 A r.v. X has density
( ) ( ) 1x0;x1;xf <<+β=β β
Obtain the ML estimate of β based on a random sample { }n21 X.....X,X of size n from
x. Solution The likelihood function is
( ) ( ) ( ) 1x0;x...xx1L in21n <<+β=β β
Taking ln and differentiating (partially) wrt ,β we get
142
( )
( )
( )nn1
n1
x.....xln1
1gives
imummaxbetoLfor0
x.......xln1
nLL1
−−=β
=
++β
=β∂
∂
which is the ML estimate for β .
So far we have considered situations where the ML estimate is got by differentiating L (and equalizing the derivative) to zero. The following example is one where the differentiation will not work. Example 12 A rv X has uniform density over [ ]β,0
(ie) The density of X is ( ) β≤≤β
=β x0;1
;xf (and 0 elsewhere)
The likelihood function based on a random sample of size n from X is
( ) ( ) ( ) ( )
β≤≤β≤≤β≤≤β
=
βββ=β
n21n
n21
x0....,,x0,x0;1
;xf.....;xf;xfL
This is a maximum when the Dr is least (ie) when β is least. But nx ii ....2,1=∀>β
Hence the least β is max { }nxx .....1 which is the MLE of β
143
Estimation of Sample proportion We have just in the above seen if p = population proportion (i.e proportion of persons, things etc. having a characteristics) then the ML estimate of p = sample proportion Now we would like to find a ( )α−1 100% confidence interval for p.
(This is treated in chapter 9 of your text book) Large Samples Suppose we have a ‘dichotomous’ universe; that is a population whose members are either “haves” on “have – nots”; that is a member has a property or not. For example we can think of a population of all bulbs produced by a factory. Any bulb is either a “have” (ie defective) or is a “have-not” (ie it is good) and p = proportion of haves = “Prob that a randomly chosen member is a “have”. As another example, we can think of a population of all females in USA. A member is a “have” ( = 0) is a blond or is a “have-not “ (=is not a blond). As a last example, consider the population of all voters in India. A member is a “have” if he follows BJP and is a “have-not” otherwise. To estimate p, we choose n members at random and count the number X of “haves”. Thus X is a rv having binomial distribution with parameters n and p!
( ) ( ) ( ) n.....2,1,0x;p1pxn
p;xfxXP xnx =−���
����
�=== −
and if n is large, we know “standardized Binomial � standard normal”
(ie) for large n , ( )p1np
npX
−−
has approx standard normal distribution. So we can say with
prob ( )α−1 that
144
( )
( ) 22
22
1
1
αα
αα
z
npp
pnx
zor
zpnp
npxz
−
−<−
<−
−<−
or ( ) ( )
npp
znx
pn
ppz
nx −+<<−− 11
22αα
In the end points, we replace ‘p’ by the MLE nX
(=sample proportion)
Thus we can say with prob ( )α−1 that
nnx
1nx
znx
pn
nx
1nx
znx
22
��
���
� −+<<
��
���
� −− αα
Hence a ( ) %1001 α− confidence interval for p is
�����
�
�
�����
�
���
���
� −+
��
���
� −− αα n
nx
1nx
znX
,n
nx
1nx
znx
22
Remark : We can say with prob ( )α−1 that the max error pnX − in approximating p by
nX
is
( )n
ppZE
−= 1
2α
We can replace p by nX
and say the
145
Max error = n
nX
1nX
Z2
��
���
� −
α
Or we note that ( ) ( )101 ≤≤− pforpp is a maximum 41
(which is obtained when 21=p )
Thus we can also say with prob ( )α−1 that the max error.
