One-sample t interval practice: 8 problems

State. Let $\mu$ be the true mean height, in centimeters, of all one-year-old red maple seedlings at this nursery. Estimate $\mu$ with a 95% confidence interval.

Plan. One-sample $t$ interval for a mean, because $\sigma$ is unknown and $s$ stands in for it. Random: the 16 seedlings are a random sample. 10%: 16 is less than 10% of all seedlings at the nursery. Normal/large sample: $n = 16 < 30$, and the problem states heights are roughly normal.

Do. Degrees of freedom $df = n - 1 = 16 - 1 = 15$, so the 95% column of the $t$-table gives $t^* = 2.131$.

Do. Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2$ cm. Margin of error: $t^* \cdot SE = 2.131 \times 2 = 4.262$ cm.

Do. Interval: $\bar{x} \pm ME = 72 \pm 4.262 = (67.738,\ 76.262)$ cm.

Conclude. We are 95% confident that the interval from 67.74 cm to 76.26 cm captures the true mean height of all one-year-old red maple seedlings at this nursery.

State. Let $\mu$ be her true mean ride time, in minutes, on this route. Estimate $\mu$ with a 90% confidence interval.

Plan. One-sample $t$ interval for a mean. Random: the 5 weekdays were chosen at random. 10%: 5 rides is less than 10% of all the rides she makes on this route. Normal/large sample: $n = 5 < 30$, and the problem states ride times are roughly normal with no outliers.

Do. Sample mean: $\bar{x} = \frac{30 + 34 + 28 + 32 + 31}{5} = \frac{155}{5} = 31$ minutes.

Do. Sample standard deviation. Deviations from 31 are $-1, 3, -3, 1, 0$, so the squared deviations are $1, 9, 9, 1, 0$ and sum to 20. Dividing by $n - 1$: $s = \sqrt{\frac{20}{5 - 1}} = \sqrt{5} = 2.2361$ minutes.

Do. Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{2.2361}{\sqrt{5}} = \frac{2.2361}{2.2361} = 1.000$ minute. With $df = 5 - 1 = 4$, the 90% column gives $t^* = 2.132$, so $ME = 2.132 \times 1.000 = 2.132$ minutes.

Do. Interval: $31 \pm 2.132 = (28.868,\ 33.132)$ minutes.

Conclude. We are 90% confident that the interval from 28.87 to 33.13 minutes captures her true mean ride time on this route.

(a) The sample mean sits at the center of the interval: $\bar{x} = \frac{612 + 688}{2} = \frac{1300}{2} = 650$ mg.

(a) The margin of error is half the width: $ME = \frac{688 - 612}{2} = \frac{76}{2} = 38$ mg. Equivalently $688 - 650 = 38$ mg, so the interval is $650 \pm 38$.

(b) Interpret the interval. We are 95% confident that the interval from 612 mg to 688 mg captures the true mean sodium content of all cans of this soup brand. The parameter estimated is a mean, and the population is every can of the brand, not the 24 sampled.

(c) Interpret the confidence level. If the lab repeated this process many times, drawing a random sample of 24 cans and building an interval from each sample, about 95% of the intervals produced would capture the true mean sodium content. The 95% describes the long-run success rate of the method, not this one interval.

(d) The blogger is describing individual cans, but the interval estimates a mean. Its width comes from the standard error $\frac{s}{\sqrt{n}}$, which is the sample-to-sample spread of $\bar{x}$ and is smaller than $s$, the spread of individual cans, by a factor of $\sqrt{24}$. An interval covering 95% of individual cans would be far wider.

Conclude. Two statements survive: the one about capturing $\mu$ in (b) and the repeated-sampling one in (c). Statements about where individual cans fall, and statements giving a 95% probability that $\mu$ lies in this computed interval, both misread the procedure.

State. Let $\mu$ be the true mean fill weight, in grams, of all boxes this machine fills. Estimate $\mu$ with a 95% confidence interval.

Plan. One-sample $t$ interval for a mean. Random: the 45 boxes are a random sample. 10%: 45 is less than 10% of all boxes produced. Normal/large sample: $n = 45 \ge 30$, so the large-sample condition is met, and the dotplot shows no strong skew or outliers.

Do. Degrees of freedom $df = 45 - 1 = 44$, which the exam table skips: it prints 40 and then 50. Use the next smaller printed value, $df = 40$, giving $t^* = 2.021$ and a slightly wider, conservative interval.

