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A random sample of size $16$ from a normal population gives the sample mean $10$ and sample standard deviation $1.6$. A $95\%$ confidence interval for the population mean is closest to:

  1. $10\pm 1.6\cdot{2.131}$
  2. $10\pm 0.4\cdot{1.96}$
  3. $10\pm 0.4\cdot{1.753}$
  4. $10\pm 1.6\cdot{1.753}$
  5. $10\pm 0.4\cdot{2.131}$

I believe the answer is $(2)$, but would appreciate it if someone could check this for me. Thank you.

Galc127
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John Rosebey
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    Checked. Yes, you are right. $z_{0.975}=1.96$ and $\frac{\sigma}{\sqrt{n}}=\frac{1.6}{4}=0.4$ – callculus42 May 17 '16 at 07:13
  • Plugging $\bar{X}=10,\sigma=1.6,\alpha=0.05,n=16$ to $\left(\bar{X}-z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}},\bar{X}+z_{1-\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\right)$ we get answer $(2)$. – Galc127 May 17 '16 at 07:15

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