(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)X_1, X_2, X_3, X_4 and X_(5)X_5 is a random sample of size 5 taken from an Exponential distribution, then estimator T_(1)T_1 is more efficient than T_(2)T_2.
To determine whether T_(1)T_1 is more efficient than T_(2)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∼”Exponential”(lambda)X \sim \text{Exponential}(\lambda). The mean of this distribution is E[X]=(1)/(lambda)E[X] = \frac{1}{\lambda}, and the variance is “Var”(X)=(1)/(lambda^(2))\text{Var}(X) = \frac{1}{\lambda^2}.
Therefore, the variance of T_(1)T_1 is smaller than that of T_(2)T_2, indicating that T_(1)T_1 is more efficient than T_(2)T_2.
Conclusion
The statement "If X_(1),X_(2),X_(3),X_(4)X_1, X_2, X_3, X_4, and X_(5)X_5 is a random sample of size 5 taken from an Exponential distribution, then estimator T_(1)T_1 is more efficient than T_(2)T_2" is true. This conclusion is based on the comparison of their variances, where T_(1)T_1 has a lower variance than T_(2)T_2.
(ii) If T_(1)T_1 and T_(2)T_2 are two estimators of the parameter theta\theta such that Var(T_(1))=1//n\operatorname{Var}\left(T_1\right)=1 / n and Var(T_(2))=n\operatorname{Var}\left(T_2\right)=n then T_(1)T_1 is more efficient than T_(2)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)T_1 and T_(2)T_2 of the parameter theta\theta with the following variances:
Since efficiency is related to having a smaller variance, we compare (1)/(n)\frac{1}{n} and nn.
Analysis:
(1)/(n)\frac{1}{n} is typically much smaller than nn when n > 1n > 1.
For n >= 1n \geq 1, (1)/(n) <= 1\frac{1}{n} \leq 1 and n >= 1n \geq 1, hence (1)/(n) < n\frac{1}{n} < n.
Conclusion:
Since (1)/(n)\frac{1}{n} is significantly smaller than nn, T_(1)T_1 has a much smaller variance than T_(2)T_2. Therefore, T_(1)T_1 is more efficient than T_(2)T_2 because it provides estimates with less variability around the parameter theta\theta.
Justification:
For n > 1n > 1: (1)/(n) < n\frac{1}{n} < n, thus Var(T_(1)) < Var(T_(2))\operatorname{Var}(T_1) < \operatorname{Var}(T_2).
For n=1n = 1: Var(T_(1))=1\operatorname{Var}(T_1) = 1 and Var(T_(2))=1\operatorname{Var}(T_2) = 1, thus they have equal variance.
For n < 1n < 1: This scenario typically does not apply as nn usually represents the sample size, which is a positive integer.
Therefore, given that the usual context implies n > 1n > 1 or n=1n = 1 in practical situations, T_(1)T_1 is more efficient than T_(2)T_2 for any reasonable sample size nn.
Hence, the statement "If T_(1)T_1 and T_(2)T_2 are two estimators of the parameter theta\theta such that Var(T_(1))=1//n\operatorname{Var}\left(T_1\right)=1 / n and Var(T_(2))=n\operatorname{Var}\left(T_2\right)=n then T_(1)T_1 is more efficient than T_(2)T_2" is true.
(iii) A 95%95 \% confidence interval is smaller than 99%99 \% confidence interval.
Answer:
To determine the truth of the statement "A 95%95\% confidence interval is smaller than a 99%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-alpha)(1 – \alpha), the confidence interval is typically given by:
where z_(alpha//2)z_{\alpha/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%95\% confidence interval, the critical value z_(alpha//2)z_{\alpha/2} corresponds to alpha=0.05\alpha = 0.05. This gives z_(0.025)~~1.96z_{0.025} \approx 1.96.
For a 99%99\% confidence interval, the critical value z_(alpha//2)z_{\alpha/2} corresponds to alpha=0.01\alpha = 0.01. This gives z_(0.005)~~2.576z_{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%95\% confidence interval: “Width”=2xx1.96 xx”Standard Error”\text{Width} = 2 \times 1.96 \times \text{Standard Error}.
For a 99%99\% confidence interval: “Width”=2xx2.576 xx”Standard Error”\text{Width} = 2 \times 2.576 \times \text{Standard Error}.
Since 2.576 > 1.962.576 > 1.96, the multiplier for the standard error in the 99%99\% confidence interval is larger than that for the 95%95\% confidence interval.
Conclusion
The 99%99\% confidence interval has a larger critical value, resulting in a wider interval compared to the 95%95\% confidence interval, assuming the same data and variability. Therefore, the statement "A 95%95\% confidence interval is smaller than a 99%99\% confidence interval" is true.
Proof
The width of a confidence interval is proportional to the critical value z_(alpha//2)z_{\alpha/2}.
For 95%95\% confidence level, the critical value is approximately 1.96.
For 99%99\% confidence level, the critical value is approximately 2.576.
Since 2.576 > 1.962.576 > 1.96, the width of the 99%99\% confidence interval will be greater than the width of the 95%95\% confidence interval.
