Question: If the standard deviation () associated with the pdf that produced the following sample is 3.6, would it be correct to claim that is a

If the standard deviation (σ) associated with the pdf that produced the following sample is 3.6, would it be correct to claim that

3.6 20 2.611.96- 2.61 + 1.96. 3.6 20 = (1.03, 4.19)

is a 95% confidence interval for μ? Explain.

3.6 20 2.611.96- 2.61 + 1.96. 3.6 20 = (1.03, 4.19)

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