9.4 Matched or Paired Samples

Step 1 — Recognize the design. The same nine subjects are measured twice, so these are matched pairs, not independent samples.

Step 2 — Compute the differences as "after \(-\) before," so that a lower cholesterol level after the diet shows up as a negative number:

$$ \{-10,\ -3,\ -16,\ 11,\ 1,\ -15,\ -27,\ -17,\ -21\} $$

Seven of the nine dropped, one was unchanged in direction by only 1 point, and one subject (D) rose by 11.

Step 3 — Summarize. \(n = 9\), \(\overline{x}_{d} = -10.78\), and \(s_{d} = 11.86\).

Step 4 — Write the hypotheses. "Lower after the diet" means the differences should be negative:

$$ H_{0}: \mu_{d} \geq 0 \qquad H_{a}: \mu_{d} < 0 $$

This is a left-tailed test.

Step 5 — Standardize. The distribution is \(t_{8}\), since \(df = 9 - 1 = 8\):

$$ t = \frac{-10.78}{\left(\frac{11.86}{\sqrt{9}}\right)} = -2.73 $$

Step 6 — Decide. The left-tail p-value is 0.0130. Since \(0.0130 < 0.05\), reject \(H_{0}\).

Answer: At the 5% level of significance, there is sufficient evidence to conclude that the subjects' cholesterol levels are lower on average after the diet. Note that this would not reject at the 1% level — and that one subject going the wrong way by 11 points is exactly the kind of thing a nine-person study cannot absorb without cost.

Step 1 — Form the pairs. Corresponding "before" and "after" values form matched pairs. Calculate "after \(-\) before."

After Data	Before Data	Difference
6.8	6.6	0.2
2.4	6.5	−4.1
7.4	9.0	−1.6
8.5	10.3	−1.8
8.1	11.3	−3.2
6.1	8.1	−2.0
3.4	6.3	−2.9
2.0	11.6	−9.6

Step 2 — The data for the test are the differences: \(\{0.2, -4.1, -1.6, -1.8, -3.2, -2.0, -2.9, -9.6\}\). The sample mean and sample standard deviation of the differences are \(\overline{x}_{d} = -3.13\) and \(s_{d} = 2.91\). Verify these values for yourself — it is the one piece of arithmetic in this section that is genuinely worth doing by hand once.

Step 3 — Name the random variable. \(\overline{X}_{d}\), the mean difference of the sensory measurements. Let \(\mu_{d}\) be the population mean for the differences.

Step 4 — Write the hypotheses.

$$ H_{0}: \mu_{d} \geq 0 \qquad H_{a}: \mu_{d} < 0 $$

The null hypothesis is zero or positive, meaning there is the same or more pain felt after hypnotism — the subject shows no improvement. The alternative is negative, meaning there is less pain felt after hypnotism. Because a lower score means less pain, the score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.

Step 5 — Choose the distribution. A Student's t with \(df = n - 1 = 8 - 1 = 7\). Use \(t_{7}\). Notice that the test is for a single population mean.

Step 6 — Compute. The test statistic is \(t = -3.036\) and the p-value is 0.0095.

Step 7 — Compare and decide. \(\alpha = 0.05\) and the p-value is 0.0095, so \(\alpha >\) p-value. Reject \(H_{0}\). This means that \(\mu_{d} < 0\) and there is improvement.

Answer: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements are, on average, lower after hypnotism. Hypnotism appears to be effective in reducing pain.

Step 1 — Fix the direction. "Higher after the class" is the claim, so compute "after \(-\) before" and look for positive differences:

$$ \{80,\ 200,\ 280,\ -50\} $$

Step 2 — Summarize. \(n = 4\), \(\overline{x}_{d} = 127.5\), and \(s_{d} = 144.08\).

Step 3 — Write the hypotheses.

$$ H_{0}: \mu_{d} \leq 0 \qquad H_{a}: \mu_{d} > 0 $$

This is a right-tailed test, with distribution \(t_{3}\).

Step 4 — Standardize.

$$ t = \frac{127.5}{\left(\frac{144.08}{\sqrt{4}}\right)} = 1.77 $$

Step 5 — Decide. The right-tail p-value is 0.0874. Since \(0.0874 > 0.05\), do not reject \(H_{0}\).

