Sample Solution

MST-016 Solved Assignment 2024 Sample Solution

Expert Answer

(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.

(iv) If the probability density function of a random variable

X

follows

F

-distribution is

f (x) = \frac{1}{(1 + x)^{2}}, x \geq 0

then degrees of freedom of the distribution will be

(2, 2)

Answer:

The given probability density function (pdf) is:

f (x) = \frac{1}{(1 + x)^{2}}, x \geq 0

We need to determine whether this pdf corresponds to an

F

-distribution with degrees of freedom

(2, 2)

Form of the $F$ -Distribution

The pdf of an

F

-distribution with degrees of freedom

d_{1}

and

d_{2}

is given by:

f (x) = \frac{Γ (\frac{d_{1} + d_{2}}{2})}{Γ (\frac{d_{1}}{2}) Γ (\frac{d_{2}}{2})} {(\frac{d_{1}}{d_{2}})}^{d_{1} / 2} \frac{x^{d_{1} / 2 - 1}}{{(1 + \frac{d_{1}}{d_{2}} x)}^{(d_{1} + d_{2}) / 2}}, x \geq 0

To match the given pdf

f (x) = \frac{1}{(1 + x)^{2}}

with the form of the

F

-distribution, let’s compare the forms:

The given pdf does not have a term involving $x^{d_{1} / 2 - 1}$ .
The given pdf has the form $\frac{1}{(1 + x)^{2}}$ , which suggests that the exponent in the denominator should match $(d_{1} + d_{2}) / 2 = 2$ .

Determining $d_{1}$ and $d_{2}$

To determine the degrees of freedom, let’s compare the denominator term:

{(1 + \frac{d_{1}}{d_{2}} x)}^{(d_{1} + d_{2}) / 2} = (1 + x)^{2}

By comparing the exponents:

\frac{d_{1} + d_{2}}{2} = 2 ⟹ d_{1} + d_{2} = 4

Next, let’s consider the term involving

x

. For an

F

-distribution:

f (x) = \frac{1}{(1 + x)^{2}}

We do not have an

x^{d_{1} / 2 - 1}

term, which means

d_{1} / 2 - 1 = 0 ⟹ d_{1} / 2 = 1 ⟹ d_{1} = 2

Using

d_{1} = 2

d_{1} + d_{2} = 4

2 + d_{2} = 4 ⟹ d_{2} = 2

Thus, the degrees of freedom are

d_{1} = 2

and

d_{2} = 2

Conclusion

The given pdf

f (x) = \frac{1}{(1 + x)^{2}}

matches the form of an

F

-distribution with degrees of freedom

(2, 2)

Therefore, the statement "If the probability density function of a random variable

X

follows

F

-distribution is

f (x) = \frac{1}{(1 + x)^{2}}, x \geq 0

, then degrees of freedom of the distribution will be

(2, 2)

" is true.

(v) A patient suffering from fever reaches to a doctor and suppose the doctor formulate the hypotheses as

H_{0}

: The patient is a chikunguniya patient

H_{1}

: The patient is not a chikunguniya patient
If the doctor rejects

H_{0}

when the patient is actually a chikunguniya patient, then the doctor commits type II error.

Answer:

Let’s review the hypothesis testing framework first:

$H_{0}$ : The patient is a chikungunya patient.
$H_{1}$ : The patient is not a chikungunya patient.

In hypothesis testing, we have two types of errors:

Type I error (False Positive): Rejecting $H_{0}$ when $H_{0}$ is true. In this context, it means the doctor concludes that the patient does not have chikungunya when the patient actually has chikungunya.
Type II error (False Negative): Failing to reject $H_{0}$ when $H_{0}$ is false. In this context, it means the doctor concludes that the patient has chikungunya when the patient actually does not have chikungunya.

Now, let’s analyze the given statement:

"If the doctor rejects

H_{0}

when the patient is actually a chikungunya patient, then the doctor commits a type II error."

This statement is incorrect because:

Rejecting $H_{0}$ when $H_{0}$ is actually true is committing a Type I error, not a Type II error.

Thus, if the doctor rejects

H_{0}

when the patient is actually a chikungunya patient, the doctor is committing a Type I error.

(b) Describe the various forms of the sampling distribution of ratio of two sample variances.

