recalling our definition of a random sample (7.2) and then, to make it of practical, mathematical use, reformulate it as follows :
Definition: Random Sample. If x<, (i = 1, 2, 3, . . . n) is a set of n statistically independent variates, each distri- buted with the same probability density <f>{x) and if the joint probability density of the set is given by f(xlt x2, . . .
x„) = $(1kx) . (f>(x2)... 4(x„), then the set Xi, (i = f , 2, 3 , . . .
rt) is a random sample of n from a population whose
probability density is 4>(x).
Now suppose from the n variates x,, with means | i i ( x i ) , we form a new variate
n
X = + a2x3 - ) - . . . + OiXi + . . . + OnXn = 2 a{Xi
i= 1
(7.7.1) where the a's are arbitrary constants, not all zero. We have
£{X) =£( S OixA = S ai£(Xi) \ i = 1 / t = i
or ^ ' ( X ) - S Oi^'iXi) (7.7.2)
i « 1
If we put at = 1 /n, for all i, and subject the t o the con- dition t h a t they all have the same probability density, we have
£•(*!> = £(X2) = . • • = £(xn) = £(x) = (x,
138 s t a t i s t i c s
sample of n froih a population <j>{x) a n d mean (x. So (7.7.2) becomes
6(x) = I S €(xi) = i . n£(x) = . (7.7.3) = i n
T h u s
the m e a n value of the m e a n of all possible samples of n is the m e a n of the population s a m p l e d : or
the m e a n of the sampling distribution of x is the population m e a n .
If n = 1, t h a t this is so is obvious, for, t a k i n g all possible samples of 1, t h e distribution of t h e sample m e a n is identical w i t h t h e distribution of t h e individual items in t h e population. (7.7.3) is also t h e justification of t h e common-sense view t h a t t h e average of a n u m b e r of m e a s u r e m e n t s of t h e " length " of a rod, say, is a b e t t e r e s t i m a t e of t h e " length " of t h e rod t h a n a n y one measurement.
W h a t of t h e variance of x ? W e write jx,- for t h e m e a n of Xi, (xx for t h a t of X, a n d a,-2 a n d aj2 for t h e variances of
a n d X, respectively. Also let pg be t h e coefficient of t h e correlation between a n d Xj (i ^ /), assuming such correlation t o exist. ,
T h e n
(X - (xx)2 = (.ZMxi - Hi)) . or, if i * j,
n
n n
= 2 ai*(xi - (x,)
2+ 2 2
Oi<ij(Xi - \n)(Xj - w) i= 1 So- (x^)
2] = 2 afE^Xi — (X()
2]
* = 1 n n +2 2
0{(lj£[(Xi- (Xi)(*
f—
W)]
B u t£{(Xi
- |X<)«] =
£(Xi>- 2(Xi*< + (Xi
2) =
e(Xi')- |Xi
2= <Ji
2Also, when i j,
£[{*• - V-i)(Xj - w)] = £(xtx, — [nXj - \ijXi + (XiW) = €{XiXj) ' (XiW
= c o v fax,) = o f l f a Hence
s a m p l e a n d p o p u l a t i o n . ii 139
This is an i m p o r t a n t formula in its own right. If, however, the variates are independent, then py = 0, for all i, j, and
a*2 = W + «2V + . . . + O n V = S chW
i= l
(7.7.4 (a)) Again p u t t i n g a{ = 1 /«, for all i, and subjecting t h e x's t o the condition t h a t t h e y all have the same probability density <f>(x), so t h a t X = x, the mean of a sample of n from </>(x), and ox2 = a2 a = . . . = CT„2 = aa, say, we have, since Xi and
Xj(i ^ j) are independent,
ox a = o / = oa/7l or 0ja = - 4 = • (7.7.5)
V I Thus :
The variance of the distribution of the sample mean is 1 /nth that of the variance of the parent population, n being the size of the sample.
In other words, t h e larger the sample, t h e more closely the sample means cluster about the population mean value.