Answer: At the 5% level of significance, there is not sufficient evidence to conclude that SAT scores are higher on average after the prep class. Three of the four students improved, two of them substantially — and the test still fails to reject. With only four pairs, \(df = 3\), and the standard deviation of 144 is inflated by the one student who dropped 50 points. Four subjects is simply not enough to separate a real effect from this much noise.

Step 1 — Record the differences. Subtract the amount lifted prior to the class from the amount lifted after completing it, so an increase is positive: \(\{90, 11, -8, -8\}\). Assume the differences have a normal distribution.

Step 2 — Summarize. \(\overline{x}_{d} = 21.3\) and \(s_{d} = 46.7\).

Step 3 — Identify the test. Using the difference data, this becomes a test of a single mean.

Step 4 — Name the random variable. \(\overline{X}_{d}\), the mean difference in the maximum lift per player. The distribution for the hypothesis test is \(t_{3}\).

Step 5 — Write the hypotheses.

$$ H_{0}: \mu_{d} \leq 0 \qquad H_{a}: \mu_{d} > 0 $$

Step 6 — Compute. The p-value is 0.2150.

Step 7 — Decide. If the level of significance is 5%, the decision is not to reject the null hypothesis, because \(\alpha <\) p-value.

Answer: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.

Step 1 — Compute the differences as "dominant \(-\) off-hand": \(\{15, 2, 37, 29, 26\}\). Every player threw farther with the dominant hand.

Step 2 — Summarize. \(n = 5\), \(\overline{x}_{d} = 21.8\), and \(s_{d} = 13.59\).

Step 3 — Write the hypotheses. The players believe the distances are the same, and "is the difference significant" gives no direction, so this is two-tailed:

$$ H_{0}: \mu_{d} = 0 \qquad H_{a}: \mu_{d} \neq 0 $$

Step 4 — Standardize. The distribution is \(t_{4}\):

$$ t = \frac{21.8}{\left(\frac{13.59}{\sqrt{5}}\right)} = 3.59 $$

Step 5 — Decide. The two-tailed p-value is 0.0230. Since \(0.0230 < 0.05\), reject \(H_{0}\).

Answer: At the 5% level of significance, there is sufficient evidence to conclude that there is a difference in the distances the players throw with their dominant and off-hands. The players' belief is not supported.

Why five pairs was enough here and four was not in Try It Now 9.4.2. Nothing about the sample size changed much — what changed is consistency. All five differences here are positive, so \(s_{d} = 13.6\) against a mean of 21.8. In Try It Now 9.4.2 one of four differences went the wrong way, and \(s_{d} = 144\) against a mean of 127.5. A paired test rewards agreement across pairs far more than it rewards raw sample size.

Step 1 — Record the differences. Subtract the distance with the weaker hand from the distance with the dominant hand: \(\{2, 12, 7, -1, 2, 0, 4\}\). The differences have a normal distribution.

Step 2 — Summarize. \(\overline{x}_{d} = 3.71\) and \(s_{d} = 4.5\).

Step 3 — Name the random variable. \(\overline{X}_{d}\), the mean difference in the distances between the hands. The distribution for the hypothesis test is \(t_{6}\).

Step 4 — Write the hypotheses. The students thought the distances were equal, and no direction is claimed, so the test is two-tailed:

$$ H_{0}: \mu_{d} = 0 \qquad H_{a}: \mu_{d} \neq 0 $$

Step 5 — Compute. The test statistic is \(t = 2.18\) and the p-value is 0.0716.

Step 6 — Decide. Assume \(\alpha = 0.05\). Since \(\alpha <\) p-value, do not reject \(H_{0}\).

Answer: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children's weaker and dominant hands in pushing the shot-put.

A near miss worth noticing. The p-value of 0.0716 sits just above 0.05, and five of the seven students did push farther with the dominant hand. Compare this with Try It Now 9.4.3, where the same two-tailed question rejected on only five pairs — the difference is that student 4 here went backwards and student 6 was exactly even, and two non-confirming pairs out of seven are enough to keep the p-value above the line. Do not read "do not reject" as "the hands are the same." It means seven children could not settle it.