Answer:

Sampling Distribution of the Ratio of Two Sample Variances

When dealing with the ratio of two sample variances, we’re typically interested in understanding the behavior of this ratio under repeated sampling from normally distributed populations. This leads us to the concept of the F-distribution. Here’s a detailed explanation of the various forms and properties of the sampling distribution of the ratio of two sample variances.

1. Basic Concept

Suppose we have two independent random samples from two normally distributed populations. Let:

$S_{1}^{2}$ be the variance of the first sample of size $n_{1}$ .
$S_{2}^{2}$ be the variance of the second sample of size $n_{2}$ .

The ratio of the two sample variances

F

is given by:

F = \frac{S_{1}^{2} / σ_{1}^{2}}{S_{2}^{2} / σ_{2}^{2}}

where

σ_{1}^{2}

and

σ_{2}^{2}

are the population variances of the two samples.

2. F-Distribution

Under the null hypothesis that the population variances are equal (i.e.,

σ_{1}^{2} = σ_{2}^{2}

), the ratio

F

follows an F-distribution with

(n_{1} - 1)

and

(n_{2} - 1)

degrees of freedom. The F-distribution is defined as:

F \sim F_{(n_{1} - 1), (n_{2} - 1)}

3. Properties of the F-Distribution

Non-Negative: Since variances are always non-negative, the ratio $F$ is always non-negative.
Skewness: The F-distribution is positively skewed, especially for smaller sample sizes.
Mean: For an F-distribution with $d_{1}$ and $d_{2}$ degrees of freedom, the mean is given by $\frac{d_{2}}{d_{2} - 2}$ for $d_{2} > 2$ .
Mode: The mode of the F-distribution is given by $\frac{d_{1} - 2}{d_{1} (d_{2} + 2)}$ for $d_{1} > 2$ .
Variance: The variance of the F-distribution is given by: $Var (F) = \frac{2 d_{2}^{2} (d_{1} + d_{2} - 2)}{d_{1} (d_{2} - 2)^{2} (d_{2} - 4)}$ for $d_{2} > 4$ .

4. Applications

The F-distribution is commonly used in:

Analysis of Variance (ANOVA): To compare the variances across multiple groups.
Regression Analysis: To test the overall significance of a regression model.
Hypothesis Testing: For testing the equality of two variances.

5. Example

Suppose we have two independent samples with the following properties:

Sample 1: Size $n_{1} = 10$ , variance $S_{1}^{2}$ .
Sample 2: Size $n_{2} = 15$ , variance $S_{2}^{2}$ .

To test whether the population variances are equal, we calculate the F-statistic:

F = \frac{S_{1}^{2}}{S_{2}^{2}}

We then compare this calculated F-statistic to the critical value from the F-distribution table with

(n_{1} - 1)

and

(n_{2} - 1)

degrees of freedom. If the calculated F-statistic is greater than the critical value, we reject the null hypothesis that the population variances are equal.

6. Assumptions

For the F-distribution to be a valid approximation of the distribution of the ratio of two sample variances, the following assumptions must hold:

The samples are independent.
Each sample is drawn from a normally distributed population.

7. Limitations

Normality Assumption: The F-distribution relies on the assumption that the populations from which samples are drawn are normally distributed. If this assumption is violated, the distribution of the ratio may not follow an F-distribution.
Sensitivity to Outliers: Sample variances are sensitive to outliers. Thus, the F-statistic can be heavily influenced by extreme values.

In conclusion, the ratio of two sample variances follows an F-distribution under the assumption of normality and independence of samples. The F-distribution is characterized by its positive skewness and is widely used in various statistical tests, including ANOVA and hypothesis testing for equality of variances. Understanding the properties and limitations of this distribution is crucial for correctly interpreting the results of such tests.

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MST-016 Solved Assignment 2024

Sample Solution

Expert Answer

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

Form of the F F FFF-Distribution

Determining d 1 d 1 d_(1)d_1d1 and d 2 d 2 d_(2)d_2d2

Conclusion

Sampling Distribution of the Ratio of Two Sample Variances

1. Basic Concept

2. F-Distribution

3. Properties of the F-Distribution

4. Applications

5. Example

6. Assumptions

7. Limitations

Mean and Variance of $T_{1}$

Mean and Variance of $T_{2}$

Form of the $F$ -Distribution

Determining $d_{1}$ and $d_{2}$