Hence, a 95%95\% confidence interval is indeed smaller than a 99%99\% confidence interval, making the statement true.
(iv) If the probability density function of a random variable X\mathrm{X} follows F\mathrm{F}-distribution is
f(x)=(1)/((1+x)^(2)),x >= 0f(x)=\frac{1}{(1+x)^2}, x \geq 0
then degrees of freedom of the distribution will be (2,2)(2,2).
Answer:
The given probability density function (pdf) is:
f(x)=(1)/((1+x)^(2)),quad x >= 0f(x) = \frac{1}{(1+x)^2}, \quad x \geq 0
We need to determine whether this pdf corresponds to an FF-distribution with degrees of freedom (2,2)(2,2).
Form of the FF-Distribution
The pdf of an FF-distribution with degrees of freedom d_(1)d_1 and d_(2)d_2 is given by:
To match the given pdf f(x)=(1)/((1+x)^(2))f(x) = \frac{1}{(1+x)^2} with the form of the FF-distribution, let’s compare the forms:
The given pdf does not have a term involving x^(d_(1)//2-1)x^{d_1/2 – 1}.
The given pdf has the form (1)/((1+x)^(2))\frac{1}{(1+x)^2}, which suggests that the exponent in the denominator should match (d_(1)+d_(2))//2=2(d_1 + d_2)/2 = 2.
Determining d_(1)d_1 and d_(2)d_2
To determine the degrees of freedom, let’s compare the denominator term:
Next, let’s consider the term involving xx. For an FF-distribution:
f(x)=(1)/((1+x)^(2))f(x) = \frac{1}{(1+x)^2}
We do not have an x^(d_(1)//2-1)x^{d_1/2 – 1} term, which means d_(1)//2-1=0Longrightarrowd_(1)//2=1Longrightarrowd_(1)=2d_1/2 – 1 = 0 \implies d_1/2 = 1 \implies d_1 = 2.
Using d_(1)=2d_1 = 2 in d_(1)+d_(2)=4d_1 + d_2 = 4:
Thus, the degrees of freedom are d_(1)=2d_1 = 2 and d_(2)=2d_2 = 2.
Conclusion
The given pdf f(x)=(1)/((1+x)^(2))f(x) = \frac{1}{(1+x)^2} matches the form of an FF-distribution with degrees of freedom (2,2)(2,2).
Therefore, the statement "If the probability density function of a random variable X\mathrm{X} follows F\mathrm{F}-distribution is f(x)=(1)/((1+x)^(2)),x >= 0f(x)=\frac{1}{(1+x)^2}, x \geq 0, then degrees of freedom of the distribution will be (2,2)(2,2)" is true.
(v) A patient suffering from fever reaches to a doctor and suppose the doctor formulate the hypotheses as H_(0)\mathrm{H}_0 : The patient is a chikunguniya patient H_(1)\mathrm{H}_1 : The patient is not a chikunguniya patient
If the doctor rejects H_(0)\mathrm{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)\mathrm{H}_0: The patient is a chikungunya patient.
H_(1)\mathrm{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)\mathrm{H}_0 when H_(0)\mathrm{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)\mathrm{H}_0 when H_(0)\mathrm{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)\mathrm{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)\mathrm{H}_0 when H_(0)\mathrm{H}_0 is actually true is committing a Type I error, not a Type II error.
Thus, if the doctor rejects H_(0)\mathrm{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)S_1^2 be the variance of the first sample of size n_(1)n_1.
S_(2)^(2)S_2^2 be the variance of the second sample of size n_(2)n_2.
The ratio of the two sample variances FF is given by:
where sigma_(1)^(2)\sigma_1^2 and sigma_(2)^(2)\sigma_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., sigma_(1)^(2)=sigma_(2)^(2)\sigma_1^2 = \sigma_2^2), the ratio FF follows an F-distribution with (n_(1)-1)(n_1 – 1) and (n_(2)-1)(n_2 – 1) degrees of freedom. The F-distribution is defined as:
Non-Negative: Since variances are always non-negative, the ratio FF is always non-negative.
Skewness: The F-distribution is positively skewed, especially for smaller sample sizes.
Mean: For an F-distribution with d_(1)d_1 and d_(2)d_2 degrees of freedom, the mean is given by (d_(2))/(d_(2)-2)\frac{d_2}{d_2 – 2} for d_(2) > 2d_2 > 2.
Mode: The mode of the F-distribution is given by (d_(1)-2)/(d_(1)(d_(2)+2))\frac{d_1 – 2}{d_1 (d_2 + 2)} for d_(1) > 2d_1 > 2.
Variance: The variance of the F-distribution is given by:“Var”(F)=(2d_(2)^(2)(d_(1)+d_(2)-2))/(d_(1)(d_(2)-2)^(2)(d_(2)-4))\text{Var}(F) = \frac{2 d_2^2 (d_1 + d_2 – 2)}{d_1 (d_2 – 2)^2 (d_2 – 4)}for d_(2) > 4d_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:
To test whether the population variances are equal, we calculate the F-statistic:
F=(S_(1)^(2))/(S_(2)^(2))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)(n_1 – 1) and (n_(2)-1)(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.