nZ
41
2α=
This last equation tell us that to assert with prob ( )α−1 that the max error is E, n must be
2
2
E
Z
41
���
�
���
�
� α
Example 13 In a random sample of 400 industrial accidents, it was found that 231 were due at least partially to unsafe working conditions. Construct a 99% confidence interval for the corresponding true proportion p. Solution Here ( ) 99.01,231x,400n =α−==
so that 575.2005.02 2
== αα
Zhence
Thus a 99% confidence interval for p will be
�����
�
�
�����
�
���
���
� −+
��
���
� −− αα n
nx
1nx
Znx
nnx
1nx
Znx
22
146
( )6411.0,5139.0
400400231
1400231
575.2400231
400400231
1400231
575.2400231
=
�����
�
�
�����
�
���
���
� −+
��
���
� −−=
Example 14 In a sample survey of the ‘safety explosives’ used in certain mining operations, explosives containing potassium mitrate were found to be used in 95 out of 250 cases. If
38.025095 = is used as an estimate of the corresponding true proportion, what can we say
with 95% confidence about the maximum error? Solution Here n = 250, X = 95, 95.01 =−α
so that 025.02
=α ; hence 96.1
2
=αZ
Hence we can say with 95% confidence that the max. error is n
nx
nx
ZE��
���
� −=
1
2α
0602.0
25062.038.0
96.1
=
××=
Example 15: Among 100 fish caught in a large lake, 18 were inedible due to the pollution of the
environment. If we use 18.010018. = as an estimate of the corresponding true proportion,
with what confidence can we assert that the error of this estimate is atmost 0.065?
147
Solution Here 065.0Eerrormax18X,100n ====
We note 100
82.18.Z
nnX
1nX
ZE22
×=��
���
� −= αα
69.103842.0
065.0
03842.0
2
2
==∴
×=
α
α
Z
Z
Hence 0455.09545.012
=−=α
9190.010910.0 =−=∴ αα or
So we can assert with ( ) %9.91%1001 =×−α confidence that the error is at most 0.065.
Example 16 What is the size of the smallest sample required to estimate an unknown proportion to within a max. error of 0.06 with at least 95% confidence? Solution
Here 025.02
or95.01;06.0E =α=α−=
96.1025.0
2
==∴ ZZα
Hence the smallest sample size n is
148
77.266
06.096.1
41
41
22
2
=
��
��
�=���
�
���
�
�
=E
Z
nα
Since n must be an integer, we take the size to be 267. Remark Read the relevant material in your text on pages 279-281 of finding the confidence interval for the proportion in case of small samples. Tests of Statistical Hypothesis In many problems, instead of estimating the parameter, we must decide whether a statement concerning a parameter is true of false. For instance one may like to test the truth of the statement: The mean life length of a bulb is 500 hours. In fact we may even have to decide whether the mean life is 500 hours or more (!) In such situations, we have a statement whose truth or falsity we want to test. We then say we want to test the null hypothesis H0 = the mean life lengths is 500 hours (Here onwards, when we say we want to test a statement, it shall mean we want to test whether the statement is true). We then have another (usually called alternative) hypothesis. Make some ‘experiment’ and on the basis of that we will ‘decide’ whether to accept the null hypothesis or reject it. (When we reject the null hypothesis we automatically accept the alternative hypothesis). Example Suppose we wish to test the null hypothesis H0 = The mean life length of a bulb is 500 hours against the alternative H1 = The mean life length is > 500 hours. Suppose we take a random sample of 50 bulbs and found that the sample mean is 520 hours. Should we accept H0 or reject H0 ? We have to note that even though the population mean is 500 hours the sample mean could be more or less. Similarly even though the population mean is > 500 hours, say 550 hours, even then the sample mean could be less than 550 hours. Thus whatever decision we may make, there is a possibility of making an error. That is
149
falsely rejecting H0 (when it should have been accepted) and falsely accepting H0 (when it should have been rejected). We put this in a tabular form as follows:
Accept H0 Reject H0
H0 is true Correct Decision Type I error
H0 is false Type II Error Correct Decision
Thus the type I error is the error of falsely rejecting H0 and the type II error is the error of falsely accepting H0. A good decision ( ≡ test) is one where the prob of making the errors is small. Notation The prob of committing a type I error is denoted by α . It is also referred to as the size of the test or the level of significance of the test. The prob of committing Type II error is denoted by β .
