Question

Let the independent random variables \(Y\) and \(Z\) be \(N\left(\mu_{1}, 1\right)\) and \(N\left(\mu_{2}, 1\right)\), respectively. Let \(\theta=\mu_{1}-\mu_{2}\). Let us observe independent observations from each distribution, say \(Y_{1}, Y_{2}, \ldots\) and \(Z_{1}, Z_{2}, \ldots\) To test sequentially the hypothesis \(H_{0}: \theta=0\) against \(H_{1}: \theta=\frac{1}{2}\), use the sequence \(X_{i}=Y_{i}-Z_{i}, i=1,2, \ldots\) If \(\alpha_{a}=\beta_{a}=0.05\), show that the test can be based upon \(\bar{X}=\bar{Y}-\bar{Z} .\) Find \(c_{0}(n)\) and \(c_{1}(n)\)

Short answer

The test for the hypothesis regarding the difference in the means can be based on the mean of \(X_i, \overline{X}=\overline{Y}-\overline{Z}.\) The lower and upper boundaries of the acceptance region are \(c_0(n)=-1.96/\sqrt{n}\) and \(c_1(n) = 1.96/\sqrt{n}\) respectively.

Step-by-step solution

Define Hypothesis

The null hypothesis \(H_0\) is \(\theta=0,\) which means that means of \(Y\) and \(Z\) are equal (\(\mu_1=\mu_2\)). The alternative hypothesis \(H_1\) is \(\theta=\frac{1}{2},\) which means that the distribution \(Y\) has a mean that is greater by 1/2 than the mean of \(Z\). Thus, the test is based on the difference between sample means, and represented as a sequence \(X_i=Y_i-Z_i, i=1,2,...\)

Show That Test is Based on Sample Mean Difference

Under the null hypothesis \(H_0,\) we have \(Y_i-Z_i=X_i∼N(\0,\2). For large \(n,\) \(\overline{X}∼N(0,\frac{1}{\sqrt{n}}). So, the z-score is \(\frac{\sqrt{n}*\overline{X}}{\sigma}\) where \(\sigma=1. When the null hypothesis is true, the z-score should follow a standard normal distribution. Therefore, the decision rule is to reject \(H_0\) if the absolute value of the z-score is greater than the critical value \(z_{\0.025}=1.96.\)

Finding Critical Values \(c_0(n)\) and \(c_1(n)\)

Let \(c_0(n)\) and \(c_1(n)\) represent the lower and upper boundaries of the acceptance region respectively. For \(\alpha_a=\beta_a=0.05\), we accept \(H_0\) when the z-score is in between -1.96 and 1.96. Therefore, we have \(c_0(n)=-1.96*\frac{\sigma}{\sqrt{n}}=-1.96*\frac{1}{\sqrt{n}}\) and \(c_1(n)=1.96*\frac{\sigma}{\sqrt{n}}=1.96*\frac{1}{\sqrt{n}}\).

Final Remarks

So, if for any observed \(\overline{X}\), it falls in the range between \(c_0(n)\) and \(c_1(n)\), we fail to reject \(H_0.\) Otherwise, we reject \(H_0\) and accept \(H_1.\)

Key concepts

Normal Distribution

The normal distribution, often referred to as the Gaussian distribution, is a fundamental concept in statistics. It is a continuous probability distribution that is symmetric around its mean, depicting a bell-shaped curve. Most data points cluster around the central peak, and it shows how much variation or "spread" there is from the mean. This feature makes it incredibly useful for statisticians and researchers.

In a normal distribution:

• The mean, median, and mode of the distribution are all equal.
• About 68% of the data falls within one standard deviation (\( +1\sigma \) or \( -1\sigma \)) from the mean.
• Approximately 95% is within two standard deviations (\( +2\sigma \) or \( -2\sigma \)).
• Near to 99.7% exists within three standard deviations (\( +3\sigma \) or \( -3\sigma \)).

Normal distributions are crucial in hypothesis testing, baseline setting, and in making predictions in various fields. In this specific exercise, the distributions of both \(Y\) and \(Z\) are assumed to be normal with a mean of \(\mu_1\) and \(\mu_2\), aiding in determining whether the means of these two variables are statistically different.

Type I and Type II Error

In hypothesis testing, making errors is a part of the game. The two main types of errors you can encounter are Type I and Type II errors. These errors describe the incorrect rejection or acceptance of hypotheses.

A **Type I Error** occurs when the null hypothesis \(H_0\) is true, but mistakenly rejected. This is often known as a "false positive". The probability of committing a Type I error is denoted by \(\alpha\), also called the significance level. In our exercise, \(\alpha_a\) is set at 0.05, indicating a 5% chance of rejecting the null hypothesis when it's actually true.

On the other hand, a **Type II Error** happens when the null hypothesis is false, yet it is erroneously accepted. This is referred to as a "false negative". The probability of this error is represented by \(\beta\). In the problem, \(\beta_a\) is also 0.05, which suggests a 5% chance of failing to reject the null hypothesis when the alternative hypothesis is true.

Understanding these errors helps in designing tests with appropriate risk levels for making incorrect decisions, which is crucial when testing for the mean difference between \(Y\) and \(Z\).

Z-score

The z-score is a statistical measure that helps us understand how far away a given observation is from the mean in terms of standard deviations. It plays a vital role in hypothesis testing, especially with the normal distribution.

To calculate the z-score, you use the following formula:

• \[ z = \frac{X - \mu}{\sigma}\]

Where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. The z-score tells us how unusual or typical the data point is within the context of the distribution.

In the context of the sequential hypothesis testing problem presented here, the z-score helps determine whether the observed difference between means \(\overline{X}\) should lead to rejecting or accepting \(H_0\). If the z-score lies within the critical values (between -1.96 and 1.96), then \(H_0\) is accepted. If it falls outside this range, \(H_0\) is rejected, and \(H_1\) is accepted, signaling a significant difference between the means of \(Y\) and \(Z\).