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 1 IGNOU MEC 003 Solved Assignment 202223
 1.1 1. (i) “Analysis of economic relationships often requires inclusion of a large number of interrelated variables into the model. Linear algebra provides tools of solving simultaneous equations with the aid of matrices and determinants means for their concise presentation and solution mechanism”. In the light of this statement, explain the concept of matrix and determinants. (ii) Discuss the properties of determinants. Use examples to elaborate your response. 2. We can derive some fundamental distributions with the help of Normal distribution. Enlist these fundamental distributions and explain each one of them in detail.
 1.2 3. Under the simple regression analysis there are different methods of estimation. One of them is the Ordinary Least Square (OLS) method of estimation. Explain this method in detail.
 1.3 4. A good estimator should satisfy certain criteria. Discuss these criteria with illustration. 5. (i) Enlist the properties of a binomial distribution. Find the probability of getting 8 heads and 5 tails in 13 tosses of an unbiased coin. (ii) In a steel factory, on an average, there are 6 defects per 10 square feet of steel sheet produced. If we assume a poisson distribution, what is the probability that and 18 square feet of steel sheet will have at least 7 defects? (iii) The height distribution of a group of 10,000 men is normal with mean height 63.5″ and standard deviation 3.5. Find the number of men whose height is (a) less than 54.5″ and (b) more than 74.5″. 6. Explain the concept of indefinite integral with the help of examples. Discuss any two economic applications of indefinite integral.
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 2 IGNOU MEC 003 Solved Assignment 202223
IGNOU MEC 003 Solved Assignment 202223
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Important Note – IGNOU MEC 003 Solved Assignment 202223 You may be aware that you need to submit your assignments before you can appear for the Term End Exams. Please remember to keep a copy of your completed assignment, just in case the one you submitted is lost in transit.
Submission Date :
 31st March 2033 (if enrolled in the July 2033 Session)
 30th Sept, 2033 (if enrolled in the January 2033 session).
SECTION – A
SECTION – B
3. Under the simple regression analysis there are different methods of estimation. One of them is the Ordinary Least Square (OLS) method of estimation. Explain this method in detail.
4. A good estimator should satisfy certain criteria. Discuss these criteria with illustration.
5. (i) Enlist the properties of a binomial distribution. Find the probability of getting 8 heads and 5 tails in 13 tosses of an unbiased coin.
(ii) In a steel factory, on an average, there are 6 defects per 10 square feet of steel sheet produced. If we assume a poisson distribution, what is the probability that and 18 square feet of steel sheet will have at least 7 defects?
(iii) The height distribution of a group of 10,000 men is normal with mean height 63.5″ and
standard deviation 3.5. Find the number of men whose height is (a) less than 54.5″ and (b) more than 74.5″.
6. Explain the concept of indefinite integral with the help of examples. Discuss any two economic applications of indefinite integral.
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IGNOU MEC 003 Solved Assignment 202223
7. Write short notes on the following: (3 X 4 = 12)
(i) Types of Sampling
Sampling is a method used in statistical analysis in which a decided number of considerations are taken from a comprehensive population or a sample survey. For sampling, the methodology used from an extensive population depends on the type of study being conducted; but may involve simple random sampling or systematic sampling.
Methods of Sampling
Random sampling
(A) Random sampling  ● Random sampling method refers to a method in which every item in the universe has an equal chance of being selected.
● It is also known as probability sampling or representative sampling. ● There is no room for discrimination in random sampling. 
(B) The merits of random sampling are as follows:


(1) No personal bias  ● The selection of various items in the sample remains free from the personal bias of the investigator. 
(2) Based on probability  ● Due to the random character of the sample, the rules of probability are applicable. 
(3) Increasing representative of the population  ● As the size of a random sample increases, it becomes more and more representative of the population. 
(4) Accuracy can be assessed  ● The accuracy can be assessed with the help of the magnitude of sampling errors. 
(c) The demerits of random sampling are as follows:  
(1) Not suitable for small samples  ● If the sample is small, it may not reflect the true characteristics of the population. 
(2) Difficult to prepare sampling frame  ● The selection of a random sample requires the preparation of a sampling frame, which may be difficult for a large or an infinite population. 
(ii) Critical values
Ans. A critical value is a point on the distribution of the test statistic under the null hypothesis that defines a set of values that call for rejecting the null hypothesis. This set is called critical or rejection region. Usually, onesided tests have one critical value and twosided test have two critical values. The critical values are determined so that the probability that the test statistic has a value in the rejection region of the test when the null hypothesis is true equals the significance level (denoted as α or alpha).
Examples of calculating critical values
In hypothesis testing, there are two ways to determine whether there is enough evidence from the sample to reject H0 or to fail to reject H0. The most common way is to compare the pvalue with a prespecified value of α, where α is the probability of rejecting H0 when H0 is true. However, an equivalent approach is to compare the calculated value of the test statistic based on your data with the critical value. The following are examples of how to calculate the critical value for a 1sample ttest and a OneWay ANOVA.
Calculating a critical value for a 1sample ttest
Suppose you are performing a 1sample ttest on ten observations, have a twosided alternative hypothesis (that is, H1 not equal to), and are using an alpha of 0.10:
Select Calc > Probability Distributions > t.
Select Inverse cumulative probability.
In Degrees of freedom, enter 9 (the number of observations minus one).
In Input constant, enter 0.95 (one minus onehalf alpha).
This gives you an inverse cumulative probability, which equals the critical value, of 1.83311. If the absolute value of the tstatistic value is greater than this critical value, then you can reject the null hypothesis, H0, at the 0.10 level of significance.
