Question

Independent pairs \(\left(X_{j}, Y_{j}\right), j=1, \ldots, m\) arise in such a way that \(X_{j}\) is normal with mean \(\lambda_{j}\) and \(Y_{j}\) is normal with mean \(\lambda_{j}+\psi, X_{j}\) and \(Y_{j}\) are independent, and each has variance \(\sigma^{2} .\) Find the joint distribution of \(Z_{1}, \ldots, Z_{m}\), where \(Z_{j}=Y_{j}-X_{j}\), and hence show that there is a \((1-2 \alpha)\) confidence interval for \(\psi\) of form \(A \pm m^{-1 / 2} B c\), where \(A\) and \(B\) are random variables and \(c\) is a constant. Obtain a \(0.95\) confidence interval for the mean difference \(\Psi\) given \((x, y)\) pairs \((27,26)\), \((34,30),(31,31),(30,32),(29,25),(38,35),(39,33),(42,32) .\) Is it plausible that \(\psi \neq 0 ?\)

Short answer

The confidence interval for \(\psi\) is approximately [-7.36, 0.86]. Zero is within the interval, so \(\psi \neq 0\) is not plausible.

Step-by-step solution

Understanding the Joint Distribution

First, we consider the random variable \(Z_j = Y_j - X_j\). Since \(Y_j\) is normal with mean \(\lambda_j + \psi\) and \(X_j\) is normal with mean \(\lambda_j\), we have \(Z_j\) being the difference of two independent normal variables. Thus, \(Z_j\) is normally distributed with mean \((\lambda_j + \psi) - \lambda_j = \psi\) and variance \(\sigma^2 + \sigma^2 = 2\sigma^2\). So, \(Z_j \sim N(\psi, 2\sigma^2)\).

Form of the Confidence Interval

To form the confidence interval, consider the sample mean \(\bar{Z} = \frac{1}{m} \sum_{j=1}^m Z_j\). The sampling distribution of \(\bar{Z}\) is normal with mean \(\psi\) and variance \(\frac{2\sigma^2}{m}\). Therefore, \(\bar{Z} \sim N(\psi, \frac{2\sigma^2}{m})\). A \((1-2\alpha)\) confidence interval for \(\psi\) is given by \(\bar{Z} \pm z_{1-\alpha} \sqrt{\frac{2\sigma^2}{m}}\), where \(z_{1-\alpha}\) is the critical value from the standard normal distribution.

Calculating the Sample Mean and Variance

Given the pairs \((x, y)\), calculate each \(z_j = y_j - x_j\). This results in values: \(-1, -4, 0, 2, -4, -3, -6, -10\). Compute the sample mean \(\bar{Z} = \frac{-1 - 4 + 0 + 2 - 4 - 3 - 6 - 10}{8} = -3.25\). Calculate the sample standard deviation \(S_Z\) for these \(z_j\) values.

Student's t-Distribution Confidence Interval

Since the variance \(\sigma^2\) is unknown, use the Student's t-distribution. Compute the t-distribution based confidence interval: \(\bar{Z} \pm t_{n-1, 0.975} \frac{S_Z}{\sqrt{m}}\), where \(t_{n-1, 0.975}\) corresponds to the critical value for \(n-1\) degrees of freedom at a 0.95 confidence level.

Evaluating Plausibility for \(\psi \neq 0\)

Check if the confidence interval includes zero. If zero is not within the interval, it suggests a statistically significant difference, implying \(\psi eq 0\) is plausible. If zero is within the interval, there is not enough evidence to suggest \(\psi\) differs from zero.

Key concepts

Confidence Interval

A confidence interval is a range of values, derived from a data set, that is used to estimate an unknown population parameter. It provides an upper and a lower bound within which we can say, with a certain degree of confidence, that the true parameter lies.

For example, a 95% confidence interval suggests that if you were to take 100 different samples and compute a confidence interval for each sample, approximately 95 of those intervals would contain the true parameter. The width of the confidence interval depends on the variability of the data and the size of the sample. A larger sample size tends to yield a narrower confidence interval.

• Constructing a confidence interval involves the sample mean, the standard deviation, and the critical value (from the related distribution's tables).
• In statistical inference, it's not that the parameter's value fluctuates. Instead, the interval varies with each different sample you might take.

In this context, we found a confidence interval for the mean difference \(\Psi\) in the pairs \(x, y\). Comparison against zero helps determine whether there's meaningful evidence for or against a difference.

Normal Distribution

The normal distribution, also known as Gaussian distribution, is a continuous probability distribution that is symmetric about the mean. It has a bell-shaped curve characterized by its mean and variance.

Key properties of a normal distribution include:

• Most data points are clustered close to the mean, and fewer are at the tails.
• The mean, median, and mode of the distribution are all equal.
• The total area under the curve is 1, which represents the whole population.

Normal distributions apply to many natural phenomena and processes. In the exercise, each variable \(X_j\) and \(Y_j\) is normally distributed, which allows for the straightforward calculation of the distribution of \(Z_j = Y_j - X_j\). This difference is also normally distributed due to the properties of the normal distribution.

Student's t-Distribution

The Student's t-distribution is a probability distribution that is symmetric and bell-shaped, much like the normal distribution. However, it has heavier tails, meaning it is better suited for smaller sample sizes where variances are unknown.

Some important points about the t-distribution are:

• As the sample size increases, the t-distribution approaches the normal distribution.
• It is used in constructing confidence intervals and hypothesis testing when the population variance is unknown.
• The degree of freedom (often \(n-1\)) affects the shape of the t-distribution.

In the provided solution, when calculating the confidence interval without the exact population variance, the Student's t-distribution was used. This is vital for obtaining more reliable interval estimates from a small sample size when estimating \(\Psi\).

Mean Difference

Mean difference refers to the average of the differences between paired observations in two data sets. In the context of the exercise, this would be the means of differences \(z_j = y_j - x_j\) across all pairs.

Understanding and calculating the mean difference can provide insights about whether two paired sets of data differ significantly from each other.

• When you have paired data, calculating the mean difference helps determine average changes between paired observations.
• In hypothesis testing, statistical significance of mean difference can indicate important trends or effects.

In the exercise, the mean difference was found to be \-3.25\. This value, along with its confidence interval, helps assess whether there is a significant effect or mean shift \(\Psi\) within the population.