Example 1 Suppose we want to test the null hypothesis 80=µ against the alternative hyp 83=µ on
the basis of a random sample of size n = 100 (assume that the population s.d. 4.8=σ )
The null hyp. is rejected if the sample mean 82>x ; otherwise is is accepted. What is the prob of typeI error; the prob of type II error? Solution
We know that when 80=µ (and 4.8=σ ) the r.v.
n
Xσ
µ− has a standard normal
distribution. Thus, P (Type I error) =P (Rejecting the null hyp when it is true)
150
( )
( )
( ) 0087.9913.0138.2ZP1
38.2ZP
104.88082
n
XP
80given82XP
=−=≤−=
>=
����
�
�
����
�
�
−>σ
µ−=
=µ>=
Thus in roughly about 1% of the cases we will be (falsely) rejecting H0. Recall this is also called the size of the test or level of significance of the test. P (Type II error) = P (Falsely accepting H0)
= P (Accepting H0 when it is false)
( )
( )
1170.08830.01)19.1Z(P1
19.1ZP
104.88382
n
XP
83given82XP
=−=≤−=
≤=
����
�
�
����
�
�
−≤σ
µ−=
=µ≤=
Thus roughly in 12% of the cases we will be falsely accepting H0. Definition (Critical Region)
In the previous example we rejected the null hypothesis when 82>x (i.e.) when x lies in the ‘region’ x>82 (of the x axis). This portion of the horizontal axis is then called the critical region and denoted by C. Thus the critical region for the above situation is
{ }82>= xC and remember we reject H0 when the (test) statistic X lies in the critical
151
region (ie takes a value > 82). So the size of the critical region ( ≡ prob that X lies in C) is the size of the test or level or significance. The shaded portion is the critical region. The portion �is the region of false acceptance of H0. Critical regions for Hypothesis Concerning the means Let X be a rv having a normal distribution with (unknown) mean µ and (known) s.d. σ .
Suppose we wish to test the null hypothesis 0µµ = .
The following tables given the critical regions (criteria for rejecting H0) for various alternative hypotheses. Null hypothesis : 0µµ = (Normal population σ known)
n
xZ σ
µ0−=
Alternative Hypothesis H1
Reject H0 if Prob of Type I error Prob of type II error
( )01 µµµ <= αZZ −< α
����
�
�
����
�
�
−−
− ασµµ
Z
n
F 101
0µµ < αZZ −< α
01 µµµ >= αZZ > α
����
�
�
����
�
�
+−
ασµµ
Z
n
F 10
0µ>µ αZZ > α
0µµ ≠
2
2
α
α
ZZor
ZZ
>
−< α
...
152
F(x) = cd f of standard normal distribution. Remark: The prob of Type II error is blank in case H1 (the alternative hypothesis) is one of the following three things = 000 ,, µµµµµµ ≠>< . This is because the Type II error can
happen in various ways and so we cannot determine the prob of its occurrence. Example 2: According to norms established for a mechanical aptitude test, persons who are 18 years old should average 73.2 with a standard deviation of 8.6. If 45 randomly selected persons averaged 76.7 test the null hypothesis 2.73=µ against the alternative 2.73>µ at the
0.01 level of significance. Solution Step I Null hypothesis 2.73:H 0 =µ
Alternative hypothesis 2.73:H1 >µ
(Thus here 2.730 =µ )
Step II The level of significance
01.0== α Step III Reject the null hypothesis if 33.201.0 ==> ZZZ α
Step IV Calculations
73.2
45
6.82.737.760 =−=
−=
n
xZ σ
µ
Step V Decision net para since 33.273.2 =>= αZZ
we reject H0 (at 0.01 level of significance) (i.e) we would say 2.73>µ (and the prob of falsely saying this is 01.0≤ ).
