2
$\begingroup$

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:

enter image description here

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.

$\endgroup$
2
  • $\begingroup$ No one? Would really appreciate some feedback. $\endgroup$
    – Katja
    Oct 22 '20 at 20:09
  • $\begingroup$ Thanks for showing what you have done with this (+1), Seems you are on the right track, but not exactly giving the same information that one sees in computer output for such a test. Many details in my Answer; some useful, I hope. $\endgroup$
    – BruceET
    Oct 24 '20 at 1:41
1
$\begingroup$

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 $T.$

enter image description here

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")

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
$\endgroup$
4
  • 2
    $\begingroup$ The "acceptance region" for the sample mean $\bar x$ for the hypothesis $$H_0 : \mu_1 = \mu-2 \quad \text{vs.} \quad H_1 : \mu_1 \ne \mu_2$$ is, I presume, the set of values for $\bar x$ under the assumption of a fixed $s = 6.82$ for which $H_0$ is not rejected. This is in contrast to the critical value which is the $97.5$ percentile of the student $t$ distribution with $\nu = 9$ degrees of freedom. I agree that it is unusual to express a rejection or "acceptance" region of a test in terms of the sample mean, especially when the population SD is not known. $\endgroup$
    – heropup
    Oct 24 '20 at 1:50
  • 2
    $\begingroup$ @heropup. Author of OP's text may think there is some initial pedagogical advantage in talking about an acceptance region in terms of $\bar X$ (perhaps better for z-tests in the 1980s), but my goal is to shift the discussion to be compatible with computer output: first to acceptance region in terms of test statistic and second to P-values. // OP asked for confirmation his task was done correctly and I gave it, but I felt an obligation to move the focus into the 21st century. // You know this stuff well, if you have an alternative helpful narrative please post an answer along those lines. $\endgroup$
    – BruceET
    Oct 24 '20 at 5:39
  • 2
    $\begingroup$ Indeed. I think pretty much every student of statistics these days needs to be able to connect the theory and the mechanics of inference with current software, since it is not always straightforward to interpret the output different programs produce. $\endgroup$
    – heropup
    Oct 24 '20 at 5:50
  • 1
    $\begingroup$ Wow this helps a lot! Thank you! $\endgroup$
    – Katja
    Oct 24 '20 at 11:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.