Tutors correcting tests - Confidence intervals

Question

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 $[-t_c \cdot \dfrac{\bar{\sigma}}{\sqrt{n}}, t_c \cdot \dfrac{\bar{\sigma}}{\sqrt{n}}] = [-4.874, 4.874].$

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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.

Answer 1

I put your data into R, with the following results, which you can compare with your work.

d = c(6, 8, -7, 4, 15, 4, 7, -2, 12, -3)
summary(d);  sd(d)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  -7.00   -0.50    5.00    4.40    7.75   15.00 
[1] 6.818276  # sample SD

t.test(d)

    One Sample t-test

data:  d
t = 2.0407, df = 9, p-value = 0.07168
alternative hypothesis: 
   true mean is not equal to 0
95 percent confidence interval:
 -0.4775009  9.2775009
sample estimates:
mean of x 
      4.4 

Because the P-value $0.07168 > 0.05 = 5\%,$ you cannot reject $H_0$ (no difference) at the 5% level.

Your 95% CI is in substantial agreement with the CI from R (maybe you could have carried an extra decimal place throughout your computations).

You never show your $T$-statistic explicitly. Usually The rejection region of a two-sided test is given in terms of critical values from the t distribution. By that method you would reject at the 5% level, if $|T| \ge 2.262.$ That is, the critical values are $\pm 2.262.$

qt(.975, 9)
[1] 2.262157

Can you find 2.262 on line DF - 9 of a printed table of Student's t distributions?

It may be useful to express acceptance and rejection regions in terms of $\bar X$ (somehow considering $S = 6.818$ fixed), but that is not the usual practice. [See @heropup's Comment below.] Maybe that's why you haven't gotten a response before now.

The P-value is the probability beyond $\pm T$ in both tails of the relevant t distribution. Typically, you can't find exact P-values in printed tables. P-values are, however, widely used in computer printouts. The P-value can be found in R, where 'pt` is the CDF of a t-distribution.

2 * pt(-2.0407, 9)
[1] 0.07168392

In the figure below, the density function of $\mathsf{T}(df=9)$ is shown (black curve) along with the critical values (vertical dotted red lines), the observed value of $T$ (heavy vertical line). Critical values cut probability $0.025 = 2.5\%$ (total 5%) from each tail of this t distribution.

The P-value is the sum of the areas in both tails outside the vertical black lines); here, it is defined as the probability under $H_0$ of seeing a t-statistic as far or farther from $0$ (in either direction) than the observed $T.$

R code to make figure:

curve(dt(x, 9), -4, 4, ylab="PDF", xlab="t", 
      main="Density of T(df=9)")
 abline(v = c(-2.262, 2.262), col="red", lty="dotted")
 abline(v = 2.0407, lw=2)
 abline(v = -2.0407, lw=2, lty="dashed")

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In case it is of any use to you, I am also showing output for this t test from a recent release of Minitab. Notice that is shows sample, mean and SD, $T$-statistic, DF, a 95% CI for $\mu,$ and P-value. (Minitab is well-known for its concise output.)

One-Sample T 

Test of μ = 0 vs ≠ 0

 N  Mean  StDev  SE Mean      95% CI        T      P
10  4.40   6.82     2.16  (-0.48, 9.28)  2.04  0.072