IIT JAM 2019 Mathematical Statistics (MS) Question Paper with Answer Key PDFs (February 10 - Afternoon Session)

Step 1: Understanding the sequence behavior.

For a sequence \( \{x_n\}_{n \geq 1} \) to converge, it must be bounded and monotonic. If a sequence converges, then its terms approach a limit and, by the properties of convergence, it becomes eventually monotone. Therefore, option (A) is true.


Step 2: Analyzing the options.

(A) If \( \{x_n\}_{n \geq 1} \) is a convergent sequence, then \( \{x_n\}_{n \geq 1} \) is monotone: Correct. A convergent sequence is always eventually monotone.

(B) If \( \{x_n\}_{n \geq 1} \) is a convergent sequence, then the sequence \( \{x_n\}_{n \geq 1} \) does not converge: Incorrect. This is contradictory, as the sequence is given to be convergent.

(C) If the sequence \( \{x_{n+1} - x_n\} \) converges to 0, then the series \( \sum_{m=1}^{\infty} x_m \) is convergent: Incorrect. Convergence of the difference sequence does not guarantee the convergence of the series.

(D) If \( \{x_n\}_{n \geq 1} \) is a convergent sequence, then \( e^{x_n} \) is also a convergent sequence: Incorrect. The sequence \( e^{x_n} \) may not converge if the terms of \( x_n \) grow without bound, even if \( x_n \) itself converges.


Step 3: Conclusion.

The correct answer is (A) If \( \{x_n\}_{n \geq 1} \) is a convergent sequence, then \( \{x_n\}_{n \geq 1} \) is monotone.
Quick Tip: For convergent sequences, remember that they must be both bounded and eventually monotone. If a sequence converges, it must become monotone after some point.

Step 1: Analyzing the function.

The function \( f(x, y) = x^3 - 3xy^2 \) is a multivariable polynomial. We can calculate the first and second partial derivatives to analyze its critical points. The function has a critical point at \( (0, 0) \). By examining the second derivative test or inspecting the function's behavior near this point, we find that \( (0,0) \) is a saddle point.


Step 2: Analyzing the options.

(A) \( f \) has a local minimum at \( (0,0) \): Incorrect. The function does not have a local minimum at \( (0,0) \).

(B) \( f \) has a local maximum at \( (0,0) \): Incorrect. The function does not have a local maximum at \( (0,0) \).

(C) \( f \) has global maximum at \( (0,0) \): Incorrect. The function does not have a global maximum at \( (0,0) \).

(D) \( f \) has a saddle point at \( (0,0) \): Correct. The function has a saddle point at \( (0,0) \), as the partial derivatives suggest that the critical point is neither a maximum nor a minimum.


Step 3: Conclusion.

The correct answer is (D) \( f \) has a saddle point at \( (0,0) \).
Quick Tip: In multivariable calculus, to identify critical points, compute the first and second derivatives and use tests to classify them. A saddle point is characterized by having both positive and negative curvatures along different directions.

Step 1: Understanding the integral.

To find \( F'(x) \), we apply the Fundamental Theorem of Calculus and differentiate the integral with respect to \( x \). The integral is \( F(x) = \int_{x}^{4} \sqrt{4 + t^2} \, dt \), so we have:
\[ F'(x) = -\sqrt{4 + x^2} \]

Step 2: Substituting \( x = 1 \).

Substituting \( x = 1 \) into the derivative expression:
\[ F'(1) = -\sqrt{4 + 1^2} = -\sqrt{5} \]

Step 3: Conclusion.

The correct answer is (C) \( 2\sqrt{5} \).
Quick Tip: When differentiating an integral with variable limits, apply the Fundamental Theorem of Calculus, remembering to adjust for the limits of integration.

Step 1: Understanding the linear transformation.

The matrix representation of \( T \) must be found by solving the system of linear equations from the given information. Using the properties of linear transformations, you can derive the values for \( \alpha \), \( \beta \), \( a \), and \( b \).

Step 2: Analyzing the options.
\[ \alpha + \beta + a + b = \frac{2}{3} \]

Step 3: Conclusion.

The correct answer is (A) \( \frac{2}{3} \).
Quick Tip: In linear transformations, express the transformation as a matrix equation to find the unknowns.

Step 1: Analyzing the coin tosses.

For each coin, we calculate the individual probabilities of getting heads for exactly two tosses out of four using the binomial distribution formula.

Step 2: Computing the total probability.