The s t a n d a r d deviation of the sampling distribution of x is usually called t h e Standard Error of the Mean, and, in general, the standard deviation of t h e sampling distribution of a n y statistic is called t h e Standard Error of t h a t statistic.
The above results hold for any population, no m a t t e r how t h e variate is distributed. However, it is known t h a t , whatever the population sampled, as the sample size increases, t h e distribution of x tends towards n o r m a l i t y ; while, even for relatively small values of n, there is evidence t h a t the ^-distribution is approximately normal.
7.8. The Distribution of x when the Population Sampled is Normal. Consider a normal population defined by
</>(x) = exp [ - (* -
Here, we recall, [/. is the population mean and <j!, t h e popula-
tion variance.
The mean-moment generating function for a normal distribu- tion of variance oa is Mm(t) = exp (|aa<2), (5.4.2), and the
function generating t h e moments about t h e origin is M(t) ^ Mm(t) exp (n<) = exp (ytt + &H*).
Now, remembering t h a t the m.g.f. for a n y distribution is M(t) = f ( e x p xt), t h e m.g.f. of the mean, x, of t h e n independent
140 s t a t i s t i c s variates X{, (t = ' 1, 2, 3, . . . n), each with probability function 4>(x), is
n , n i
£ ( e x p xt) — £ (exp S Xit/n) = £ ( II exp Xit/n) «=l „ /
= II [£ (exp Xit/n)] i = l
But, since (xv x2, . . . xn) is a sample from <f>(x), £ (exp xt) = (£ (exp xt/n))n = (M(tjn))»
= exp { n ^ / n + iaV/n1)},
i.e., t h e m.g.f. for x is
exp (at + i ( aal n ) t " ) . . . . (7.8.1)
B u t this is t h e m.g.f. of a normal population with mean jj. a n d variance a* In. Hence—
The mean of samples of n from a normal population (|x, a) is itself normally distributed about |x as mean with standard deviation (error) a/Vn.
This is in agreement with (7.7.3) and (7.7.5), b u t a d d s t h e important information t h a t t h e distribution of x, when t h e population sampled is normal, is itself normal. Actually this is a particular case of a more general theorem :
If xlt . . . xn are n independent variates normally distributed about a common mean (which may be taken at zero), with variances c^2, a22, . . . a„2, then any linear function of these n variates is itself normally distributed.
n
T h e proof is simple : Let X = S OiXi. The m.m.g.f. of t = i
X{, M{(t) is exp (Joi2**) a n d therefore
Mx(t) = £ (exp Xt) = £ ( e x p ( S OiXit) j = £ ( II (exp «(*<<)). B u t t h e *'s are independent, and, so,
Mx(t) = n £ (exp OiXit) = n £ (exp x((ait))
= U M i i o i t ) = exp S ^ ' c ' j / ' J which is t h e m.g.f. of a normal distribution with variance
s a m p l e a n d p o p u l a t i o n . ii 141
i3i
Consequently,(i) if n — 2 and = a2 = 1, the distribution of the sum,
*i + *2> the two normal variates xlt x2, is normal
about the common mean with variance ctj2 + a2 l; and
(ii) if n — 2 and ax = 1, a2 = — 1, the distribution of the
difference, — x2, of the two normal variates, xlt x2, is
normal about the common mean with variance <ji2 + <r2a.
Now in m a n y of the problems we encounter, n is large, and so we m a y assume t h a t x, the mean of a sample of n from an infinite population of mean jx and variance a2, is approxi-
mately normally distributed about (jl as mean with variance a2In, although we do n o t know t h a t t h e population sampled is normal. This being t h e case the variate t = (x — y.)/(a/Vn) will be approximately normally distributed about zero with unit variance. I n other words:
The probability that the sample mean, X, will differ numerically from the population mean, p, by less than a n amount d (measured in units of the standard error of the sample mean) is given approximately by
I t m u s t be emphasised, however, t h a t here we are sampling a population whose variance is assumed known. When this is not t h e case, the problem is complicated somewhat and will be dealt with later.
