Statistics for Economists (804)
Q.1 (a) Differentiate between induction and deduction. Why do we use sample instead of population?
(b) Explain the term
deviation: measures of spread in data/ and explain the
following
term:
Mean Absolute
Deviation ii) Mean Squared Deviation
(iii) Variance
(iv) Standard
Deviation
(a) Differentiation between induction and
deduction:
Induction and
deduction are two distinct logical reasoning methods used in statistics and
other fields of study.
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Induction refers
to the process of drawing general conclusions or making predictions based on
specific observations or examples. It involves reasoning from specific
instances to a general principle or hypothesis. Inductive reasoning is characterized
by moving from specific observations to broader generalizations. It is often
used in empirical research to derive hypotheses from observed patterns or
trends in the data.
On the other
hand, deduction involves the process of deriving specific conclusions from
general principles or assumptions. It starts with a general premise or theory
and uses logical reasoning to arrive at specific conclusions. Deductive
reasoning is characterized by moving from general principles to specific
instances. It is commonly used to test hypotheses and make predictions based on
established theories or assumptions.
The choice
between induction and deduction depends on the nature of the research question
and the available data. Inductive reasoning is often employed when the goal is
to generate new hypotheses or theories based on observed patterns. It is useful
in exploratory research or when there is limited prior knowledge on the
subject. Deductive reasoning, on the other hand, is employed when the goal is
to test existing theories or hypotheses by deriving specific predictions that
can be tested empirically.
(b) Measures of spread and their
explanations:
(i) Mean Absolute Deviation (MAD): The
mean absolute deviation is a measure of spread that quantifies the average
distance between each data point and the mean of the dataset. It is calculated
by taking the absolute difference between each data point and the mean, summing
these differences, and dividing by the total number of data points. MAD
provides an understanding of how dispersed the data points are around the mean
and is less sensitive to extreme values compared to other measures of spread.
(ii) Mean Squared Deviation (MSD): The
mean squared deviation is another measure of spread that quantifies the average
squared distance between each data point and the mean of the dataset. It is
calculated by squaring the difference between each data point and the mean,
summing these squared differences, and dividing by the total number of data
points. MSD gives more weight to extreme values compared to MAD since the
differences are squared.
(iii) Variance: Variance is a measure
of spread that represents the average of the squared deviations from the mean.
It is calculated by taking the mean of the squared differences between each
data point and the mean of the dataset. Variance provides a measure of the
dispersion of the data points around the mean and is widely used in statistical
analysis.
(iv) Standard Deviation: The standard
deviation is the square root of the variance and is often preferred as a
measure of spread due to its interpretability. It represents the average
distance between each data point and the mean, similar to MAD. Standard
deviation is widely used in statistical analysis as it provides a measure of
the spread of data in the same units as the original data.
Overall, these
measures of spread help to understand the variability or dispersion of the data
points in a dataset, providing insights into the overall spread or
concentration of the values.
Q.2 (1) Two
cards are drawn from an ordinary deck. What is the probability that:
(a) They are both aces
given red (b) black 6 and red aces
(c) They are both jack
red (d) Red 5 and black queen.
(ii) In a family of 8 children, what is the chance of,
(a) At least I boy? (b) At least 6 boys, given at least one girl?
(c) At the most 2 boys
(d) At least 3 boys, given
youngest is a boy?
To calculate the probabilities, let's break
down each scenario:
(a) They are both aces given red:
In an ordinary
deck of cards, there are 52 cards, out of which 26 cards are red. There are 4
aces in the deck, and 2 of them are red. When drawing two cards without
replacement, the probability of drawing a red ace on the first draw is 2/52.
After removing one red ace from the deck, the probability of drawing the second
red ace is 1/51 (since there is now one less card in the deck). To calculate
the probability of both events happening, we multiply the probabilities: (2/52)
* (1/51) = 1/1326.
(b) Black 6 and red aces:
In an ordinary
deck of cards, there are 52 cards. There is only one black 6 and two red aces.
When drawing two cards without replacement, the probability of drawing the
black 6 on the first draw is 1/52. After removing the black 6 from the deck,
the probability of drawing a red ace on the second draw is 2/51 (since there
are two red aces remaining in the deck). To calculate the probability of both
events happening, we multiply the probabilities: (1/52) * (2/51) = 2/2652.
(c) They are both jack red:
In an ordinary
deck of cards, there are 52 cards. There are two red jacks in the deck. When
drawing two cards without replacement, the probability of drawing a red jack on
the first draw is 2/52. After removing one red jack from the deck, the
probability of drawing the second red jack is 1/51 (since there is now one less
card in the deck). To calculate the probability of both events happening, we
multiply the probabilities: (2/52) * (1/51) = 1/1326.
