# Confidence Level with error

We measure four times in a row mercury in the same food sample. Every time there is a small measurement error the four measurements give:

$X_1 = 13, X_2 = 7, X_3 = 10, X_4=10$

Test on a 95%-significance level that there would be more than 14 units of Mercury in the sample. Assuming the standard deviation of the measurement error being given and equal to 2.

Let $\mu$ designate the true level of Mercury. The assumption is that $X_i = \mu +\alpha_i$, where $\alpha_i$ is the $i^{th}$ measurement error. We assume it to be normal and have 0 expectation, which means $E[\alpha_i] = 0$ which implies

$$E[X_i] = E[\mu + \alpha_i] = \mu + E[\alpha_i] = \mu$$

The sample mean is our estimate for μ. So:

$$\hat \mu = \frac{X_1 + ... + X_4}{4} = 10$$ This is less than the hypothesis would be saying, so let us see the probability of this when have the hypothesis

$$P_{\mu = 14}(\frac{X_1 + ... + X_4}{4} \le 10) = P(\frac{X_1 + ... + X_4}{4}-14 \le -4)$$ Now let $Z = (\frac{X_1 + ... + X_4}{4})$, then $E[Z] = E[X_i] = \mu$. Also, $\frac{\sigma x_i}{2}=1$. Thus, $P(N(0,1) \le -4)$ which is approxiametly close to zero, which means we can reject the hypothesis.

My question is how were they able to find $\frac{\sigma x_i}{2}=1$? Also, is there a simpler way to do this?

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The variance of $$\frac{1}{n}(X_1+X_2+\cdots +X_n),\tag{1}$$ if the $X_i$ are independent identically distributed, each with variance $\sigma^2$, is equal to $$\frac{\sigma^2}{n}.\tag{2}$$ In our case, we have $n=4$. And since the standard deviation of each $X_i$ is $2$, we have that $\sigma^2=(2)^2=4$. It follows from $(2)$ that the variance of $\dfrac{1}{4}(X_1+X_2+X_3+X_4)$ is $\dfrac{4}{4}=1$. Taking the square root of $1$ does nothing, so the standard deviation of $\dfrac{1}{4}(X_1+X_2+X_3+X_4)$ is $1$.
Formally, we are testing the null hypothesis $H_0$ that the mean is $14$, versus the alternate hypothesis $H_1$ that the mean is less than $14$. And at our chosen significance level, we reject the null hypothesis.