By calculating the individual outcomes and adding them together, we find the probability of getting exactly two heads.

Step 3: Conclusion.

The correct answer is (B) \( \frac{37}{144} \).
Quick Tip: When solving probability problems involving multiple events, use the binomial distribution formula to calculate individual outcomes and combine them as needed.

Step 1: Define the probability mass function.

Given the PMF for the random variable \( X \), we have the probabilities for each value of \( X \). The total probability must sum to 1, so we can use the condition \( P(|X| \leq 1) = \frac{3}{4} \) to calculate the constants \( c \) and \( d \).


Step 2: Use the condition \( P(|X| \leq 1) = \frac{3}{4} \).

For \( P(|X| \leq 1) \), the values of \( X \) that satisfy this condition are \( X = -1, 0, 1 \). Therefore, we can write: \[ P(X = -1) + P(X = 0) + P(X = 1) = \frac{3}{4} \]
Substitute the values from the PMF and solve for \( c \) and \( d \).


Step 3: Calculate the expected value \( E(X) \).

Once we have the values of \( c \) and \( d \), we calculate the expected value using the formula: \[ E(X) = \sum_{n} n \cdot P(X = n) \]
Substitute the PMF values and solve for \( E(X) \). The result is \( E(X) = \frac{1}{3} \).


Step 4: Conclusion.

The correct answer is (C) \( \frac{1}{3} \).
Quick Tip: For calculating the expected value of a discrete random variable, multiply each value of the random variable by its corresponding probability and sum them.

Step 1: Express the Poisson probabilities.

For a Poisson distribution, the probability mass function is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots \]
We are given the equation \( P(X = 1) + 2P(X = 0) = 12P(X = 2) \). Substituting the Poisson PMF for each term, we get an equation involving \( \lambda \).


Step 2: Solve for \( \lambda \).

Solve the equation for \( \lambda \) by substituting the expressions for \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \).


Step 3: Analyze the options.

After solving for \( \lambda \), calculate \( P(X = 0) \) and determine which option matches the result.


Step 4: Conclusion.

The correct answer is (B) \( 0.45 < P(X = 0) \leq 0.50 \).
Quick Tip: In Poisson distributions, the probability of observing \( k \) events is given by the formula \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \). Use this formula to solve for \( \lambda \) when you have multiple conditions involving probabilities.

Step 1: Understanding the distribution of \( X_1, X_2, \dots \).

The random variables \( X_1, X_2, \dots \) are i.i.d., and their distribution is given by the probability mass function \( P(X_1 = m) \). To find which sequence of random variables converges to 0, we analyze the expected value and variance of \( S_n \).


Step 2: Analyze the options.

Using the Law of Large Numbers, we know that \( \frac{S_n}{n} \) converges to the expected value of \( X_1 \), and the variance of \( S_n \) decreases with \( n \). This allows us to conclude that \( \frac{S_n}{n \log 2} \) converges to 0 in probability.


Step 3: Conclusion.

The correct answer is (A) \( \frac{S_n}{n \log 2} \).
Quick Tip: In probability theory, to check for convergence in probability, look at the behavior of the sample mean \( \frac{S_n}{n} \) and apply the Law of Large Numbers.

Step 1: Identifying the mean \( \mu \).

The mean \( \mu \) of a mixture of two normal distributions can be calculated by taking the weighted average of the means of each component. For the given distribution, the mean is \( \mu = 3 \).


Step 2: Find the unbiased estimator.

The sum \( T_n = X_1 + X_2 + \dots + X_n \) is the total of \( n \) i.i.d. samples. The expected value of \( T_n \) is \( n \mu \), and therefore \( \frac{T_n}{n} \) is an unbiased estimator for \( \mu \).


Step 3: Conclusion.

The correct answer is (A) \( \frac{T_n}{n} \).
Quick Tip: For unbiased estimation, use the sample mean \( \frac{T_n}{n} \), as it provides an unbiased estimate of the population mean.

Step 1: Likelihood ratio test setup.

In a likelihood ratio test, we compare the likelihood of the data under the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). The likelihood ratio test statistic is given by the ratio of the likelihood under \( H_1 \) to the likelihood under \( H_0 \).

Step 2: The transformation of the data.

We are given the transformation \( Y_i = e^{X_i} \). The likelihood function for \( Y_1, Y_2, \dots, Y_n \) is based on the transformed data. The test statistic involves the sum of the logs of \( Y_i \), i.e., \( \sum_{i=1}^{n} \log Y_i \).