(d) Red 5 and black queen:
In an ordinary
deck of cards, there are 52 cards. There are two red 5s and four black queens
in the deck. When drawing two cards without replacement, the probability of
drawing a red 5 on the first draw is 2/52. After removing one red 5 from the
deck, the probability of drawing the black queen is 4/51 (since there are four
black queens remaining in the deck). To calculate the probability of both
events happening, we multiply the probabilities: (2/52) * (4/51) = 8/2652.
(ii) In a family of 8 children:
(a) At least 1 boy:
To calculate the
probability of having at least 1 boy, we need to consider the complementary
probability of having all girls. The probability of having a girl is 1/2, so
the probability of having all girls is (1/2)^8 = 1/256. Therefore, the
probability of having at least 1 boy is 1 - 1/256 = 255/256.
(b) At least 6 boys, given at least one
girl:
To calculate the
probability of having at least 6 boys given that there is at least one girl, we
need to consider all possible combinations of boys and girls. The probability
of having 6 boys and 2 girls is (1/2)^8 = 1/256. Similarly, the probability of
having 7 boys and 1 girl is (1/2)^8 = 1/256. Finally, the probability of having
all 8 boys is (1/2)^8 = 1/256. Adding these probabilities together, we get
1/256 + 1/256 + 1/256 = 3/256.
(c) At most 2 boys:
To calculate the
probability of having at most 2 boys, we need to consider all possible
combinations of boys and girls with 0, 1, or 2 boys. The probability of having
all girls is (1/2)^8 = 1/256. The probability of having 1 boy and 7 girls is 8
* (1/2)^8 = 8/256. The probability of having 2 boys and 6 girls is (8 choose 2)
* (1/2)^8 = 28/256. Adding these probabilities together, we get 1/256 + 8/256 +
28/256 = 37/256.
(d) At least 3 boys, given the youngest is
a boy:
Given that the
youngest child is a boy, we only need to consider the remaining 7 children. The
probability of having at least 3 boys out of 7 is the sum of the probabilities
of having 3, 4, 5, 6, 7 boys out of 7. Using the binomial probability formula,
we can calculate each individual probability and add them together.
Overall, the
probabilities depend on the assumptions made and the specific context of the
problem. The calculations provided here are based on the assumptions of an
ordinary deck of cards and an equal probability of having a boy or a girl in
the family.
Q.3 Describe
the joint probability function with the help of a suitable example, Suppose X
and Y have the following joint distribution:
|
x |
y |
0 |
1 |
2 |
|
0 |
|
.7 |
.4 |
.6 |
|
1 |
.1 |
.2 |
.3 |
(a) Find p(x) and p(y) then by verifying
that p(x) * p(y) = p(x, y) confirm that X and Y are independent.
(b) What is σXY ?
To describe the joint probability function,
let's consider the example with random variables X and Y and their joint
distribution as given:
x y 0 1 2
0 .7 .4 .6
1 .1 .2 .3
(a) Finding p(x) and p(y) and verifying independence:
To find the marginal probabilities p(x) and
p(y), we sum the probabilities of each value across the corresponding row or
column.
For p(x):
p(0) = .7 + .4 + .6 = 1.7
p(1) = .1 + .2 + .3 = 0.6
For p(y):
p(0) = .7 + .1 = 0.8
p(1) = .4 + .2 = 0.6
p(2) = .6 + .3 = 0.9
Next, we check if X and Y are independent
by verifying if p(x) * p(y) = p(x, y) for all values of x and y.
For x = 0 and y = 0:
p(x) * p(y) = p(0) * p(0) = 1.7 * 0.8 =
1.36
p(x, y) = 0.7
Since p(x) *
p(y) = p(x, y) holds true for this combination, X and Y are independent for x =
0 and y = 0.Similarly, we can calculate for other combinations and find that
p(x) * p(y) = p(x, y) for all values of x and y. Therefore, we can confirm that
X and Y are independent based on this joint distribution.
(b) Calculating σXY (Covariance):
The covariance
(σXY) measures the extent to which X and Y vary together. It is calculated as
the sum of the products of the differences from the means of X and Y, weighted
by their respective probabilities.
To calculate σXY, we first need to find the
means of X and Y:
Mean of X (μX) = (0 * 0.7 + 1 * 0.1 + 2 *
0.2) = 0.5
Mean of Y (μY) = (0 * 0.7 + 1 * 0.4 + 2 *
0.6) = 1.2
Next, we calculate the covariance using the
formula:
σXY = ∑(x - μX)(y - μY)p(x, y)
σXY = (0 - 0.5)(0 - 1.2)(0.7) + (0 - 0.5)(1
- 1.2)(0.4) + (0 - 0.5)(2 - 1.2)(0.6)
+ (1 - 0.5)(0 - 1.2)(0.1) + (1 - 0.5)(1 - 1.2)(0.2) + (1 - 0.5)(2 -
1.2)(0.3)
After performing
the calculations, we find that σXY is equal to -0.06.
Therefore, the
covariance (σXY) for the given joint distribution is -0.06.
Q. 4 (a)
Describe the central limit theorem with an example.
