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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.

Answer:

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.

You ask whether there is a simpler way to do it. I will interpret "do it" as "solve the problem." The answer, sort of, is yes. The author has tried simultaneously to solve the hypothesis testing problem and indicate the reasoning behind the calculation. Once this sort of calculation becomes routine, it can be done semi-mechanically, without explaining the details of the probability calculation.

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Thank you Andre!!! –  Q.matin Oct 30 '12 at 4:44
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