153
Example 3 It is desired to test the null hypothesis 100=µ against the alternative hypothesis
100<µ on the basis of a random sample of size n = 40 from a population with .12=σ
For what values of x must the null hypothesis be rejected if the prob of Type I error is to be ?01.0=α Solution
33.201.0 == ZZα . Hence from the table we reject H0 if αZZ −< =-2.33 where
33.2
40
121000 −<−=
−= x
n
xZ σ
µ gives
58.9540
1233.2100 =×−<x
Example 4 To test a paint manufacturer’s claim that the average drying time of his new “fast-drying” paint is 20 minutes, a ‘random sample’ of 36 boards is painted with his new paint and his
claim is rejected if the mean drying time 50.20isx > minutes. Find
(a) The prob of type I error (b) The prob of type II error when 21=µ minutes.
(Assume that 4.2=σ minutes) Solution Here null hypothesis 20:H 0 =µ
Alt hypothesis 20:H1 >µ
P (Type I error) = P (Rejecting H0 when it is true) Now when H0 is true, 20=µ and hence
154
( )20X4.2
6
36
4.220X
n
X −=−=σ
µ− is standard normal.
Thus P (Type I error)
P= ( 50.20X > given that 20=µ )
( ) ( ) ( )
1056.08944.01
25.1F125.1ZP125.1ZP36
4.22050.20
n
XP
=−=
−=≤−=>=
����
�
�
����
�
�
−>σ
µ−=
(b) P (Type II error when 21=µ )
=P (Accepting H0 when 21=µ )
( )
( ) ( )
1056.0
25.1ZP25.1ZP
36
4.22150.20
n
XP
21when50.20XP
=
>=−≤=����
�
�
����
�
�
−≤σ
µ−=
=µ≤=
155
Example 5 It is desired to test the null hypothesis 100=µ pounds against the alternative hypothesis
100<µ pounds on the basis of a random sample of size n=40 from a population with
.12=σ For what values of x must the null hypothesis be rejected if the prob of type I error is to be ?01.0=α Solutions We want to test the null hypothesis 100:H 0 =µ against the alt hypothesis 100:H1 <µ
given .50n,12 ==σ
Suppose we reject H0 when .Cx < Thus P (Type I error) = P (Rejecting H0 when it is true)
( )100givenCXP =µ<=
����
�
�
����
�
�
−<=����
�
�
����
�
�
−<σ
µ−=
50
12100C
ZP
50
12100C
n
XP
01.0
50
12100 =
����
�
�
����
�
�
−= CF
implies 33.2
50
12100 −=−C
Or 05.9633.250
12100 =×−=C
Thus reject H0 if 05.96X <
156
Example 6
Suppose that for a given population with 24.8 in=σ , we want to test the null hypothesis 20.80 in=µ against the alternative hypothesis 20.80 in<µ on the basis of a random
sample of size n = 100.
(a) If the null hypothesis is rejected for 20.78 inx < and otherwise it is accepted,
what is the probability of type I error?
(b) What is the answer to part (a) if the null hypothesis is 280 in≥µ instead of 20.80 in=µ
Solution (a) null hypothesis 80:H 0 =µ
Alt hypothesis 80:H1 <µ
Given 100,4.8 == nσ
P (Type I error) = P (Rejecting H0 when it is true)
( )
��
���
� −<=����
�
�
����
�
�
−<σ
µ−=
=µ<=
2.410
1ZP
100
4.80.800.78
n
XP
80given0.78XP
( )38.212.4
101 FZP −=�
�
���
� <−=
=1-0.9913 =.0087
(b) In this case we define the type I error as the max prob of rejecting H0 when it is
true ( )0.800.78 ≥<= numberaisgivenxP µ
Now ( )µismeanpopulationthewhenxP 0.78<
157
( )��
���
� −<=����
�
�
����
�
�
−<−= µµσ
µ78
4.810
100
4.80.78
ZP
n
xP
( )( )µ−= 7819.1F
We note that cdf of Z, viz F(z) is an increasing function of Z. Thus when
( )( )µµ −≥ 7819.1,80 F is largest when µ is smallest i.e. .80=µ Hence P (Type I
error)
( )( ) ( )( )
0087.080
807819.17819.1
=≥
−×=−=µ
µ FFMax
Example 7 If the null hypothesis 0µµ = is to be tested against the one-sided alternative hypothesis
( )00 µµµµ >< or and if the prob of Type I error is to be α and the prob of Type II
error is to be β when 1µµ = , it can be shown that this is possible when the required
sample size is
( )( )2
01
22 ZZn
µ−µ
+σ= βα
where 2σ is the population variance.
