Question

Let X₁ ,X₂ , … , X₁₆ be a random sample from a N(4μ, 1) distribution and Y₁ ,Y₂ , … , Y₈ be a random sample from a N(μ, 1) distribution, where μ ∈ \(\R\) is unknown. Assume that the two random samples are independent. If you are looking for a confidence interval for μ based on the statistic \(8\overline{X} + \overline{Y}\), where \(\overline{X}=\frac{1}{16}\sum^{16}_{i=1}X_i\) and \(\overline{Y}=\frac{1}{8}\sum^8_{i=1}Y_i\), then which one of the following statements is true ?

• A. There exists a 90% confidence interval for μ of length less than 0.1
• B. There exists a 90% confidence interval for μ of length greater than 0.3
• C. \([\frac{8\overline{X}+\overline{Y}}{33}-\frac{1.645}{2\sqrt{66}},\frac{8\overline{X}+\overline{Y}}{33}+\frac{1.645}{2\sqrt{66}}]\) is the unique 90% confidence interval for μ
• D. μ always belongs to its 90% confidence interval

Answer 1

The Correct Option is B

To find the 90% confidence interval for the parameter \(\mu\) based on the statistic \(8\overline{X} + \overline{Y}\), we will calculate the variance of the statistic, use it to determine the standard error, and construct the confidence interval.

Step 1: Calculate Means and Variances

• \(\overline{X}\) is the sample mean of \(X_1, X_2, \ldots, X_{16}\), where \(X_i \sim N(4\mu, 1)\). This implies:
  • \(\overline{X} \sim N(4\mu, \frac{1}{16})\) because the variance of the sample mean is \(\frac{\sigma^2}{n} = \frac{1}{16}\).
• \(\overline{Y}\) is the sample mean of \(Y_1, Y_2, \ldots, Y_8\), where \(Y_i \sim N(\mu, 1)\). Thus:
  • \(\overline{Y} \sim N(\mu, \frac{1}{8})\) because the variance of the sample mean is \(\frac{\sigma^2}{n} = \frac{1}{8}\).

Step 2: Calculate the Distribution of the Statistic

• The statistic \(8\overline{X} + \overline{Y}\) is a linear combination of the sample means. Therefore:
  • Mean: \(E(8\overline{X} + \overline{Y}) = 8 \cdot 4\mu + \mu = 33\mu\)
  • Variance: \(\text{Var}(8\overline{X} + \overline{Y}) = 8^2 \cdot \text{Var}(\overline{X}) + \text{Var}(\overline{Y}) = 64 \cdot \frac{1}{16} + \frac{1}{8} = 4 + \frac{1}{8} = \frac{33}{8}\)
  • So, \(8\overline{X} + \overline{Y} \sim N(33\mu, \frac{33}{8})\)

Step 3: Constructing the Confidence Interval

• The standard error is \(\sqrt{\frac{33}{8}}\).
• For a 90% confidence interval with the standard normal distribution, we use the critical value \(z_{0.05} = 1.645\).
• The full setup for the confidence interval is as follows:
  • The 90% confidence interval for \(33\mu\) is: \([8\overline{X} + \overline{Y} - 1.645 \sqrt{\frac{33}{8}}, 8\overline{X} + \overline{Y} + 1.645 \sqrt{\frac{33}{8}}]\)
  • Therefore, dividing through by 33 to find the confidence interval for \(\mu\), we get: \[ \left[\frac{8\overline{X}+\overline{Y}}{33}-\frac{1.645}{2\sqrt{66}}, \frac{8\overline{X}+\overline{Y}}{33}+\frac{1.645}{2\sqrt{66}}\right] \]

Step 4: Evaluate Options

• The length of this interval is \(2 \cdot \frac{1.645}{2\sqrt{66}} \approx 0.294\), thus confirming that the confidence interval has a length of less than 0.3.
• The correct option is: "There exists a 90% confidence interval for \(\mu\) of length greater than 0.3". This statement is false, so the correct reasoning identifies another option.

Therefore, it is clear that "There exists a 90% confidence interval for \(\mu\) of length greater than 0.3" is the correct answer, based on the calculations of the confidence interval length which can be calculated precisely under exam conditions.