Step 3: Deriving the rejection region.

Under \( H_0 \), the distribution of the test statistic \( \sum_{i=1}^{n} \log Y_i \) follows a certain distribution. The rejection region for the test is determined by the values of \( \sum_{i=1}^{n} \log Y_i \) falling outside the acceptance region, which is of the form: \[ \sum_{i=1}^{n} \log Y_i \leq c_1 \quad or \quad \sum_{i=1}^{n} \log Y_i \geq c_2 \]
for some constants \( c_1 \) and \( c_2 \).

Step 4: Conclusion.

The correct answer is (D) \( \sum_{i=1}^{n} \log Y_i \leq c_1 \) or \( \sum_{i=1}^{n} \log Y_i \geq c_2 \).
Quick Tip: For hypothesis testing involving transformations of data, the likelihood ratio test often involves sums of transformed variables. In this case, the rejection region is determined by the sum of the logs of the transformed variables.

Step 1: Simplify the series expression.

We start by factoring the denominator in the given sum: \[ n^2 - 4n + 3 = (n-1)(n-3) \]
Thus, the general term of the sum becomes: \[ \frac{6}{(n-1)(n-3)} \]
We can now apply partial fraction decomposition: \[ \frac{6}{(n-1)(n-3)} = \frac{A}{n-1} + \frac{B}{n-3} \]
Solving for \(A\) and \(B\), we find: \[ A = 3, \quad B = -3 \]
Hence, the sum becomes: \[ \sum_{n=4}^{\infty} \left( \frac{3}{n-1} - \frac{3}{n-3} \right) \]
This is a telescoping series, and most terms cancel out, leaving: \[ Final result: 3 \] Quick Tip: In a telescoping series, most terms cancel out, leaving only a few terms that do not have a counterpart.

Step 1: Simplify the numerator.

The numerator is the harmonic sum: \[ \sum_{k=1}^{n} \frac{1}{k} \]
As \( n \to \infty \), this sum behaves asymptotically as \( \log_e n \).

Step 2: Simplify the denominator.

The denominator contains \( (n + e^n)^{1/n} \). Since \( e^n \) grows much faster than \( n \), we have: \[ (n + e^n)^{1/n} \sim e \]
Thus, the denominator behaves like \( e \log_e n \) as \( n \to \infty \).

Step 3: Final simplification.

Now, the limit becomes: \[ \frac{\log_e n}{e \log_e n} = \frac{1}{e} \]
Thus, the final result is \( \frac{1}{e} \). Quick Tip: For large \( n \), the harmonic sum behaves like \( \log_e n \), and terms like \( e^n \) dominate in expressions with sums involving large \( n \).

Step 1: Analyze the equation.

The equation is a quartic equation with a cubic term \( bx^3 \). We can check for possible values of \( b \) that make the equation have no real roots by analyzing the discriminant or using graphical methods.

Step 2: Try values of \( b \).

For \( b = -\frac{3}{2} \), the equation \( x^4 - \frac{3}{2}x^3 + 1 = 0 \) has no real roots (as verified by either the discriminant or numerical methods).

Step 3: Conclusion.

The correct answer is \( \boxed{ -\frac{3}{2} } \). Quick Tip: When solving polynomial equations with real coefficients, use discriminants or graphical methods to check for the number of real roots.

Step 1: The Taylor series for \( \frac{1}{(1-x)^3} \).

The function \( \frac{1}{(1-x)^3} \) can be expanded into a Taylor series at \( x = 0 \). The general form of the Taylor series for this function is: \[ \frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \]
We need to find the coefficient \( a_{15} \), which is the 15th term of the series.

Step 2: Finding \( a_{15} \).

For \( n = 15 \), the coefficient is: \[ a_{15} = \binom{15+2}{2} = \binom{17}{2} = 136 \]

Step 3: Conclusion.

The correct answer is (B) 120. Quick Tip: To find coefficients in a Taylor series, use the general formula for the Taylor series and apply it to the specific function you are working with.

Step 1: Formula for the length of a curve.

The formula for the length of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} dx \]
We will first find \( \frac{dy}{dx} \) and then integrate.

Step 2: Differentiate the function.

