Interpreting confidence intervals: practice problems

Name the parameter first. Here it is $p$, the true proportion of all attendees at this county fair who arrived by public transit. The sample proportion (p-hat, written $\hat{p}$) was $\frac{100}{500} = 0.20$, but the interval is a statement about $p$, not about $\hat{p}$.

Use the required opening: we are 95% confident that the interval from one endpoint to the other contains the parameter.

Attach the population by name. 'All attendees at this county fair' is specific enough; 'people' is not.

Keep the endpoints as given, 0.165 and 0.235, or state them as 16.5% and 23.5%. Either is fine as long as they match the interval.

Do not claim the interval holds 95% of the attendees, and do not attach a probability to a finished interval. Both of those lose the point.

(a) The confidence level describes the method across many samples, not the one interval in front of you. The template is: if the sampling were repeated many times with the same sample size, about 90% of the intervals built this way would contain the true parameter.

Name the parameter and the population inside that sentence, exactly as you would for an interval interpretation.

A useful check: your confidence level sentence should still make sense if you had never computed 0.412 and 0.528, because it is about the procedure.

(b) Once the sample is drawn and the arithmetic is done, both endpoints are fixed numbers and the true survival proportion is a fixed number. This interval either contains it or it does not, so the probability is 0 or 1, never 0.90.

The randomness lived in the sampling step, before the data arrived. That is why the wording is '90% confident' rather than '90% probability'.

The 90% attaches to the long-run capture rate of the procedure, so about 1 interval in 10 built this way misses the true proportion completely.

First get your bearings: $\hat{p} = \frac{270}{900} = 0.30$, and the interval is centered there.

Statement 1 is wrong. This is the '95% of the data' error. A confidence interval estimates a population parameter, not the spread of individual observations. Each senior answered yes or no, so an individual response is not even a number that could sit inside $(0.270,\ 0.330)$.

Statement 2 is wrong. The interval is already computed, so its endpoints are fixed numbers and the true proportion $p$ is a fixed number. The probability that $p$ lies between them is 0 or 1, not 0.95. The randomness belonged to the sampling step, which is why the wording is '95% confident' rather than '95% probability'.

Statement 3 is correct. It gives the confidence level, says the interval contains the parameter, and identifies the parameter as the proportion of all seniors in this district with a part-time job.

Statement 4 is correct. It is the confidence level interpretation: repeat the sampling at the same sample size and about 95% of the resulting intervals capture the parameter.

Statement 5 is wrong. The sampling distribution of $\hat{p}$ is centered at the unknown $p$, not at your one sample proportion 0.30. About 95% of sample proportions land within about 1.96 standard deviations of $p$, while this interval collects the values within 1.96 standard errors of $\hat{p} = 0.30$. Those two ranges coincide only in the lucky case where $\hat{p}$ landed exactly on $p$.

The rule: a confidence interval is a set of plausible values for the parameter. A claimed value inside the interval stays plausible, so the data give no convincing evidence against it. A claimed value outside the interval is not plausible at that confidence level, so the data do give convincing evidence against it.

(a) Is 0.35 inside $(0.312,\ 0.388)$? Yes, because $0.312 < 0.35 < 0.388$. There is no convincing evidence against the company's claim.

(b) Is 0.25 inside? No, because $0.25 < 0.312$, so it sits below the entire interval. Every plausible value exceeds 0.25, so there is convincing evidence against the analyst's claim, and the evidence points to a higher repeat rate.

(c) Failing to rule out a value is not the same as proving it. The interval contains many other values, including 0.32 and 0.38, and on this evidence each is as plausible as 0.35.

Write conclusions in that language: 'plausible' or 'not plausible', never 'proved' or 'exactly equal to'.

(a) The Brand A and Brand B intervals overlap on the stretch from 0.86 to 0.89. Any value in that overlap is plausible for both brands, so a single on-time proportion shared by the two is still consistent with the data.

Overlap therefore does not establish a difference, and the manager's reasoning is not valid on this evidence.

Watch the converse trap as well. Overlapping intervals do not prove the two proportions are equal; they only fail to rule equality out.

(b) The Brand A interval tops out at 0.89 and the Brand C interval starts at 0.95, so no single value is plausible for both. Two intervals that do not overlap are strong evidence that the true proportions really differ.

