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:
- $95\%$ confidence, $n=20$.
- $99\%$ confidence, $n=20$.
- $95\%$ confidence, $n=40$.
- $99\%$ 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.