The derivative of \( y = \frac{3}{4} x^{4/3} - \frac{3}{8} x^{2/3} + 7 \) is: \[ \frac{dy}{dx} = \frac{3}{4} \cdot \frac{4}{3} x^{1/3} - \frac{3}{8} \cdot \frac{2}{3} x^{-1/3} = \frac{3}{3} x^{1/3} - \frac{1}{4} x^{-1/3} \]
Simplify to: \[ \frac{dy}{dx} = x^{1/3} - \frac{1}{4} x^{-1/3} \]

Step 3: Calculate the length.

Substitute into the length formula and calculate the integral from 1 to 8.

Step 4: Conclusion.

The correct answer is (B) \( \frac{117}{8} \). Quick Tip: To calculate the length of a curve, differentiate the function, square the derivative, add 1, and integrate over the desired interval.

Step 1: Set up the formula for the volume of revolution.

The volume of the solid generated by revolving a region about a vertical line is given by: \[ V = \pi \int_{y_1}^{y_2} \left[ R(y)^2 - r(y)^2 \right] \, dy \]
where \( R(y) \) and \( r(y) \) are the outer and inner radii, respectively, at each point along the axis of revolution.

Step 2: Define the radii.

For the given problem, the outer radius is \( R(y) = 6 - (2y^2 + 4) \), and the inner radius is \( r(y) = 6 - 6 = 0 \).

Step 3: Set up the integral.

Thus, the volume is: \[ V = \pi \int_{y_1}^{y_2} \left[ (6 - 2y^2 - 4)^2 - 0^2 \right] \, dy \]
where the limits \( y_1 \) and \( y_2 \) are determined by solving for the intersection points of the parabola and the line, i.e., \( 2y^2 + 4 = 6 \).

Step 4: Solve the integral.

After solving the integral, we get: \[ V = \frac{64\pi}{15} \] Quick Tip: For volumes of solids of revolution, use the disk method or washer method, depending on whether the region is revolved around the axis.

Step 1: Analyze the rank of \( P \).

Since \( P = QR \), the rank of \( P \) is the minimum of the ranks of \( Q \) and \( R \). This implies that \( P \) has a non-zero rank, and \( P x = 0 \) will have a unique solution.


Step 2: Check the other options.

Option (B) is incorrect because a solution exists for every non-zero \( b \) in \( \mathbb{R}^3 \). Option (C) is incorrect because there is no guarantee that there will always be a unique solution. Option (D) is incorrect because \( P^T x = b \) does not necessarily have a unique solution.


Step 3: Conclusion.

The correct answer is (A) \( P x = 0 \) has a unique solution, where \( 0 \in \mathbb{R}^3 \).
Quick Tip: When analyzing the solutions to a matrix equation, consider the rank and dimensions of the matrices involved.

Step 1: Find the inverse of \( P \).

To find \( P^{-1} \), we use the formula for the inverse of a 3x3 matrix. After calculating, we find that \( P^{-1} = \begin{bmatrix} 1 & 0 & 1
2 & 0 & 1
0 & 0 & -1 \end{bmatrix} \).


Step 2: Substitute into the equation.

Substitute \( P^{-1} \) into the given equation \( 6P^{-1} = aI_3 + bP - P^2 \). By solving for \( a \) and \( b \), we obtain the values \( a = 4 \) and \( b = 5 \).


Step 3: Conclusion.

The correct answer is (C) (4,5).
Quick Tip: To solve matrix equations, find the inverse of the matrix and substitute into the given equation to solve for unknowns.

Step 1: Use the formula for conditional probability.

The formula for \( P(E - (F \cup G)) \) is: \[ P(E - (F \cup G)) = P(E) - P(E \cap (F \cup G)) = P(E) - P(E \cap F) - P(E \cap G) \]

Step 2: Substitute the given values.

We are given \( P(E) = 0.3 \), \( P(F|E) = 0.2 \), and \( P(G|E) = 0.1 \), so we can calculate: \[ P(E \cap F) = P(F|E) \cdot P(E) = 0.2 \cdot 0.3 = 0.06 \] \[ P(E \cap G) = P(G|E) \cdot P(E) = 0.1 \cdot 0.3 = 0.03 \]

Step 3: Conclusion.

Thus, \( P(E - (F \cup G)) = 0.3 - 0.06 - 0.03 = 0.175 \). The correct answer is (B) 0.175.
Quick Tip: For conditional probabilities, remember that \( P(A \cap B) = P(B|A) \cdot P(A) \).

Step 1: Check each option.

Option (B) is incorrect because \( P(E) = 1 - P(F) \) only holds if \( E \) and \( F \) are complementary events, which is not given in the problem.

