# IES ISS 2023 ECONOMICS PAPER 1 COMPLETE SOLUTION WITH DETAILED ANSWERS QUESTION- 1a, b, 2b, 3c, 6b.

**1. a. In a two-good world, show that both the goods cannot be inferior. **

In a two-good world, let's consider two goods: Good A and Good B. To show that both goods cannot be inferior, we need to understand the concept of inferior goods.

An inferior good is a type of good where demand decreases as consumer income increases. This means that as individuals' income rises, they tend to consume less of the inferior good and shift their consumption towards other goods.

Now, suppose both Good A and Good B are inferior goods. This implies that as income increases, the demand for both goods would decrease.

However, if both goods are inferior, it would mean that individuals would reduce their consumption of both Good A and Good B as their income rises. As a result, the demand for both goods would decrease simultaneously.

This contradicts the assumption that individuals' income has increased. If individuals' income has increased, it would generally lead to an increase in demand for most goods, especially normal goods.

Therefore, if both goods are considered inferior goods, it would imply a situation where individual's income increase leads to a decrease in demand for both goods, which is not a typical characteristic of inferior goods.

Hence, it can be concluded that in a two-good world, both goods cannot be inferior. At least one of the goods must be a normal good or a superior good, where the demand increases as income rises.

**b. Calculate the elasticity of substitution for the following production function: **

**Q= (L^ρ+K^ρ)^ y/ρ**

To calculate the elasticity of substitution (σ) for the given production function, we need to differentiate the logarithm of the production function with respect to the logarithm of the input ratio (K/L) and then evaluate it at the desired input ratio.

Let's proceed with the calculation:

ln(Q) = (y/ρ) * ln(L^ρ + K^ρ)

Taking the derivative of ln(Q) with respect to ln(K/L):

∂ln(Q)/∂ln(K/L) = (y/ρ) * ∂ln(L^ρ + K^ρ) / ∂ln(K/L)

Using the chain rule, we can rewrite the right-hand side of the equation as:

∂ln(Q)/∂ln(K/L) = (y/ρ) * (∂ln(L^ρ + K^ρ) / ∂(L^ρ + K^ρ)) * (∂(L^ρ + K^ρ) / ∂ln(K/L))

The first term in parentheses (∂ln(L^ρ + K^ρ) / ∂(L^ρ + K^ρ)) represents the elasticity of substitution between labor and capital inputs, which is denoted by σ. The second term (∂(L^ρ + K^ρ) / ∂ln(K/L)) can be calculated as follows:

∂(L^ρ + K^ρ) / ∂ln(K/L) = (∂(L^ρ + K^ρ) / ∂(K^ρ/L^ρ)) * (∂(K^ρ/L^ρ) / ∂ln(K/L))

Simplifying the equation, we find:

∂(L^ρ + K^ρ) / ∂ln(K/L) = ρ(K^ρ/L^ρ) * (L^ρ/K^ρ)

Now, substituting this value back into the equation for the derivative, we have:

∂ln(Q)/∂ln(K/L) = (y/ρ) * σ * (ρ(K^ρ/L^ρ) * (L^ρ/K^ρ))

Simplifying further, we find:

∂ln(Q)/∂ln(K/L) = y * (K^ρ/L^ρ)

This expression represents the elasticity of substitution (σ) for the given production function. It is equal to the ratio of the capital input raised to the power of ρ and the labor input raised to the power of ρ.

Therefore, the elasticity of substitution (σ) is given by:

σ = y * (K^ρ/L^ρ)

Please note that this result holds for the specific production function provided, but may vary for different functional forms of the production function.

**2. B. The following estimated equation was obtained by OLS with sample size 80:**

**Y hat_i= 2.2 +0.11x_1i+3.48x_2i+0.34x_3i**

** (3.4) (0.005) (2.2) (0.15)**

**Figures in Parentheses indicate standard errors. The explained sum of square was 112.5 and the residual sum of square was 19.5. **

**i Calculate the values of R^2 and R bar ^2. **

**ii. Test the significance of the slope coefficients by using t statistics at 5% level of significance. **

**[given that t_0.95, 76= 1.98]**

**iii. test for overall significance of the model at 5% level of significance. **

**[Given that F_0.95,76= 1.35]**

i. To calculate the values of R^2 and R bar ^2, we need to use the following formulas:

R^2 = Explained Sum of Squares (ESS) / Total Sum of Squares (TSS)

R^2 = ESS / (ESS + RSS)

R bar ^2 = 1 - [(1 - R^2) * (n - 1) / (n - k - 1)]

where n is the sample size and k is the number of independent variables.

