At the end of the semester, two tutors Albert and Ben are correcting an exam with $10$ tasks. They share the $100$ written exams and measure the time needed to correct a task in minutes. The difference $x_i$ of the correction times (Ben's time $-$ Albert's time) for task $i$ is given in the following table:
The sample mean $\bar{x} = 4.4$ and the sample standard deviation $\bar{\sigma} = 6.82$. We assume that the values $x_1, x_2, ..., x_{10}$ are realizations of $10$ independent and identically normally distributed random variables.
For the significance level $\alpha = 0.05$, find a confidence interval for the difference $x_i$ and determine the acceptance region for $\bar{x}.$
Since the population standard deviation $\sigma$ is not given, we will use the $t-$distribution (or Student-$t$-distribution) to find the confidence interval for the population mean $\mu$.
First we calculate our acceptance thresholds $t_c$ and $-t_c$:
Since we know that $\alpha = 0.05$, the area of the region right to $t_c$ $= 0.025 = $ the area left to $-t_c$.
We also know that we have $n-1 = 10-1 = 9$ degrees of freedom.
Using the $t-$distribution values table, we find $t_c = 2.26$ and $-t_c = -2.26.$
Now we find our test statistic $T_s$:
$T_s = \dfrac{\bar{x} - \mu}{\dfrac{\bar{\sigma}}{\sqrt{n}}}$ $= \dfrac{4.4 - \mu}{\dfrac{6.82}{\sqrt{10}}}$.
We know that $P(-t_c \leq T_s \leq t_c) = 1- \alpha = 0.95.$ Substituting then gives us:
$$\bar{x} - t_c \cdot \dfrac{\bar{\sigma}}{\sqrt{n}} \leq \mu \leq \bar{x} + t_c \cdot \dfrac{\bar{\sigma}}{\sqrt{n}}$$
$$4.4 -2.26 \cdot \dfrac{6.82}{\sqrt{10}} \leq \mu \leq 4.4 +2.26 \cdot \dfrac{6.82}{\sqrt{10}}$$
$$-0.474 \leq \mu \leq 9.274$$
So we know that $-0.474 \leq \mu \leq 9.274$ with $95\%$ confidence.
The acceptance region for $\bar{x}$ would be $[-t_c \cdot \dfrac{\bar{\sigma}}{\sqrt{n}}, t_c \cdot \dfrac{\bar{\sigma}}{\sqrt{n}}] = [-4.874, 4.874].$
Did I do this correctly? I'm very unsure about my work and don't know how to interpret the negative values in the confidence interval.