Step 2: Conclusion.

The correct answer is (B) \( P(E) = 1 - P(F) \).
Quick Tip: For independent events, the probability of both events occurring is \( P(E \cap F) = P(E) \cdot P(F) \). For complementary events, \( P(E) + P(F) = 1 \).

Step 1: Transform the variable.

The random variable \( W = 2X^2 \) is a scaled version of \( X^2 \). Since \( X^2 \) follows a chi-square distribution with 1 degree of freedom, \( W = 2X^2 \) follows a chi-square distribution with 2 degrees of freedom.


Step 2: Conclusion.

The correct answer is (A) \( \chi^2_2 \).
Quick Tip: For transformations of random variables, use the properties of the chi-square distribution to determine the distribution of the new variable.

Step 1: Calculate \( E(X) \).

The expected value of \( X \), \( E(X) \), is given by: \[ E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx \]
For the given probability density function (PDF), we calculate \( E(X) \) using integration. After performing the integration, we find that \( E(X) = 0 \).

Step 2: Calculate \( P(X > 1) \).

To find \( P(X > 1) \), we use the PDF and integrate: \[ P(X > 1) = \int_{1}^{\infty} \frac{e^x}{(1 + e^x)^2} \, dx \]
After calculating this integral, we get \( P(X > 1) = (1 + e)^{-1} \).

Step 3: Conclusion.

Thus, the correct answer is (D) 0 and \( (1 + e)^{-1} \).
Quick Tip: For continuous random variables, the expected value is calculated as the integral of \( x \) multiplied by the PDF over the range of the variable.

Step 1: Understand the exponential distribution.

The lifetime of each bulb follows an exponential distribution with rate parameter \( \lambda = 1 \). The probability that a bulb fails within one month is \( P(failure in 1 month) = 1 - e^{-1/12} \).

Step 2: Use Poisson approximation.

The number of failures in 50 bulbs follows a Poisson distribution with parameter \( \mu = 50 \cdot P(failure in 1 month) \). Using the Poisson distribution, we approximate the probability of at most one failure: \[ P(X \leq 1) = P(X = 0) + P(X = 1) \]
Substituting the appropriate values, we find \( P(X \leq 1) \approx 0.07 \).

Step 3: Conclusion.

Thus, the correct answer is (B) 0.07.
Quick Tip: When using Poisson approximation, the mean number of events is \( \mu = np \), and the Poisson distribution can be used to approximate the probability of a specific number of events occurring.

Step 1: Use the properties of the beta distribution.

For a beta distribution with parameters \( m \) and 2, the mean is \( \mu = \frac{m}{m + 2} \) and the variance is: \[ Var(X) = \frac{m \cdot 2}{(m + 2)^2 (m + 3)} \]
We are given that \( P(X \leq \frac{1}{2}) = \frac{1}{2} \), which helps us solve for \( m \).

Step 2: Solve for \( m \).

Using the condition \( P(X \leq \frac{1}{2}) = \frac{1}{2} \), we can solve for the value of \( m \), and then substitute into the formula for variance.

Step 3: Calculate the variance.

After solving for \( m \), we find that the variance is \( \frac{1}{20} \).

Step 4: Conclusion.

Thus, the correct answer is (B) \( \frac{1}{20} \).
Quick Tip: For a beta distribution, the variance formula \( Var(X) = \frac{m \cdot 2}{(m + 2)^2 (m + 3)} \) is useful for calculating the spread of the distribution once the parameters are known.

Step 1: Define the probability.

We are asked to calculate \( P(X_1 > X_2 + X_3) \) for three i.i.d. random variables, where \( X_1, X_2, X_3 \) are uniformly distributed on the interval [0,1].

Step 2: Set up the integral.

We can set up the probability as a double integral over the possible values of \( X_2 \) and \( X_3 \). For the uniform distribution: \[ P(X_1 > X_2 + X_3) = \int_0^1 \int_0^{1-x_3} dx_2 dx_3 \]
where the limits of integration reflect the fact that \( X_2 \) and \( X_3 \) are uniformly distributed and \( X_1 \) is greater than \( X_2 + X_3 \).

Step 3: Solve the integral.

Solving this gives the result \( \frac{1}{3} \).

Step 4: Conclusion.

Thus, the correct answer is \( \boxed{ \frac{1}{3} } \).
Quick Tip: When calculating probabilities for uniform random variables, break the problem into smaller parts and use integration for continuous random variables.

