Inference for means: 8 mixed t procedure problems

(a) One-sample $t$ interval for $\mu$. There is one group, no claimed value to test against, and the word 'estimate' with a confidence level. The interval is $t$ rather than $z$ because $\sigma$ is unknown and $s$ stands in for it, with $df = 39$.

(b) Paired $t$ procedure on the differences. Each volunteer produces two ratings, so volunteer 12's two scores belong together. Collapse to 25 differences and run a one-sample procedure with $df = 24$. Whether it is an interval or a test depends on what is asked next, but the pairing is fixed by the design.

(c) Two-sample $t$ test for $\mu_1 - \mu_2$. Two separate groups of students with no link between any senior and any first-year, and the question asks whether means differ, which is a test rather than an interval. By hand, $df = \min(30, 30) - 1 = 29$. Equal group sizes here are a coincidence, not pairing.

(d) One-sample $t$ test for $\mu$. One group measured against a fixed claimed value, $\mu_0 = 500$ hours, with $df = 17$. The 500 is a number, not a second sample, so no two-sample procedure is involved.

Notice the pattern. Nothing in the arithmetic decided any of these. The number of groups, the presence or absence of a link between the columns, and the verb in the question did all the work.

Choose the procedure. One group of 20 prints compared against a fixed advertised value, and $\sigma$ is unknown, so this is a one-sample $t$ test for a mean. There is no second sample to compare with.

State. Let $\mu$ be the true mean print time, in minutes, for this benchmark model on this machine. $H_0: \mu = 45$ versus $H_a: \mu \ne 45$, with $\alpha = 0.05$. The alternative is two-sided because 'differs' names no direction.

Plan. Random: the 20 prints are a random sample. 10%: 20 is less than 10% of all prints this machine could produce. Normal/large sample: $n = 20 < 30$, so lean on the dotplot, which is roughly symmetric with no outliers.

Compute the standard error and the test statistic. $SE = \frac{s}{\sqrt{n}} = \frac{5.4}{\sqrt{20}} = \frac{5.4}{4.4721} = 1.2075$ minutes, so $t = \frac{47.6 - 45}{1.2075} = \frac{2.6}{1.2075} = 2.15$ with $df = 20 - 1 = 19$.

Find the p-value. Two-sided, so use $|t| = 2.15$. In the $df = 19$ row it falls between $2.093$ (tail $0.025$) and $2.539$ (tail $0.01$), so one tail holds between $0.01$ and $0.025$ and the two-sided p-value is between $0.02$ and $0.05$. Technology gives $0.0444$.

Conclude. The p-value $0.0444$ is below $0.05$, so reject $H_0$. There is convincing evidence that this machine's true mean print time differs from the advertised 45 minutes, running longer in this sample. The margin is thin: at $\alpha = 0.01$ the same data would not have rejected, so the evidence is real but not overwhelming.

(a) Why pairing fails. Pairing requires a link between one specific observation in each group, created by measuring the same individual twice or by matching individuals deliberately. Here the two samples are different butterflies caught at different sites, chosen independently, so butterfly 7 at Meadow has no partner at Woodland. Equal sample sizes are a necessary consequence of pairing, never a cause of it. The correct procedure is a two-sample $t$ test for a difference of means.

(b) State. Let $\mu_1$ be the true mean forewing length at Meadow Site and $\mu_2$ the true mean at Woodland Site, in millimeters. $H_0: \mu_1 - \mu_2 = 0$ versus $H_a: \mu_1 - \mu_2 \ne 0$, with $\alpha = 0.05$.

(b) Plan. Two independent random samples, one from each site. 10%: 15 is less than 10% of the butterflies at each site. Normal/large sample: both samples are under 30, so assume forewing lengths at each site are roughly normal with no strong skew or outliers. Independence: the samples were drawn independently of each other.

