AP Statistics · Free-response skill

Build and use a paired t interval

The move. Build a t interval for a mean difference from paired data, check its conditions, interpret it in context, and use it to judge a claim.

How to do it

  1. Name the procedure from the design. If each unit gives two measurements, the data are paired: work with the differences and say which order you subtracted in.
  2. Define the parameter in words: the true mean difference for the whole population, with the order of subtraction and the units.
  3. Check every condition with a number or a reason: Random (how units were chosen), 10% (n compared with the population), and Normal/Large Sample (with n under 30, say what the differences look like).
  4. Show the work: df = n - 1, t∗t^*, the standard error sdn\frac{s_d}{\sqrt{n}}, and the interval. Then interpret it: we are C% confident the interval captures the true mean difference, in context.
  5. To judge a claim, ask whether the claimed value is inside the interval. Values in the interval are plausible; values outside it are not. Remember that a before-and-after design cannot rule out other changes over the same time.

Practice prompt Try it

A city transit planner wants to know whether a new signal system, which gives buses priority at traffic lights, shortens trips. The city runs 140 bus routes. The planner selects 10 routes at random and records the mean trip time on each route for one month before the system starts and for one month after. The table gives each route's difference, before minus after, in minutes; a positive value means trips got shorter. The 10 differences have mean 1.20 minutes and standard deviation 0.88 minutes.

Mean trip time before minus after, for 10 randomly chosen routes, in minutes

RouteDifference (minutes)
11.4
20.6
32.1
4-0.3
51.8
60.9
72.6
80.2
91.5
101.2

Use the information above to answer each part. A graphing calculator is allowed.

  1. Explain why the planner should analyze the 10 differences rather than treat the before times and after times as two separate samples. Then say in words what parameter the planner is estimating.
  2. Check whether the conditions for a t interval for a mean difference are met. Use numbers from the passage and the table.
  3. Work out a 95% confidence interval for the parameter. Show the degrees of freedom, the critical value and the standard error you used.
  4. Say what your interval tells the planner about the city's bus routes.
  5. The company that sold the system says it cuts the mean trip time by at least 2 minutes. Does your interval support that claim? Does it give evidence that the system is linked to shorter trips at all? Explain both answers using the interval.

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Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 9 points

Model answer, annotated

(a) The same 10 routes give both a before time and an after time, so each before time is linked to an after time on the same route. The two sets are paired, not independent, so the planner should analyze each route's difference. The parameter is μd\mu_d, the true mean difference in trip time, before minus after, for all 140 of the city's bus routes.

  • Point 1 Points to the design: the same routes are measured twice, which is what makes the data paired.
  • Point 2 Names the parameter with its population, its order of subtraction and its context.

(b) Random: the 10 routes were selected at random. 10%: 10 routes is no more than 10% of the 140 routes, since 10 is less than 14. Normal/Large Sample: n = 10 is under 30, so look at the differences. The quartiles are 0.6 and 1.8, so the IQR is 1.2 and the outlier fences are 0.6 - 1.8 = -1.2 and 1.8 + 1.8 = 3.6. Every difference, from -0.3 to 2.6, lies inside the fences, and the values spread fairly evenly around the median of 1.3, so there are no outliers or strong skew and a t interval is reasonable.

  • Point 3 Both checks are stated with the actual numbers, not just named.
  • Point 4 Because n is small, the answer checks the 10 differences against the 1.5 IQR fences and finds no outliers.

(c) df = 10 - 1 = 9, so t∗=2.262t^* = 2.262. The standard error is 0.8810≈0.278\frac{0.88}{\sqrt{10}} \approx 0.278, so the margin of error is 2.262×0.278≈0.632.262 \times 0.278 \approx 0.63. The interval is 1.20±0.631.20 \pm 0.63, from about 0.57 to 1.83 minutes.

  • Point 5 Shows every piece of the formula: df, the critical value and the standard error.
  • Point 6 Finishes with the interval itself, with its unit.

(d) We are 95% confident that the interval from 0.57 to 1.83 minutes captures the true mean difference in trip time, before minus after, for all 140 of the city's bus routes. In other words, the mean trip time across all routes plausibly fell by between about 0.57 and 1.83 minutes after the system started.

  • Point 7 Uses all three parts of an interpretation: confidence level, interval, and the parameter in context.

(e) The interval does not support the company's claim. Every value in it is below 2 minutes; the upper end is about 1.83, so a mean cut of 2 minutes or more is not plausible. The interval does give evidence that trips got shorter, because the whole interval is above 0, so a mean difference of 0, no change, is not plausible. Since this is a before-and-after comparison, other changes during those months could also play a part.

  • Point 8 Compares the claimed value of 2 with the interval's upper end.
  • Point 9 Checks whether 0 is in the interval to decide whether there is evidence of any change.