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Que. Consider a normal population with unknown mean $\mu$ and variance $\sigma^2=9$. To test $H_{0}:\mu=0 $against $H_{1}:\mu\ne0$, a random sample of size 100 is taken. Based on this sample, the test of the form $|\bar{X}_n|>K$ rejects the null hypothesis at 5% level of significance. Then, which of the following is a possible 95% confidence interval for $\mu$?

  1. (-0.488,0.688)
  2. (-1.96,1.96)
  3. (0.422,1.598)
  4. (0.588,1.96)

Generally whenever I encountered a problem related to finding the confidence limits for parameter (like$\mu$) I used to find the sample mean and then use the formula $\left(\bar{x}-z_{\alpha}\frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+z_{\alpha}\frac{\sigma}{\sqrt{n}}\right)$ where $\bar{x},z_{\alpha}$ are sample mean and standard normal value for given $\alpha$ here 0.05, I also know how to find size and power of the test and may be related to those things as for $|\bar{X}_n|>K$, I know I can't solve these questions with these methods what do I need to know to solve such problems?

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The test rejects $H_0$ $\mu=0$ so $0$ cannot be in the confidence interval.

The range of the confidence interval is $(\bar{x}+z\frac{\sigma}{\sqrt{n}})-(\bar{x}-z\frac{\sigma}{\sqrt{n}})= 2z \frac{\sigma}{\sqrt{n}} $

You have $z=1.96$ and $\sigma=3$ so the range is a function of $n$ only

$R= 2\cdot 1.96 \frac{3}{\sqrt{n}} $

Which of the confidence intervals has a range which is possible for integer values of $n$?

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  • $\begingroup$ Thanks for the help I was wondering what is the significance of $|\bar{X}_n|>K$ in the question. $\endgroup$ – Onix Jan 31 '16 at 19:53

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