(c) Comparing two separate one-sample intervals is a rough screen, not a test. Build a confidence interval for the difference $p_1 - p_2$ (the true proportion for the first brand minus the true proportion for the second), or run a two-proportion $z$-test.

Read that difference interval the same way: if it contains 0, a difference of zero stays plausible and nothing is established; if it excludes 0, there is convincing evidence of a real difference, and its sign tells you which brand is higher.

$\hat{p} = \frac{320}{1600} = 0.20$. Conditions: the sample is random, it is under 10% of all prescriptions, and large counts hold since $n\hat{p} = 320 \geq 10$ and $n(1 - \hat{p}) = 1280 \geq 10$.

$SE = \sqrt{\frac{0.20(0.80)}{1600}} = \sqrt{\frac{0.16}{1600}} = \sqrt{0.0001} = 0.01$.

(a) At 90% confidence the critical value (z-star, written $z^*$) is $z^* = 1.645$, so $ME = 1.645 \times 0.01 = 0.01645$ and the interval is $0.20 \pm 0.01645 = (0.18355,\ 0.21645)$, about $(0.184,\ 0.216)$.

(b) At 99% confidence $z^* = 2.576$, so $ME = 2.576 \times 0.01 = 0.02576$ and the interval is $0.20 \pm 0.02576 = (0.17424,\ 0.22576)$, about $(0.174,\ 0.226)$.

Compare widths: $2 \times 0.01645 = 0.0329$ against $2 \times 0.02576 = 0.05152$. The 99% interval is $\frac{0.05152}{0.0329} = 1.566$ times as wide, the same ratio as $\frac{2.576}{1.645}$.

(c) What improved is the capture rate of the method: 99 of every 100 intervals built this way would contain the true late-pickup proportion, against 90 of every 100. What got worse is precision: the 99% interval leaves a wider set of plausible values, so it pins the proportion down less tightly.

Note what did not change. The sample never changed, so the standard error is 0.01 in both, and both intervals are centered at 0.20. Only the critical value moved.

$\hat{p} = \frac{60}{250} = 0.24$.

Conditions: the sample is random, it is under 10% of all station-mornings, and large counts hold since $n\hat{p} = 250(0.24) = 60 \geq 10$ and $n(1 - \hat{p}) = 250(0.76) = 190 \geq 10$.

$SE = \sqrt{\frac{0.24(0.76)}{250}} = \sqrt{\frac{0.1824}{250}} = \sqrt{0.0007296} = 0.027011$.

For 95% confidence $z^* = 1.96$, so $ME = 1.96 \times 0.027011 = 0.052942$.

(a) Interval: $0.24 \pm 0.052942 = (0.187058,\ 0.292942)$, about $(0.187,\ 0.293)$.

(b) The interval interpretation names the parameter and the population, and says the interval contains that parameter. Do not say 95% of mornings.

(c) The confidence level interpretation describes the method across repeated samples, not this one interval, so it mentions many samples of the same size and the share of intervals that capture the parameter.

(a) The sample mean (x-bar, written $\bar{x}$) is 42 and the sample standard deviation is $s = 6$, with $n = 16$. The parameter is $\mu$ (mu), the true mean rotation time at this shop.

Degrees of freedom: $df = n - 1 = 16 - 1 = 15$, which is the row that gives the critical value (t-star, written $t^*$) $t^* = 2.131$ at 95% confidence.

$SE = \frac{s}{\sqrt{n}} = \frac{6}{\sqrt{16}} = \frac{6}{4} = 1.5$ minutes.

$ME = t^* \cdot SE = 2.131 \times 1.5 = 3.1965$ minutes.

Interval: $42 \pm 3.1965 = (38.8035,\ 45.1965)$, about $(38.80,\ 45.20)$ minutes.

(b) Interpret it as a statement about $\mu$ and about this shop, with the units attached.

(c) Check whether 40 is inside: $38.80 < 40 < 45.20$, so it is. The promised mean stays plausible and the interval gives no convincing evidence against it.

(d) The interval estimates $\mu$, the mean of all rotation times, not the spread of individual rotations. A range holding 95% of individual times would be much wider, because it would have to absorb the car-to-car variation instead of the much smaller variation of a sample mean. That job belongs to a prediction interval, a different tool.