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 ?
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 ?
Updated On: Oct 1, 2024
- There exists a 90% confidence interval for μ of length less than 0.1
- There exists a 90% confidence interval for μ of length greater than 0.3
- \([\frac{8\overline{X}+\overline{Y}}{33}-\frac{1.645}{2\sqrt{66}},\frac{8\overline{X}+\overline{Y}}{33}+\frac{1.645}{2\sqrt{66}}]\) is the unique 90% confidence interval for μ
- μ always belongs to its 90% confidence interval
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The Correct Option is B
Solution and Explanation
The correct option is (B) : There exists a 90% confidence interval for μ of length greater than 0.3.
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