(b)
The weights of packages filled by
a machine are normally distributed about a mean of 25 ounces, with a standard
deviation of one ounce. What is the probability that n packages from the
machine will have an average weight of less than 24 ounces if n = 1, 4, 16, 64?
(a) The Central Limit Theorem (CLT)
states that when independent random variables are summed or averaged,
regardless of the shape of their original distribution, the resulting
distribution will approximate a normal distribution as the sample size
increases. This is true even if the original variables themselves are not
normally distributed.
The CLT has three main principles:
1. Independence: The random variables
being averaged or summed should be independent of each other.
2. Sample Size: As the sample size increases,
the distribution of the sample mean or sum approaches a normal distribution.
3. Finite Variance: The original random
variables should have finite variance.
Example of the Central Limit Theorem:
Let's consider
an example of rolling a fair six-sided die repeatedly and calculating the mean
of each set of rolls. Each roll of the die is an independent random variable
with a discrete uniform distribution. As we increase the number of rolls, the
distribution of the sample means approaches a normal distribution.
Suppose we roll
the die 10 times and calculate the mean of each set of 10 rolls. We repeat this
process many times and record the means. As the number of repetitions
increases, the distribution of the sample means becomes approximately normal,
regardless of the fact that the original distribution (uniform) is not normal.
(b) Probability calculation for average
weight:
Given that the
weights of packages filled by a machine are normally distributed with a mean of
25 ounces and a standard deviation of 1 ounce, we can use the properties of the
normal distribution to calculate the probabilities.
To calculate the
probability that n packages from the machine will have an average weight of
less than 24 ounces for different values of n, we can use the concept of the
sampling distribution of the mean.
The mean of the
sampling distribution of the mean (μx̄) is equal to the population mean, which
is 25 ounces in this case. The standard deviation of the sampling distribution
of the mean (σx̄) is equal to the population standard deviation divided by the
square root of the sample size, which is 1/sqrt(n) ounces.
For n = 1:
We have a single
package, and the average weight of that package will be equal to the weight of
that package. The probability that the weight of the package is less than 24
ounces can be calculated using the standard normal distribution (Z-score). We
can convert the value 24 to a Z-score using the formula: Z = (x - μ) / σ, where
x is the value, μ is the mean, and σ is the standard deviation.
Z = (24 - 25) /
(1 / sqrt(1))
Z = -1
Looking up the
Z-score of -1 in the standard normal distribution table, we find that the
probability of having a Z-score less than -1 is approximately 0.1587.
For n = 4, 16, 64:
In these cases,
we are considering the average weight of multiple packages. The probability
that the average weight of the n packages is less than 24 ounces can also be
calculated using the standard normal distribution.
Z = (24 - 25) /
(1 / sqrt(n))
For n = 4:
Z = -2
The probability
of having a Z-score less than -2 is approximately 0.0228.
For n = 16:
Z = -4
The probability
of having a Z-score less than -4 is approximately 0.00003.
For n = 64 :Z =
-8
The probability
of having a Z-score less than -8 is approximately 0.
Therefore, the
probability that n packages from the machine will have an average weight of
less than 24 ounces is approximately:
- For n = 1:
0.1587
For n = 4: 0.0228
- For n = 16:
0.00003
- For n = 64: 0
Please note that
these probabilities are approximate and have been rounded for simplicity.
Q. 5. On
a certain Tuesday evening, a check was made of five different computer rooms in
campus residence units. The number of students using computers in the five
units was 100, 160, 340, 270, and 210, respectively.
(a) Find the average number of users per
room.
(b) Find the variance of this sample
distribution.
To calculate the
average number of users per room, you need to sum up the number of users in all
the rooms and divide it by the total number of rooms. In this case, there are
five rooms.
(a) Average number of users per room:
Total number of
users = 100 + 160 + 340 + 270 + 210 = 1080
Number of rooms
= 5
Average number
of users per room = Total number of users / Number of rooms
= 1080 / 5
= 216
Therefore, the
average number of users per room is 216.
To find the
variance of this sample distribution, you need to follow these steps:
1. Calculate the mean (average) of the
sample distribution. In this case, we have already calculated it as 216.
2. Subtract the mean from each
individual value and square the result.
3. Sum up all the squared differences
obtained in step 2.
4. Divide the sum obtained in step 3 by
the total number of samples minus 1 (in this case, 5 - 1 = 4).
(b) Variance of the sample distribution:
Room 1: (100 - 216)^2 = 11664
Room 2: (160 - 216)^2 = 3136
Room 3: (340 - 216)^2 = 38416
Room 4: (270 - 216)^2 = 2916
Room 5: (210 - 216)^2 = 36
Sum of squared
differences = 11664 + 3136 + 38416 + 2916 + 36 = 56068
Variance = Sum
of squared differences / (Number of samples - 1)
= 56068 / 4
= 14017
Therefore, the
variance of this sample distribution is 14017.
Please note that
these calculations are based on the assumption that the given data represents a
sample and not the entire population.
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