(a) It is desired to test the null hypothesis 40=µ against the alternative hypothesis
40<µ on the basis of a large random sample from a population with .4=σ
If the prob of type I error is to be 0.05 and the prob of Type II error is to be 0.12 for ,38=µ find the required size of the sample.
(b) Suppose we want to test the null hypothesis 64=µ against the alternative
hypothesis 64<µ for a population with standard deviation .2.7=σ How large a
158
sample must we take if α is to be 0.05 and β is to be 0.01 for ?61=µ Also for
what values of x will the null hypothesis have to be rejected? Solution (a) Hence 4,38,4012.0,05.0 10 ===== σµµβα
175.1,645.1 12.005.0 ==== ZZZZ βα
Thus the required sample size
( )( )
.3289.314038
175.1645.1162
2
≥∴=−+= n
(b) Here 2.7,61,64,01.0,05.0 10 ===== σµµβα
( ) ( )( )
9201.916461
33.2645.12.72
22
≥∴=−
+≥∴ nn
We reject 76.62Xor645.1
92
2.764X
ieZZifH 0 <−<−−< α
Tests concerning mean when the sample is small
If X is the sample mean and S the sample s.d. of a (small) random sample of size n from
a normal population (with mean 0µ ) we know that the statistic
n
SX
t 0µ−= has a t-
distribution with (n-1) degrees of freedom. Thus to test the null hypothesis 00 :H µ=µ
against the alternative hypothesis 01 :H µ>µ , we note that when 0H is true, (ie) when
( ) αµµ α =>= − ,10 , nttP
Thus if we reject the null hypothesis when α,1−> ntt (ie) when n
StX ,1n0 α−+µ> we
shall be committing a type I error with prob α .
159
The corresponding tests when the alternative hypothesis is ( )00 & µµµµ ≠< are
described below. Note: If n is large, we can approximate αα Zbytn ,1− in these tests.
Critical Regions for Testing 00 :H µ=µ (Normal population, unknownσ )
Alt Hypothesis Reject Null hypothesis if
0µµ < α,1−−< ntt
0µµ > α,1−> ntt
0µµ ≠ 2,1
2,1
α
α
−
−
>
−<
n
n
tt
ortt
( )sizesamplen
ns
Xt 0 →
µ−=
In each case P(Type I error) = α
Example 8 A random sample of six steel beams has a mean compressive strength of 58,392 psi (pounds per square inch) with a s.d. of 648 psi. Use this information and the level of significance 05.0=α to test whether the true average compressive strength of the steel from which this sample came is 58,000 psi. Assume normality. Solution 1. Null Hypothesis 000,580 == µµ
Alt hypothesis ( )!000,58 why>µ
2. Level of significance 05.0=α 3. Criterion : Reject the null hypothesis if 015.205.0,5,1 ==> − ttt n α
4. Calculations
160
48.16.
