Question:
Let \(X_1, X_2, \ldots, X_{10}\) be a random sample from the \(N(3,4)\) distribution, and let \(Y_1, Y_2, \ldots, Y_{15}\) be a random sample from the \(N(-3,6)\) distribution. Assume that the two samples are drawn independently. Define:
\[ \bar{X} = \frac{1}{10} \sum_{i=1}^{10} X_i \]
\[ \bar{Y} = \frac{1}{15} \sum_{j=1}^{15} Y_j \]
\[ S = \sqrt{\frac{1}{9} \sum_{i=1}^{10} (X_i - \bar{X})^2} \]
Then the distribution of \( U = \frac{\sqrt{5}(\bar{X} + \bar{Y})}{S} \) is:
Let \(X_1, X_2, \ldots, X_{10}\) be a random sample from the \(N(3,4)\) distribution, and let \(Y_1, Y_2, \ldots, Y_{15}\) be a random sample from the \(N(-3,6)\) distribution. Assume that the two samples are drawn independently. Define:
\[ \bar{X} = \frac{1}{10} \sum_{i=1}^{10} X_i \]
\[ \bar{Y} = \frac{1}{15} \sum_{j=1}^{15} Y_j \]
\[ S = \sqrt{\frac{1}{9} \sum_{i=1}^{10} (X_i - \bar{X})^2} \]
Then the distribution of \( U = \frac{\sqrt{5}(\bar{X} + \bar{Y})}{S} \) is:
\[ \bar{X} = \frac{1}{10} \sum_{i=1}^{10} X_i \]
\[ \bar{Y} = \frac{1}{15} \sum_{j=1}^{15} Y_j \]
\[ S = \sqrt{\frac{1}{9} \sum_{i=1}^{10} (X_i - \bar{X})^2} \]
Then the distribution of \( U = \frac{\sqrt{5}(\bar{X} + \bar{Y})}{S} \) is:
Updated On: Oct 1, 2024
- \(N\left(0, \frac{4}{5}\right)\)
- \(\chi^2_9\)
- \(t_9\)
- \(t_{23}\)
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The Correct Option is C
Solution and Explanation
The correct option is (C): \(t_9\)
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