Calculating a critical value for an analysis of variance (ANOVA)
Suppose you are performing a oneway ANOVA on twelve observations, the factor has three levels, and you are using an alpha of 0.05:
Choose Calc > Probability Distributions > F.
Select Inverse cumulative probability.
In Numerator degrees of freedom, enter 2 (the number of factor levels minus one).
In Denominator degrees of freedom, enter 9 (the degrees of freedom for error).
In Input constant, enter 0.95 (one minus alpha).
This gives you an inverse cumulative probability (critical value) of 4.25649. If the Fstatistic is greater than this critical value, then you can reject the null hypothesis, H0, at the 0.05 level of significance.
(iii) Probability Density Function
Ans. In probability theory, a probability density function (PDF), or density of a continuous random variable, is a function whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative likelihood that the value of the random variable would be close to that sample. Probabilility density is the probability per unit length, in other words, while the absolute likelihood for a continuous random variable to take on any particular value is 0 (since there is an infinite set of possible values to begin with), the value of the PDF at two different samples can be used to infer, in any particular draw of the random variable, how much more likely it is that the random variable would be close to one sample compared to the other sample.
In a more precise sense, the PDF is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value. This probability is given by the integral of this variable’s PDF over that range—that is, it is given by the area under the density function but above the horizontal axis and between the lowest and greatest values of the range. The probability density function is nonnegative everywhere, and its integral over the entire space is equal to 1.
The terms “probability distribution function” and “probability function” have also sometimes been used to denote the probability density function. However, this use is not standard among probabilists and statisticians. In other sources, “probability distribution function” may be used when the probability distribution is defined as a function over general sets of values or it may refer to the cumulative distribution function, or it may be a probability mass function (PMF) rather than the density. “Density function” itself is also used for the probability mass function, leading to further confusion. In general though, the PMF is used in the context of discrete random variables (random variables that take values on a countable set), while the PDF is used in the context of continuous random variables.
Suppose bacteria of a certain species typically live 4 to 6 hours. The probability that a bacterium lives exactly 5 hours is equal to zero. A lot of bacteria live for approximately 5 hours, but there is no chance that any given bacterium dies at exactly 5.00… hours. However, the probability that the bacterium dies between 5 hours and 5.01 hours is quantifiable. Suppose the answer is 0.02 (i.e., 2%). Then, the probability that the bacterium dies between 5 hours and 5.001 hours should be about 0.002, since this time interval is onetenth as long as the previous. The probability that the bacterium dies between 5 hours and 5.0001 hours should be about 0.0002, and so on.
In this example, the ratio (probability of dying during an interval) / (duration of the interval) is approximately constant, and equal to 2 per hour (or 2 hour−1). For example, there is 0.02 probability of dying in the 0.01hour interval between 5 and 5.01 hours, and (0.02 probability / 0.01 hours) = 2 hour−1. This quantity 2 hour−1 is called the probability density for dying at around 5 hours. Therefore, the probability that the bacterium dies at 5 hours can be written as (2 hour−1) dt. This is the probability that the bacterium dies within an infinitesimal window of time around 5 hours, where dt is the duration of this window. For example, the probability that it lives longer than 5 hours, but shorter than (5 hours + 1 nanosecond), is (2 hour−1)×(1 nanosecond) ≈ 6×10−13 (using the unit conversion 3.6×1012 nanoseconds = 1 hour).
(iv) Cramer’s rule
In linear algebra, Cramer’s rule is a specific formula used for solving a system of linear equations containing as many equations as unknowns, efficient whenever the system of equations has a unique solution. This rule is named after Gabriel Cramer (1704–1752), who published the rule for an arbitrary number of unknowns in 1750. This is the most commonly used formula for getting the solution for the given system of equations formed through matrices. The solution obtained using Cramer’s rule will be in terms of the determinants of the coefficient matrix and matrices obtained from it by replacing one column with the column vector of the righthand sides of the equations.
Cramer’s Rule Definition
Cramer’s rule is one of the important methods applied to solve a system of equations. In this method, the values of the variables in the system are to be calculated using the determinants of matrices. Thus, Cramer’s rule is also known as the determinant method.
Cramer’s Rule Formula
Consider a system of linear equations with n variables x₁, x₂, x₃, …, xₙ written in the matrix form AX = B.
Here,
A = Coefficient matrix (must be a square matrix)
X = Column matrix with variables
B = Column matrix with the constants (which are on the right side of the equations)
Now, we have to find the determinants as:
D = A, Dx1, Dx2, Dx3,…, Dxn
Here, Dxi for i = 1, 2, 3,…, n is the same determinant as D such that the column is replaced with B.
Thus,
x1 = Dx1/D; x2 = Dx2/D; x3 = Dx3/D; ….; xn = Dxn/D {where D is not equal to 0}
Let’s have a look at the formulas of Cramer’s rule for 2×2 and 3×3 matrices.
Cramer’s Rule Formula
Consider a system of linear equations with n variables x₁, x₂, x₃, …, xₙ written in the matrix form AX = B.
Here,
A = Coefficient matrix (must be a square matrix)
X = Column matrix with variables
B = Column matrix with the constants (which are on the right side of the equations)
Now, we have to find the determinants as:
D = A, Dx1, Dx2, Dx3,…, Dxn
Here, Dxi for i = 1, 2, 3,…, n is the same determinant as D such that the column is replaced with B.
Thus,
x1 = Dx1/D; x2 = Dx2/D; x3 = Dx3/D; ….; xn = Dxn/D {where D is not equal to 0}
Let’s have a look at the formulas of Cramer’s rule for 2×2 and 3×3 matrices.
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