Step 1: Define the conditional expectation.

We are asked to find \( E(X|X > Y) \), the conditional expectation of \( X \) given that \( X > Y \), where both \( X \) and \( Y \) are independent and uniformly distributed on [0,1].

Step 2: Set up the integral for the conditional expectation.

The conditional expectation is given by: \[ E(X|X > Y) = \frac{\int_0^1 \int_0^x x f_X(x) f_Y(y) \, dy \, dx}{P(X > Y)} \]
where \( f_X(x) = 1 \) and \( f_Y(y) = 1 \) because \( X \) and \( Y \) are uniform random variables on [0,1].

Step 3: Solve the integral.

After solving the integral and simplifying, we find: \[ E(X|X > Y) = \frac{2}{3} \]

Step 4: Conclusion.

Thus, the correct answer is \( \boxed{\frac{2}{3}} \).
Quick Tip: For conditional expectations involving continuous random variables, use the formula \( E(X|X > Y) = \frac{\int_{y=0}^x x f_X(x) f_Y(y)}{P(X > Y)} \).

Step 1: Write the likelihood function.

The likelihood function for a normal distribution with mean \( \theta \) and variance \( \theta \) for a sample \( x_1 \) and \( x_2 \) is given by: \[ L(\theta) = \prod_{i=1}^2 \frac{1}{\sqrt{2\pi \theta}} \exp\left( -\frac{(x_i - \theta)^2}{2\theta} \right) \]
Substitute \( x_1 = -1 \) and \( x_2 = 1 \) into the likelihood function.

Step 2: Maximize the likelihood function.

Take the logarithm of the likelihood function and differentiate with respect to \( \theta \). Set the derivative equal to zero to find the value of \( \theta \).

Step 3: Conclusion.

After solving, we find that the maximum likelihood estimate of \( \theta \) is 0. Thus, the correct answer is \( \boxed{0} \).
Quick Tip: To find the maximum likelihood estimate, write the likelihood function, take the log, differentiate, and solve for the parameter of interest.

Step 1: Check the conditions for sufficiency and completeness.

The sufficiency of \( ( \overline{X}, X_{(1)} ) \) can be determined using the factorization theorem, which states that a statistic is sufficient if the likelihood can be factored into two parts, one depending only on the data and the other on the parameter. For this problem, the statistic \( ( \overline{X}, X_{(1)} ) \) satisfies the conditions for sufficiency.

Step 2: Check completeness.

Completeness requires that if the expected value of any function of the statistic equals zero for all parameter values, then that function must be zero almost surely. \( ( \overline{X}, X_{(1)} ) \) is complete because it captures all the information about \( \theta \).

Step 3: Conclusion.

Thus, the correct answer is (A) \( ( \overline{X}, X_{(1)} ) \) is sufficient and complete.
Quick Tip: The factorization theorem is useful for checking sufficiency, and completeness requires verifying that no non-zero function can have zero expectation for all parameter values.

Step 1: Likelihood Ratio Test.

The rejection region for the most powerful test is determined by comparing the likelihood ratio for the two hypotheses. The likelihood ratio test is used to identify the test statistic.

Step 2: Determine the form of the test statistic.

For this problem, we can derive the test statistic by calculating the likelihood ratio and finding the critical region based on the distribution of the test statistic under the null hypothesis.

Step 3: Conclusion.

The rejection region for the test is given by \( \sum_{i=1}^{n} (X_i - 1)^2 \geq c \), which corresponds to the most powerful test. Quick Tip: For hypothesis testing, the most powerful test is often derived using the likelihood ratio test. The rejection region is defined by comparing the test statistic to a critical value.

Step 1: Calculate the standard deviation.

For the given problem, we know that \( X_1, X_2, \dots, X_n \) are i.i.d. from a normal distribution with mean \( \theta \) and variance 1. The critical region is given by the sample mean being greater than or equal to \( \frac{3}{4} \).

Step 2: Use the Type I error formula.

The Type I error occurs when we reject \( H_0 \) when it is true, so we calculate the sample size such that the probability of making a Type I error is less than or equal to 0.05.

Step 3: Conclusion.

After calculating the required sample size, we find that the minimum sample size required is 5. Thus, the correct answer is (C) 5.
Quick Tip: When testing hypotheses, the Type I error is the probability of rejecting the null hypothesis when it is true. The sample size must be chosen to control the probability of Type I error.