(b) Standard error. $\frac{s_1^2}{n_1} = \frac{37.21}{15} = 2.4807$ and $\frac{s_2^2}{n_2} = \frac{53.29}{15} = 3.5527$, so $SE = \sqrt{2.4807 + 3.5527} = \sqrt{6.0334} = 2.4563$ mm. Note that variances are added, never standard deviations.

(b) Test statistic and p-value. $t = \frac{62.4 - 68.1}{2.4563} = \frac{-5.7}{2.4563} = -2.32$, with conservative $df = \min(15, 15) - 1 = 14$. Two-sided, so use $|t| = 2.32$, which falls between $2.145$ (tail $0.025$) and $2.624$ (tail $0.01$), putting the two-sided p-value between $0.02$ and $0.05$. Technology gives $0.0359$.

(b) Conclude. The p-value is below $0.05$, so reject $H_0$. There is convincing evidence that true mean forewing length differs between the two sites, with Woodland longer in these samples.

(c) Scope. Both samples were randomly selected, so the conclusion generalizes to the butterflies at each of these two sites. Site was not assigned by the biologist, so this is observational: the difference cannot be attributed to the habitat itself, since anything that differs between the sites, food supply, predation, or which subspecies happens to live there, could produce the same gap.

(a) Identify the design. Each water sample is measured by both kits, so the two readings on sample 5 belong together and their difference is meaningful. That link makes this a matched-pairs design, and the correct procedure is a paired $t$ interval, which is a one-sample $t$ interval built on the 12 differences with $df = 11$. There are 24 readings but only 12 pairs, and $n = 12$.

(b) Plan. Random: the 12 water samples are a random sample. 10%: 12 pairs is under 10% of the water samples the lab could test. Normal/large sample: $n = 12 < 30$, so assume the differences are roughly normal with no strong skew or outliers. Conditions are checked on the differences, not on the two columns of readings.

(b) Standard error. $SE = \frac{s_d}{\sqrt{n}} = \frac{0.58}{\sqrt{12}} = \frac{0.58}{3.4641} = 0.16743$ ppm.

(b) Critical value and margin of error. With $df = 11$, the 95% column gives $t^* = 2.201$, so $ME = 2.201 \times 0.16743 = 0.3685$ ppm.

(b) Interval. $0.42 \pm 0.3685 = (0.0515,\ 0.7885)$ ppm.

(c) Interpret. We are 95% confident that the interval from $0.0515$ to $0.7885$ ppm captures the true mean difference in chlorine reading (Kit 1 minus Kit 2) on water samples like these.

(c) Answer the agreement question. The interval lies entirely above 0, so a true mean difference of zero is not plausible: the kits do not agree on average, and Kit 1 reads higher. A matching two-sided test confirms it, with $t = \frac{0.42}{0.16743} = 2.51$ at $df = 11$ giving a p-value of $0.0291$, below $0.05$. Note the practical caveat: the lower bound is only $0.05$ ppm, so the bias might be too small to matter for the lab's purposes even though it is statistically detectable.

Choose the procedure. Two independent random samples of different hens, and the question asks for an interval at a stated confidence level, so this is a two-sample $t$ interval for $\mu_1 - \mu_2$. The unequal sizes 14 and 16 rule out any paired procedure on sight.

State. Let $\mu_1$ and $\mu_2$ be the true mean annual egg production for hens on Feed A and Feed B. Estimate $\mu_1 - \mu_2$ with 90% confidence.

Plan. Random: two independent random samples, one per feed. 10%: 14 and 16 are each less than 10% of the hens that could be raised on that feed. Normal/large sample: both samples are under 30, so assume egg counts under each feed are roughly normal with no strong skew or outliers.

Point estimate and standard error. $\bar{x}_1 - \bar{x}_2 = 232 - 221 = 11$ eggs. Then $\frac{s_1^2}{n_1} = \frac{324}{14} = 23.1429$ and $\frac{s_2^2}{n_2} = \frac{484}{16} = 30.25$, so $SE = \sqrt{23.1429 + 30.25} = \sqrt{53.3929} = 7.307$ eggs.