Do. Standard error: $SE = \frac{s}{\sqrt{n}} = \frac{12}{\sqrt{45}} = \frac{12}{6.7082} = 1.7889$ g. Margin of error: $2.021 \times 1.7889 = 3.6153$ g.

(a) Interval: $496 \pm 3.615 = (492.385,\ 499.615)$ g. Technology using the exact $df = 44$ gives $t^* = 2.015$ and $(492.395,\ 499.605)$, the same interval to the nearest tenth of a gram.

(b) Conclude. Every value in the interval is below 500, so 500 g is not a plausible value for the true mean fill weight. There is convincing evidence that the boxes average less than the labeled 500 g, by somewhere between about 0.4 g and 7.6 g.

State. Let $\mu$ be the true mean run time, in minutes, of all batteries in this production run. Estimate $\mu$ at three confidence levels from the same sample.

Plan. One-sample $t$ interval for a mean. Random: the 25 batteries are a random sample. 10%: 25 is less than 10% of the production run. Normal/large sample: $n = 25 < 30$, and the problem states run times are roughly normal.

Do. The standard error is set by the data, not by the confidence level: $SE = \frac{s}{\sqrt{n}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2$ minutes, with $df = 25 - 1 = 24$ for every level.

Do. At 90%, $t^* = 1.711$, so $ME = 1.711 \times 2 = 3.422$ and the interval is $240 \pm 3.422 = (236.578,\ 243.422)$, a width of 6.844 minutes.

Do. At 95%, $t^* = 2.064$, so $ME = 2.064 \times 2 = 4.128$ and the interval is $240 \pm 4.128 = (235.872,\ 244.128)$, a width of 8.256 minutes.

Do. At 99%, $t^* = 2.797$, so $ME = 2.797 \times 2 = 5.594$ and the interval is $240 \pm 5.594 = (234.406,\ 245.594)$, a width of 11.188 minutes.

Conclude. The center 240 minutes and the standard error 2 minutes never move, because they come from one sample; only $t^*$ changes. Going from 90% to 99% confidence stretches the interval from 6.844 to 11.188 minutes, about 63% wider. Higher confidence that the interval captures the true mean run time is paid for with a less precise estimate of it.

State and plan. Same parameter $\mu$, the true mean run time in minutes for this production run, and the same 95% level; only $n$ changes. Random: a random sample of 100 from the run. 10%: 100 is less than 10% of the run. Normal/large sample: $n = 100 \ge 30$, so the large-sample condition covers the shape of the population.

(a) Standard error: $SE = \frac{10}{\sqrt{100}} = \frac{10}{10} = 1$ minute, exactly half the $SE = 2$ from $n = 25$, since $SE$ shrinks like $\frac{1}{\sqrt{n}}$ and $\sqrt{100} = 2\sqrt{25}$.

(a) Critical value: $df = 100 - 1 = 99$, which the table skips, since it prints 80 and then 100. Use $df = 80$, giving $t^* = 1.990$. So $ME = 1.990 \times 1 = 1.990$ minutes. Technology at the exact $df = 99$ gives $t^* = 1.984$ and $ME = 1.984$, which agrees to one decimal: 2.0 minutes either way, a difference of about 0.006 minutes. Using the $df = 80$ row is the conservative choice, since it gives the slightly wider interval.

(b) Compare. At $n = 25$ the margin was $2.064 \times 2 = 4.128$ minutes. The ratio is $\frac{1.990}{4.128} = 0.482$, so the new margin is about 48% of the old one, slightly more than a halving.

(b) Explain. Two things moved. The standard error halved exactly, which is the $\frac{1}{\sqrt{n}}$ rule at work. On top of that, $t^*$ fell from 2.064 to 1.990 because $df$ grew from 24 to 99, so the table row moves from 24 to 80, and the $t$ distribution tightens toward the standard normal as $df$ rises. Together they beat an exact halving by a small amount.

Conclude. Quadrupling the sample from 25 to 100 batteries cuts the 95% margin of error for the mean run time from about 4.1 minutes to about 2.0 minutes. Cutting it to about 1 minute would take roughly 16 times the original sample, near 400 batteries, which is the practical cost of the square root in the denominator.

