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Let $n$ samples of normal distribution with an unknown expected value and variance of $10$. The sampled mean is $3.2$. Build confidence interval for the expected value, where:

  1. $95\%$ confidence, $n=20$.
  2. $99\%$ confidence, $n=20$.
  3. $95\%$ confidence, $n=40$.
  4. $99\%$ confidence, $n=40$.

Here's the numbers (I think I got it right):

$$\left[3.2-z_{0.975}\frac{\sqrt{10}}{\sqrt{20}},3.2+z_{0.975}\frac{\sqrt{10}}{\sqrt{20}}\right]=\left[3.2-1.96\cdot\frac{\sqrt{10}}{\sqrt{20}},3.2+1.96\cdot\frac{\sqrt{10}}{\sqrt{20}}\right]\\=\left[1.814,4.585\right]$$

$$\left[3.2-z_{0.995}\frac{\sqrt{10}}{\sqrt{20}},3.2+z_{0.995}\frac{\sqrt{10}}{\sqrt{20}}\right]=\left[3.2-3.28\cdot\frac{\sqrt{10}}{\sqrt{20}},3.2+3.28\cdot\frac{\sqrt{10}}{\sqrt{20}}\right]\\=\left[0.88,5.512\right]$$

$$\left[3.2-z_{0.975}\frac{\sqrt{10}}{\sqrt{40}},3.2+z_{0.975}\frac{\sqrt{10}}{\sqrt{40}}\right]=\left[3.2-1.96\cdot\frac{\sqrt{10}}{\sqrt{40}},3.2+1.96\cdot\frac{\sqrt{10}}{\sqrt{40}}\right]\\=\left[2.22,4.18\right]$$

$$\left[3.2-z_{0.995}\frac{\sqrt{10}}{\sqrt{40}},3.2+z_{0.995}\frac{\sqrt{10}}{\sqrt{40}}\right]=\left[3.2-3.28\cdot\frac{\sqrt{10}}{\sqrt{40}},3.2+3.28\cdot\frac{\sqrt{10}}{\sqrt{40}}\right]\\=\left[1.56,4.84\right]$$

Now, I'm asked (for every one of the four options) if it is reasonable to infer that the expected value is $0$ - As far as I can tell, the answer is no for all of the options since $\mu=0$ is outside of all the intervals.

Am I missing something?

EDIT:
I fixed the numbers as suggested, but the question (any my answer) remains the same.

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    You don't seem to be missing anything to me. Maybe they meant standard deviation 10 rather than variance 10? That might make the problem more interesting. – 2017-01-19
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    I checked again - it's the variance. So I really don't get the intention of this question... – 2017-01-19
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    Actually, looking again and eyeballing your numbers, you seem to be using incorrect values for $z$'s. For instance, $z_{.975}=1.96,$ not $.835.$ – 2017-01-19
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    I would still expect that getting a sample mean that's a full standard deviation from the true on a sample size of 20/40 is pretty unlikely though. – 2017-01-19
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    @spaceisdarkgreen, that's weird. This table (for example) says $.835$ for $z_{.975}$: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf – 2017-01-19
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    Maybe I'm reading it wrong.. – 2017-01-19
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    Yes, you are. You can get it from your table by hunting for .975 in the bulk.. it's towards the middle of the 2nd page. You'll see it corresponds to $z = 1.96.$ Or you can use an inverse cdf table, easier to read for these purposes https://faculty.biu.ac.il/~shnaidh/zooloo/library/normal.3.pdf – 2017-01-19
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    Oh I see, thanks! – 2017-01-19
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    @blueplusgreen Note that $z_{.995}=2.575$. In the table you have the values $z_{.9949}=2.57$ and $z_{.9950}=2.58$. I you make a **linear interpolation** you get $\frac{.9949+0.9950}{2}=0.995$ and $\frac{2.57+2.58}{2}=2.575$ – 2017-01-19

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