Given:

ESS = 112.5

RSS = 19.5

n = 80

k = 3 (number of independent variables)

Calculating R^2:

R^2 = ESS / (ESS + RSS) = 112.5 / (112.5 + 19.5) = 0.852

Calculating R bar ^2:

R bar ^2 = 1 - [(1 - R^2) * (n - 1) / (n - k - 1)]

R bar ^2 = 1 - [(1 - 0.852) * (80 - 1) / (80 - 3 - 1)]

R bar ^2 = 0.840

Therefore, the values of R^2 and R bar ^2 are 0.852 and 0.840, respectively.

ii. To test the significance of the slope coefficients, we compare the t-statistics with the critical value of t_(0.95, 76) = 1.98 at a 5% level of significance.

Given t_(0.95, 76) = 1.98, we compare it with the t-statistics for each coefficient:

t_(x1) = 0.11 / 0.005 = 22

t_(x2) = 3.48 / 2.2 = 1.58

t_(x3) = 0.34 / 0.15 = 2.27

Since t_(x1) = 22 > t_(0.95, 76) = 1.98, the coefficient for x1 is statistically significant.

Since t_(x2) = 1.58 < t_(0.95, 76) = 1.98, the coefficient for x2 is not statistically significant.

Since t_(x3) = 2.27 > t_(0.95, 76) = 1.98, the coefficient for x3 is statistically significant.

iii. To test for the overall significance of the model, we use the F-statistic and compare it with the critical value of F_(0.95, 76) = 1.35 at a 5% level of significance.

Given F_(0.95, 76) = 1.35, we calculate the F-statistic using the formula:

F = (ESS / k) / (RSS / (n - k - 1))

F = (112.5 / 3) / (19.5 / (80 - 3 - 1))

F = 12.5

Since F = 12.5 > F_(0.95, 76) = 1.35, the overall model is statistically significant at a 5% level of significance.

In summary:

i. The values of R^2 and R bar ^2 are 0.852 and 0.840, respectively.

ii. The coefficient for x1 is statistically significant, while the coefficients for x2 and x3 are not statistically significant.

iii. The overall model is statistically significant.

**3. C. In a regression equation of Y and X, the value of X is fixed at 5. What is the regression equation look like?**

In a regression equation of Y and X, if the value of X is fixed at 5, the regression equation will still include the coefficient of X, but the variable X will be replaced by its fixed value of 5. Let's assume the original regression equation is:

Y = β₀ + β₁X + ε

If we fix X at 5, the regression equation will become:

Y = β₀ + β₁(5) + ε

Simplifying the equation, we have:

Y = β₀ + 5β₁ + ε

So, the regression equation when X is fixed at 5 would be:

Y = (β₀ + 5β₁) + ε

In this new equation, the coefficient of X (β₁) is still present, but the variable X itself is replaced by the fixed value of 5. The constant term (β₀) may also be affected depending on the specific values of the original regression equation.

**6. b. Suppose that a monopolistic competitive market consists of 11 firms with the following identical demand and cost functions: **

**P_k= 150-2q_k- 0.2 Ʃq_i **

** i=1**

** i≠k**

**c_k= 0.5 q^3_k- 20q^2_k+270q_k**

**k=1,2…11**

**Determine the maximum profit and the corresponding price and quantity for a representative form. Assume that number of firms in the industry does not change. **

To determine the maximum profit, price, and quantity for a representative firm in a monopolistic competitive market with 11 firms, we need to find the profit-maximizing output level where marginal revenue equals marginal cost.

The demand function for a firm in the market is given by:

P_k = 150 - 2q_k - 0.2Ʃq_i (i=1, i≠k)

The cost function for a firm is given by:

c_k = 0.5(q_k^3) - 20(q_k^2) + 270q_k

To find the marginal cost, we take the derivative of the cost function with respect to q_k:

MC_k = d(c_k)/dq_k = 1.5(q_k^2) - 40q_k + 270

To find the marginal revenue, we take the derivative of the demand function with respect to q_k:

MR_k = d(P_k)/dq_k = -2 - 0.2Ʃdq_i (i=1, i≠k)

Setting MR_k equal to MC_k:

-2 - 0.2Ʃdq_i = 1.5(q_k^2) - 40q_k + 270

Simplifying the equation:

1.5(q_k^2) - 40q_k + 0.2Ʃdq_i = 272

Since there are 11 firms, the sum of the quantities produced by all firms in the market is:

Ʃdq_i = 11q_k

Substituting this into the equation:

1.5(q_k^2) - 40q_k + 0.2(11q_k) = 272

Simplifying further:

1.5(q_k^2) - 40q_k + 2.2q_k = 272

1.5(q_k^2) - 37.8q_k = 272

Solving this quadratic equation will give us the optimal output level (quantity) for the representative firm. From there, we can calculate the price and maximum profit.

Note: The specific values of q_k, P_k, and profit depend on the solution of the quadratic equation, which can be found using methods such as factoring, completing the square, or the quadratic formula.

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