648000,58392,58
nS
Xt 0
=
−=µ−
=
5. Decision = 1.48 015.2≤
Since observedt
we cannot reject the null hypothesis. That is we can say the true average compressive strength is 58,000 psi. Example 9 Test runs with six models of an experimental engine showed that they operated for 24,28,21,23,32 and 22 minutes with a gallon of a certain kind of fuel. If the prob of type I error is to be at most 0.01, is this evidence against a hypothesis that on the average this kind of engine will operate for at least 29 minutes per gallon with this kind of fuel? Assume normality. Solution
1. Null hypothesis 29:H 00 =µ≥µ
Alt hypothesis: 01 :H µ<µ
2. Level of significance 01.0=≤ α 3. Criterion : Reject the null hypothesis if ( )6nNote365.3ttt 01.0,5,1n =−=−=−< α−
where
nS
Xt 0µ−
=
4. Calculations
256
223223212824X =+++++=
161
( ) ( ) ( ) ( ) ( ) ( )[ ]
34.2
66.17
2925
6.17
25222532252325212528252416
1 2222222
−=−=∴
=
−+−+−+−+−+−−
=
t
S
5. Decision Since 365.334.2t obs −≥−= , we cannot reject the null hypothesis. That is we can
say that this kind of engine will operate for at least 29 minute per gallon with this kind of fuel.
Example 10 A random sample from a company’s very extensive files shows that orders for a certain piece of machinery were filled, respectively in 10,12,19,14,15,18,11 and 13 days. Use the level of significance 01.0=α to test the claim that on the average such orders are filled in 10.5 days. Choose the alternative hypothesis so that rejection of the null hypothesis.
5.10=µ indicates that it takes longer than indicated. Assume normality.
Solution
1. Null hypothesis 5.10:H 00 =µ≥µ
Alt hypothesis : 5.10:H1 <µ
2. Level of significance 01.0=α 3. Criterion : Reject the null hypothesis if 998.201.0,7001,18,1 ==−=−< −− tttt n α
where
nS
Xt 0µ−
= ( )8n,5.10where 0 ==µ
4. Calculations
148
1311181514191210X =+++++++=
162
( ) ( ) ( ) ( ) ( )( ) ( ) ( )
09.3
829.10
5.101429.10
141314111418
14151414141914121410
181
222
222222
=−=∴
=���
���
�
−+−+−+
−+−+−+−+−−
=
t
S
5. Decision
Since 998.209.3 >=observedt , we have to reject the null hypothesis .That is we can
say on the average, such orders are filled in more than 10.5 days. Example 11 Tests performed with a random sample of 40 diesel engines produced by a large manufacturer show that they have a mean thermal efficiency of 31.4% with a sd of 1.6%. At the 0.01 level of significance, test the null hypothesis %3.32=µ against the
alternative hypothesis %3.32≠µ
Solution
1. Null hypothesis 3.320 == µµ
Alt hypothesis 3.32≠µ
2. Level of significance 01.0=α 3. Criterion : Reject
2,12,10 αα −−−<
nntortifH (ie) if 005.0,39005.0,39 tortt −< .
Now 575.2005.0005.0,39 =≈ Zt
Thus we reject 575.2575.20 >−< tortifH
where
nS
Xt 0µ−
=
4. Calculations
558.3
406.1
3.324.31 −=−=t
5. Decision Since 575.2558.3 −<−=observedt
Reject H0 ; That is we can say the mean thermal efficiency 3.32≠
163
Example 12 In 64 randomly selected hours of production, the mean and the s.d. of the number of acceptable pieces produced by an automatic stamping machine are
.146Sand038,1X == At the 0.05 level of significance, does this enable us to reject the
null hypothesis 1000=µ against the alt hypothesis ?1000>µ
Solution
1. The null hypothesis 1000: 00 == µµH
Alt hypothesis 1000:H1 >µ
2. Level of significance 05.0=α 3. Criterion : Reject 05.0,164,10 −− => tttifH n α
Now 645.105.005.0,63 =≈ Zt
Thus we reject 645.10 >tifH
4. Calculations: 082.2
64146
000,1038,1
nS
Xt 0 =−=
µ−=
5. Decision : Since 645.1082.2 >=obst
we reject 05.00 atH level of significance.
