Sample Solution

IGNOU MST-016 Solved Assignment | June 2024 to June 2025 | Statistical Inference | MSCAST Sample Solution

(a) State whether the following statements are True or False. Give reason in support of your answer:
(i) If $X_{1}, X_{2}, X_{3}, X_{4}$ and $X_{5}$ is a random sample of size 5 taken from an Exponential distribution, then estimator $T_{1}$ is more efficient than $T_{2}$ .

T_{1} = \frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}, T_{2} = \frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}

Answer:

To determine whether

T_{1}

is more efficient than

T_{2}

, we need to compare the variances of these estimators, since efficiency is related to the variance of an unbiased estimator.

First, let’s recall the properties of an Exponential distribution. Suppose

X \sim Exponential (λ)

. The mean of this distribution is

E [X] = \frac{1}{λ}

, and the variance is

Var (X) = \frac{1}{λ^{2}}

Mean and Variance of $T_{1}$

The estimator

T_{1}

is the sample mean:

T_{1} = \frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}

For the sample mean of a random sample from an Exponential distribution:

$E [T_{1}] = E [\frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}] = \frac{1}{5} (E [X_{1}] + E [X_{2}] + E [X_{3}] + E [X_{4}] + E [X_{5}]) = \frac{5}{5} \cdot \frac{1}{λ} = \frac{1}{λ}$
$Var (T_{1}) = Var (\frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}) = \frac{1}{25} (Var (X_{1}) + Var (X_{2}) + Var (X_{3}) + Var (X_{4}) + Var (X_{5})) = \frac{5}{25} \cdot \frac{1}{λ^{2}} = \frac{1}{5 λ^{2}}$

Mean and Variance of $T_{2}$

The estimator

T_{2}

is a weighted sum of the sample values:

T_{2} = \frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}

To find the expected value and variance of

T_{2}

$E [T_{2}] = E [\frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}] = \frac{1}{15} (E [X_{1}] + 2 E [X_{2}] + 3 E [X_{3}] + 4 E [X_{4}] + 5 E [X_{5}]) = \frac{1}{15} (1 + 2 + 3 + 4 + 5) \cdot \frac{1}{λ} = \frac{15}{15} \cdot \frac{1}{λ} = \frac{1}{λ}$
$Var (T_{2}) = Var (\frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}) = \frac{1}{15^{2}} (Var (X_{1}) + 4 Var (X_{2}) + 9 Var (X_{3}) + 16 Var (X_{4}) + 25 Var (X_{5})) = \frac{1}{225} (1 + 4 + 9 + 16 + 25) \cdot \frac{1}{λ^{2}} = \frac{55}{225} \cdot \frac{1}{λ^{2}} = \frac{11}{45 λ^{2}}$

Comparing Variances

Now, we compare the variances of

T_{1}

and

T_{2}

$Var (T_{1}) = \frac{1}{5 λ^{2}}$
$Var (T_{2}) = \frac{11}{45 λ^{2}}$

To determine which estimator is more efficient, we compare

\frac{1}{5}

and

\frac{11}{45}

\frac{1}{5} = \frac{9}{45}

Since

\frac{9}{45} < \frac{11}{45}

, we have:

Var (T_{1}) < Var (T_{2})

Therefore, the variance of

T_{1}

is smaller than that of

T_{2}

, indicating that

T_{1}

is more efficient than

T_{2}

Conclusion

The statement "If

X_{1}, X_{2}, X_{3}, X_{4}

, and

X_{5}

is a random sample of size 5 taken from an Exponential distribution, then estimator

T_{1}

is more efficient than

T_{2}

" is true. This conclusion is based on the comparison of their variances, where

T_{1}

has a lower variance than

T_{2}

(ii) If

T_{1}

and

T_{2}

are two estimators of the parameter

θ

such that

Var (T_{1}) = 1 / n

and

Var (T_{2}) = n

then

T_{1}

is more efficient than

T_{2}

Answer:

The efficiency of an estimator is inversely related to its variance. An estimator with a smaller variance is considered more efficient because it has less variability and, therefore, tends to be closer to the true parameter value.

Given two estimators

T_{1}

and

T_{2}

of the parameter

θ

with the following variances:

Var (T_{1}) = \frac{1}{n}

Var (T_{2}) = n

Comparison of Variances

To determine which estimator is more efficient, we compare their variances directly.

