Hi makbool!
Here are the answers to your questions.
1. In order to answer these set of questions, you must first know some
properties of the sample mean. You may want to check them in the
following page.
Sample Mean
http://www.stat.yale.edu/Courses/1997-98/101/sampmn.htm
We're particularly interested in the mean and standard deviation (SD)
of the sample mean. Given a sample of size n, we have n independent
observations, which can be considered random variables. Since all
these observations come from the same population, they are independent
and identically distributed (iid) random variables. Now, the sample
mean is defined as:
M = (X1 + X2 + X3 + ... + Xn) / n
where X1 is the first observation, X2 the second one, etc. What is the
expected value of this statistic?
E(M) = E[(X1 + X2 + X3 + ... + Xn) / n]
= (1/n) * E[(X1 + X2 + X3 + ... + Xn)
= (1/n) * [E(X1) + E(X2) + ... + E(Xn)]
= (1/n) * [ 69 + 69 + ... + 69]
= (1/n) * [ n*69 ]
= 69
I have used standard properties of the expected value (E) operator
here. You can find them at:
Expectation (check especially "properties of expectation")
http://bme.bu.edu/senlab/lecture11.ppt
Therefore the first result we have here is that the expected value of
the sample mean is equal to the population mean.
Let's see now what happens with variance of the sample mean (and thus
with SD). We know that the population variance is 3.2^2 = 10.24.
Var(M) = Var[(X1 + X2 + X3 + ... + Xn) / n]
= (1/n^2)*Var(X1 + X2 + X3 + ... + Xn)
= (1/n^2)*[Var(X1)+Var(X2)+...+Var(Xn)]
= (1/n^2)*[ 10.24 + 10.24 +...+10.24 ]
= (1/n^2)* (n*10.24)
= (1/n)*10.24
The third line has made use of the fact that the observations are
independent; so the variance of the sum is the sum of variances. Check
some properties of the variance at:
Variance - Properties
http://www.xycoon.com/properties.htm
Variance of the sum of 2 independent variables
http://people.ccmr.cornell.edu/~muchomas/8.04/Lecs/lec_statistics/node14.html#SECTION00053000000000000000
Thus, the variance of the sample mean is the variance of the
population divided by the sample size. This is consistent with the
idea that, as we take a bigger sample size, the sample mean becomes
more accurate. Since the SD is the square root of the variance, we
also have that the SD of the sample mean must be then the SD of the
population divided by the square root of n.
Now let's go to your questions:
a. The mean of the sample means would be 69 and the SD would be 3.2/2
= 1.6. So the sample mean would fluctuate around 69 inches.
b. It is not possible to answer this question because we don't know
the distribution of the population; and therefore we don't know the
distribution of the sample mean, so just knowing the mean and SD is
not enough information to tell wether 70 is a likely observation.
However, if we assume that the population follows a normal (Gaussian)
distribution, then the sample mean does too. For a normal
distribution, 95% of the observations are concentrated within 2 SD's
from the mean. So 95% of the observations fall between 69-3.2 and
69+3.2. Since 70 falls in this interval, we could conclude that 70 is
not unlikely at all. However, we must know (or make a reasonable
assumption about) the population distribution in order to answer this
question for small samples.
c. With n=16, the mean is still 69, but the SD falls to 3.2/4 = 0.8.
Therefore, now the dispersion is smaller, so the sample mean becomes
more accurate.
d. The statement is false. The "double" and "quadruple" are reversed:
by quadrupling the sample size, we find that the SD (a measurement of
"accuracy") falls by a 50%, so the estimate becomes twice as
"accurate". This is because, as the statement says, the square root in
the divisor of the SD of the sample mean, as in the formula I've shown
above.
Question 2
a. It is reasonable to assume that the sales of paint arts follow a
Poisson process. The Poisson process describes the distribution of an
integer number of "arrivals" over a continuous time. If we assume that
the artist can receive a request for a paint any time of the day, this
fits a Poisson process. The assumptions of a Poisson process (given in
the following link) imply that:
1) It may only take integer, non-negative values: this is consistent
with the trivial fact that the artist sells an integer quantity of art
paints, which must be clearly greater than or equal to zero.
2) Each "arrival" is independent: again this could be consistent with
this case: a person calling the artist to request an art paint doesn't
change the probability that another person will call. This could be
violated in this case, as some "arrivals" might be people who will
then tell everybody that this is the best artist in the world, so that
the artist will afterwards receive hundreds of requests for paintings.
In this case, an arrival changed the probability that future arrivals
occur, thus violating the assumptions of a Poisson process. For the
sake of the argument, let's assume that this is not the case and that
every "arrival" is independent.
Poisson Process
http://www.sics.se/~andersa/report/node15.html
b. This was explained in point (1) in question a. The artist can't
sell a negative amount of paintings. Neither can he sell "half" a
painting, or a quarter of a painting. Therefore the relevant sample
space are the natural numbers. On the other hand, there's no upper
limit as to how many paintings he can sell in a day: he could receive
a request for a painting every second, every milisecond, etc. Of
course, these events are highly improbable, but are part of the sample
space.
c.
a) Since he can only sell an integer number of paintings, then the
probability of selling more than 1 painting is equal to the
probability of selling 1, plus the probability of selling 2, etc. This
is consistent with the result in probability theory that says that
Prob ( A U B) = Prob(A) + Prob(B) - Prob(A I B)
where U is union, I is intersection, and A and B are two events. So,
Prob(X>=1) = Prob(X=1 U X=2 U X=3 U...) = P(X=1)+P(X=2)+...
Notice that the "intersection terms" are gone, since the probability
of selling two different number of paintings in a day is zero (you
either sell a quantity or another one).
b. Another result in probability theory is that:
P(Ac) = 1 - P(A)
where Ac is the complement of A. The complement of X>=1 is X<1,
therefore,
P(X>=1) = 1 - P(X<1)
c. Since X can only take on integer values, then X < 1 is the same
event as X=0, since X can't be 1/2, 3/4, etc.
d. 1 - 2^0*e^(-2)/0!
This formula comes directly from the formula of the Poisson
distribution (check the formula in the page given above). In your
case, using the notation of this page, lambda*t is equal to 2, since
the artist sells 2 paintings per day (this is what's called
"intensity" in this page).
e. Since 2^0 is equal to 1, and 0! is also equal to 1, the formula
clearly becomes 1 - e^(-2)
Google search terms used:
"sample mean" properties
variance properties
poisson process
I hope this helps! If you have any doubt regarding my answer, please
request a clarification before rating it. Otherwise, I await your
clarification and final rating.
Best wishes!
elmarto |
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