Critical value and margin of error. Conservative $df = \min(14, 16) - 1 = 13$. A 90% interval leaves $0.05$ in each tail, so the $df = 13$ row gives $t^* = 1.771$, and $ME = 1.771 \times 7.307 = 12.94$ eggs.

Build the interval. $11 \pm 12.94 = (-1.94,\ 23.94)$ eggs per year.

Interpret. We are 90% confident that the interval from $-1.94$ to $23.94$ eggs per year captures the true difference in mean annual egg production between hens on Feed A and hens on Feed B.

Read the zero. The interval contains 0, so a true difference of zero is a plausible value and this study gives no convincing evidence that the two feeds differ in true mean production. The interval is nearly 26 eggs wide, so 'no convincing evidence' here means the study is too small to settle the question, not that the feeds are known to perform identically.

(a) Why the two-sample idea fails. A two-sample procedure needs two sets of data, each with its own $n$, mean, and standard deviation, so that the sampling variability of both can enter the standard error. The 600-calorie guideline is a fixed published number with no sample behind it, so it contributes no variability. The correct procedure is a one-sample $t$ test for a mean, with $\mu_0 = 600$.

(b) State. Let $\mu$ be the true mean calorie content of this manufacturer's frozen dinners. Because the guideline is an upper limit and the concern is exceeding it, $H_0: \mu = 600$ versus $H_a: \mu > 600$, with $\alpha = 0.05$.

(b) Plan. Random: the 22 dinners are a random sample. 10%: 22 is less than 10% of all dinners this manufacturer produces. Normal/large sample: $n = 22 < 30$, so lean on the boxplot, which shows no outliers and little skew.

(b) Compute. $SE = \frac{s}{\sqrt{n}} = \frac{82}{\sqrt{22}} = \frac{82}{4.6904} = 17.4825$ calories, so $t = \frac{638 - 600}{17.4825} = \frac{38}{17.4825} = 2.17$ with $df = 22 - 1 = 21$.

(b) p-value and decision. Right-tailed. In the $df = 21$ row, $2.17$ falls between $2.080$ (tail $0.025$) and $2.518$ (tail $0.01$), so the p-value is between $0.01$ and $0.025$. Technology gives $0.0207$. That is below $0.05$, so reject $H_0$. There is convincing evidence that this manufacturer's dinners average more than 600 calories.

(c) Interpret the p-value. If the true mean calorie content really were exactly 600, only about 2.07% of random samples of 22 dinners would produce a sample mean of 638 calories or higher by chance alone. A result that unlikely under the guideline is what makes the evidence against it convincing. Because $H_0$ was rejected, a wrong decision here would be a Type I error: reporting that the dinners exceed the guideline when in truth they average 600.

(a) Procedure and df. Two independent random samples of different weeks at different stations, with an interval requested, so this is a two-sample $t$ interval for a difference of means. Conservative $df = \min(12, 15) - 1 = 11$, and the 95% column of the $t$-table at $df = 11$ gives $t^* = 2.201$.

(b) Center and margin. The observed difference sits at the midpoint: $\bar{x}_A - \bar{x}_B = \frac{-1.2 + 4.8}{2} = \frac{3.6}{2} = 1.8$ mm. The margin of error is half the width: $ME = \frac{4.8 - (-1.2)}{2} = \frac{6.0}{2} = 3.0$ mm. So the interval is $1.8 \pm 3.0$.

(b) Standard error. Since $ME = t^* \times SE$, invert it: $SE = \frac{ME}{t^*} = \frac{3.0}{2.201} = 1.363$ mm.

(c) The test decision. The interval contains 0, so a true difference of zero is a plausible value at 95% confidence. A two-sided test at $\alpha = 0.05$ and a 95% interval always agree, so the test fails to reject $H_0$: there is not convincing evidence that the two stations differ in true mean weekly rainfall. As a cross-check, $t = \frac{1.8}{1.363} = 1.32$ at $df = 11$ is nowhere near the $2.201$ needed to reject.