7.9. Worked Examples.
1. The net weight of i~kg boxes of chocolates has a mean of 0-51 kg and a standard deviation of 0-02 kg. The chocolates are despatched from manufacturer to wholesaler in consignments of 2,500 boxes. What proportion of these consignments can be expected to weigh more than 1,276 kg net? What proportion will weigh between 1,273 and 1,277 kg net?
Treatment: We axe here drawing samples of 2,500 from an assumed infinite population. The mean net weight of boxes in a consign- ment weighing 1,276 kg will be 1,276/2,500 kg = 0-5104 kg.
The standard error of the mean net weight for samples of 2,500 will be 0-02/V^500 = 0-0004 kg. Thus in this case t = 0-0004/ 0-0004 = 1. The probability that the sample mean will deviate
0
where fit)
= JL
exp ( - i f2) .142 s t a t i s t i c s
from the populati6n mean by more than this amount is P(t > 1) = 0-5 — P(0 < t sj 1), for this is a " one-tail " problem. P(t ^ 1) = 0-3413. Therefore P(t > 1) = 0-1587. Therefore just under 16% of the consignments of 2,500 boxes will weigh more than 1,276 kg.
If a consignment weighs 1,273 kg, the mean weight of a box in that consignment is 0-5092. The deviation from the mean is then — 0 0008, or, in standardised units, —2. If the consignment weighs 1,277 kg, the corresponding mean weight is 0-5108 kg, a deviation from the population mean of + 2 standard errors. The probability that a consignment will weigh between these two limits is then—this being a " two-tail " problem—
P ( - 2 < t < 2) = 2P(0 < l < 2 ) = 2 x 0-4772 = 0-9544 In other words, just over 95% of the batches of 2,500 boxes will lie between the given net weights.
2. The " guaranteed " average life of a certain type of electric light bulb is 1,000 hours with a standard deviation of 125 hours. It is decided to sample the output so as to ensure that 90% of the bulbs do not fall short of the guaranteed average by more than 2-5%. What must be the minimum sample size ?
Treatment: Let n be the size of a sample that the conditions may be fulfilled. Then the standard error of the mean is 125/V «. Also the deviation from the 1,000-hour mean must not be more than 25 hours, or, in standardised units, not more than 25/(125/V») = Vnj5. This is a " one-tail" problem, for we do not worry about those bulbs whose life is longer than the guaranteed average.
P(t > «i/5) = 0-1 and, so, P(0 < t < = 0-4 Using Table 5-4, we find that t = 1-281. Therefore, »i/5 = 1-281 or n = 40-96. Consequently, the required minimum sample size is 41. 3. The means of simple samples of 1,000 and 2,000 are 67-5 and 68-0 respectively. Can the samples be regarded as drawn from a single population of standard deviation 2-5 ?
Treatment: Just as the sample mean has its own distribution, so, too, does the difference between the means of two samples of a given size. If x1 and x% are two independent variates distributed about mean /i„ /i, with variances CT12,CT22i respectively, let X = x1 — x%. Then X = /x, — ft2. and, by 7.7.4 (a), aj = or,2 + <ra2. Now set up
the hypothesis that two samples of n1 and «2 are drawn from a
single population (/i, o). Then, if x1 and are the means of these two samples, X is distributed about' zero mean with variance
o-j' = a2/«! -f And, since xx and are approximately
normally distributed, so is X.
In our example, X = 0-5 and ox% = (2-5)a (1/1,000 + 1/2,000) =
0-009375 or ox = 0-0968. Thus the observed X is more than 5 times the standard error of its distribution calculated on the hypothesis that the samples are from the same population with standard deviation
s a m p l e a n d p o p u l a t i o n. ii 143 2.5. This, on the assumption that X is approximately normally distributed, is most unlikely. We therefore reject the hypothesis that the samples are from a single population of standard deviation 2.5.
7.10. Sampling Distribution of the Mean when Sampling is