Variance of $T_{1}$ :

$Var (T_{1}) = \frac{1}{n}$
Variance of $T_{2}$ :

$Var (T_{2}) = n$

Since efficiency is related to having a smaller variance, we compare

\frac{1}{n}

and

n

Analysis:

$\frac{1}{n}$ is typically much smaller than $n$ when $n > 1$ .
For $n \geq 1$ , $\frac{1}{n} \leq 1$ and $n \geq 1$ , hence $\frac{1}{n} < n$ .

Conclusion:

Since

\frac{1}{n}

is significantly smaller than

n

T_{1}

has a much smaller variance than

T_{2}

. Therefore,

T_{1}

is more efficient than

T_{2}

because it provides estimates with less variability around the parameter

θ

Justification:

For $n > 1$ : $\frac{1}{n} < n$ , thus $Var (T_{1}) < Var (T_{2})$ .
For $n = 1$ : $Var (T_{1}) = 1$ and $Var (T_{2}) = 1$ , thus they have equal variance.
For $n < 1$ : This scenario typically does not apply as $n$ usually represents the sample size, which is a positive integer.

Therefore, given that the usual context implies

n > 1

n = 1

in practical situations,

T_{1}

is more efficient than

T_{2}

for any reasonable sample size

n

Hence, the statement "If

T_{1}

and

T_{2}

are two estimators of the parameter

θ

such that

Var (T_{1}) = 1 / n

and

Var (T_{2}) = n

then

T_{1}

is more efficient than

T_{2}

" is true.

(iii) A

95 %

confidence interval is smaller than

99 %

confidence interval.

Answer:

To determine the truth of the statement "A

95 %

confidence interval is smaller than a

99 %

confidence interval," we need to understand how confidence intervals are constructed and how the confidence level affects their width.

Confidence Intervals

A confidence interval for a parameter is an interval estimate that is likely to contain the parameter with a certain level of confidence. For a given confidence level

(1 - α)

, the confidence interval is typically given by:

Estimate \pm z_{α / 2} \times Standard Error

where

z_{α / 2}

is the critical value from the standard normal distribution corresponding to the desired confidence level, and the standard error is a measure of the variability of the estimate.

Comparison of 95% and 99% Confidence Intervals

Critical Values:
- For a $95 %$ confidence interval, the critical value $z_{α / 2}$ corresponds to $α = 0.05$ . This gives $z_{0.025} \approx 1.96$ .
- For a $99 %$ confidence interval, the critical value $z_{α / 2}$ corresponds to $α = 0.01$ . This gives $z_{0.005} \approx 2.576$ .
Interval Width:
- The width of a confidence interval is determined by the product of the critical value and the standard error.
- For a $95 %$ confidence interval: $Width = 2 \times 1.96 \times Standard Error$ .
- For a $99 %$ confidence interval: $Width = 2 \times 2.576 \times Standard Error$ .

Since

2.576 > 1.96

, the multiplier for the standard error in the

99 %

confidence interval is larger than that for the

95 %

confidence interval.

Conclusion

The

99 %

confidence interval has a larger critical value, resulting in a wider interval compared to the

95 %

confidence interval, assuming the same data and variability. Therefore, the statement "A

95 %

confidence interval is smaller than a

99 %

confidence interval" is true.

Proof

The width of a confidence interval is proportional to the critical value $z_{α / 2}$ .
For $95 %$ confidence level, the critical value is approximately 1.96.
For $99 %$ confidence level, the critical value is approximately 2.576.
Since $2.576 > 1.96$ , the width of the $99 %$ confidence interval will be greater than the width of the $95 %$ confidence interval.

Hence, a

95 %

confidence interval is indeed smaller than a

99 %

confidence interval, making the statement true.

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IGNOU MST-016 Solved Assignment | June 2024 to June 2025 | Statistical Inference | MSCAST

Sample Solution

Mean and Variance of T 1 T 1 T_(1)T_1T1

Mean and Variance of T 2 T 2 T_(2)T_2T2

Comparing Variances

Conclusion

Comparison of Variances

Analysis:

Conclusion:

Justification:

Confidence Intervals

Comparison of 95% and 99% Confidence Intervals

Conclusion

Proof

Mean and Variance of $T_{1}$

Mean and Variance of $T_{2}$