(d) Correct the colleague. Failing to reject a null hypothesis is not proof that it is true. The interval runs from $-1.2$ to $4.8$ mm, so a 4 mm advantage for Station A is still a plausible value for the true difference, and so is exactly zero. An interval marks the values the data cannot rule out, not values the data support equally: the estimate is $1.8$ mm, which sits $\frac{1.8}{1.363} = 1.32$ standard errors from zero and $\frac{2.2}{1.363} = 1.61$ standard errors from 4 mm, and neither distance reaches the $2.201$ that would push a value outside the interval. The honest statement is that this study did not detect a difference, and with 12 and 15 weeks of data it lacked the precision to detect a modest one. A wider net of weeks would narrow the interval and settle the question.

(a) Match design to procedure. Design A splits 20 different students into two groups with no link between any pair of them, so it calls for a two-sample $t$ test with conservative $df = \min(10, 10) - 1 = 9$. Design B has each student read in both media, so the two scores for student 6 belong together: paired $t$ test on 10 differences, also with $df = 9$. Same $df$ here by coincidence, but very different procedures.

(b) State. Let $\mu_d$ be the true mean difference in comprehension score (paper minus screen) for students like these. $H_0: \mu_d = 0$ versus $H_a: \mu_d \ne 0$, with $\alpha = 0.05$. Two-sided, because 'affects' names no direction.

(b) Plan. Paired $t$ test on the 10 differences. Random: medium order and passage were randomly assigned within each student, which is the form the random condition takes in this experiment. The 10% condition does not apply, since the 10 students are not a sample drawn from a population of pairs. Normal/large sample: $n = 10 < 30$, so check the differences: sorted they run $-1, 0, 1, 1, 2, 2, 3, 3, 4, 5$, which is single-peaked with no strong skew, and with $Q_1 = 1$ and $Q_3 = 3$ the $1.5 \times IQR$ fences at $-2$ and $6$ flag no outliers.

(b) Compute. $\bar{d} = \frac{2 + (-1) + 3 + 4 + 0 + 2 + 5 + 1 + 3 + 1}{10} = \frac{20}{10} = 2.0$ points. The squared deviations from 2 are $0, 9, 1, 4, 4, 0, 9, 1, 1, 1$, summing to $30$, so $s_d = \sqrt{\frac{30}{9}} = \sqrt{3.3333} = 1.8257$. Then $SE = \frac{1.8257}{\sqrt{10}} = \frac{1.8257}{3.1623} = 0.5774$ and $t = \frac{2.0}{0.5774} = 3.46$ with $df = 9$.

(b) p-value and conclusion. Two-sided with $|t| = 3.46$. In the $df = 9$ row that clears $3.250$, the critical value for a tail of $0.005$, so one tail is below $0.005$ and the two-sided p-value is below $0.01$. Technology gives $0.0071$. Reject $H_0$: there is convincing evidence that reading medium affects mean comprehension, with paper scoring higher here. The matching 95% interval, $2.0 \pm 2.262 \times 0.5774 = (0.694,\ 3.306)$ points, excludes 0 and agrees.

(c) Why Design B is more sensitive. Students differ enormously from one another in reading ability, and Design A's standard error is built from all of that between-student variation. Design B measures each student against themselves, so ability cancels out of every difference and the standard error is built only from how a single student's score changes with the medium. Here $s_d = 1.83$ points, far smaller than the spread of comprehension scores across students would be, which is what makes a 2-point effect detectable with only 10 students.

(d) Scope. Medium was randomly assigned within each student, so a cause-and-effect conclusion about reading medium is justified for these students. But the 10 students were not randomly selected from any larger population, so the finding does not generalize beyond them. Random assignment buys causation; random selection buys generalization, and only the first is present.