State. Let $\mu$ be the true mean time, in minutes, from the start of the roast to first crack for all batches roasted this way. Estimate $\mu$ with a 95% confidence interval.

Plan. One-sample $t$ interval for a mean. Random: the 8 batches are a random sample. 10%: 8 is less than 10% of all batches the roaster runs. Normal/large sample: $n = 8 < 30$, so lean on the dotplot, which is roughly symmetric with no outliers.

Do. Sample mean: $\bar{x} = \frac{12 + 15 + 11 + 14 + 16 + 13 + 12 + 11}{8} = \frac{104}{8} = 13$ minutes.

Do. Sample standard deviation. Deviations from 13 are $-1, 2, -2, 1, 3, 0, -1, -2$, so the squared deviations are $1, 4, 4, 1, 9, 0, 1, 4$ and sum to 24. Dividing by $n - 1$: $s = \sqrt{\frac{24}{8 - 1}} = \sqrt{3.4286} = 1.8516$ minutes.

Do. Standard error: $SE = \frac{1.8516}{\sqrt{8}} = \frac{1.8516}{2.8284} = 0.6547$ minutes. With $df = 8 - 1 = 7$, the 95% column gives $t^* = 2.365$, so $ME = 2.365 \times 0.6547 = 1.548$ minutes.

(a) Interval: $13 \pm 1.548 = (11.452,\ 14.548)$ minutes. We are 95% confident this interval captures the true mean time to first crack.

(b) Conclude. The claimed value 15 minutes sits above the upper bound of 14.548, so 15 is not a plausible value for the true mean. The interval does not support the guide's claim, and it points to a true mean below 15 minutes. A two-sided $t$-test of $H_0: \mu = 15$ at $\alpha = 0.05$ would reject for the same reason.

(a) Center and half-width. $\bar{x} = \frac{1.0 + 5.4}{2} = \frac{6.4}{2} = 3.2$ words per minute, and $ME = \frac{5.4 - 1.0}{2} = \frac{4.4}{2} = 2.2$ words per minute. So the interval is $3.2 \pm 2.2$.

Conditions. The reported interval, and the rebuilt one in (d), rest on the usual three: the 30 students are a random sample, 30 is well under 10% of all students who complete this program, and $n = 30$ meets the large-sample threshold, so the sampling distribution of the mean change is approximately normal.

(b) Standard error. With $n = 30$, $df = 30 - 1 = 29$, and the 95% column gives $t^* = 2.045$. Since $ME = t^* \cdot SE$, invert it: $SE = \frac{ME}{t^*} = \frac{2.2}{2.045} = 1.0758$ words per minute.

(b) Sample standard deviation. From $SE = \frac{s}{\sqrt{n}}$, $s = SE \cdot \sqrt{n} = 1.0758 \times \sqrt{30} = 1.0758 \times 5.4772 = 5.892$ words per minute.

(c) Decision from the interval. The 95% interval runs from 1.0 to 5.4 and does not contain 0, so 0 is not a plausible value for the true mean improvement. A two-sided test of $H_0: \mu = 0$ at $\alpha = 0.05$ would therefore reject $H_0$: a two-sided test at level $\alpha$ and a $100(1 - \alpha)\%$ confidence interval always agree.

(d) Rebuild at 99%. The center and standard error do not change, only $t^*$: at $df = 29$ the 99% column gives $t^* = 2.756$, so $ME = 2.756 \times 1.0758 = 2.965$ and the interval is $3.2 \pm 2.965 = (0.235,\ 6.165)$. It still excludes 0, though with little room to spare, so a two-sided test at $\alpha = 0.01$ would also reject $H_0$.

Cross-check with the table. The test statistic is $t = \frac{3.2 - 0}{1.0758} = 2.97$ at $df = 29$. That clears 2.756, the last entry in the $df = 29$ row of the t-table, which carries a two-tail probability of 0.01. So the two-sided p-value is below 0.01, the same message the 99% interval sends by excluding 0 with little room to spare. Bounding a p-value between table columns rather than pinning it exactly is what the exam expects; software returns a single value.

Conclude. There is convincing evidence that the true mean change in reading speed for students who complete this program is positive, with plausible improvements running from about 1.0 to 5.4 words per minute at 95% confidence, or from about 0.2 to 6.2 at 99% confidence. Because there is no control group and students were not randomly assigned, this does not establish that the program caused the improvement.