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Course: Statistics and probability > Unit 11

Lesson 3: Estimating a population mean

Making a t interval for paired data

In some studies, we make two observations on the same individual. For instance, we might look at each student's pre-test and post-test scores in a course. In other studies, we might make an observation on each of two similar individuals. For example, some medicine trials involve pairing similar subjects so one receives the medicine and the other receives a placebo.

In both types of studies, we're working with paired data, and whenever we're working with paired data, we're typically interested in the difference between each pair—for example, the difference between the pre-test and the post-test data, or the difference between the medicine and the placebo data.

If certain conditions are met, we can construct a

t

interval to estimate the mean of these differences and draw conclusions.

In this article, we'll be going through two examples of making a

t

interval for paired data. Importantly, you'll have a chance to work through the second example on your own to ensure you've picked up on the main ideas.

Example 1

A running magazine wanted to review two watches—watch A and watch B—that use global position systems (GPS) to calculate the distance someone runs. They noticed that the watches didn't usually agree on the distance someone traveled in a given run.

The magazine took a random sample of

5

subscribers and asked them to run a

10

kilometer route wearing both watches at the same time (they all agreed to participate). At the end of their runs, the participants recorded the distance each watch said they traveled. Here are the data (all distances are in kilometers):

Runner	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍	$5$ ‍
Watch A	$9.8$ ‍	$9.8$ ‍	$10.1$ ‍	$10.1$ ‍	$10.2$ ‍
Watch B	$10.1$ ‍	$10$ ‍	$10.2$ ‍	$9.9$ ‍	$10.1$ ‍

Construct a $95 %$ ‍ confidence interval to estimate the mean difference in the distances reported by these watches. Does the interval suggest that there is a difference between the two watches?

Step 1: Calculate the differences

Even though it appears we have two sets of data—watch A and watch B—these data didn't come from two independent samples. The magazine took a single sample of

5

runners, and each runner wore both watches, so this is a matched pairs design. The one set of data we're interested in is the difference between watch A and watch B for each runner. Let's define this variable as

difference = B - A

and calculate the difference for each runner:

Runner	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍	$5$ ‍
Watch A	$9.8$ ‍	$9.8$ ‍	$10.1$ ‍	$10.1$ ‍	$10.2$ ‍
Watch B	$10.1$ ‍	$10$ ‍	$10.2$ ‍	$9.9$ ‍	$10.1$ ‍
Difference $(B - A)$ ‍	$0.3$ ‍	$0.2$ ‍	$0.1$ ‍	$- 0.2$ ‍	$- 0.1$ ‍

Key idea: When dealing with paired data, we're most interested in the distribution of the differences.

Step 2: Check conditions

We want to use these

n = 5

differences to construct a confidence interval for the mean difference. Since we don't know the population standard deviation of the differences, we'll have to use the sample standard deviation in its place. This makes it appropriate to use a

t

interval instead of a

z

interval to estimate the mean difference. Let's check the conditions for making a

t

interval.

Random: The magazine took a random sample of their subscribers.
Normal: Since our sample of $n = 5$ ‍ runners is small, we need to plot the data. The differences are roughly symmetric with no outliers, so it should be safe to proceed.

Independent: It's reasonable to assume independence between each runner's measurements. They were randomly selected, and they shouldn't influence each other's results.

Step 3: Construct the interval

Here are the data:

Runner	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍	$5$ ‍
Watch A	$9.8$ ‍	$9.8$ ‍	$10.1$ ‍	$10.1$ ‍	$10.2$ ‍
Watch B	$10.1$ ‍	$10.0$ ‍	$10.2$ ‍	$9.9$ ‍	$10.1$ ‍
Difference $(B - A)$ ‍	$0.3$ ‍	$0.2$ ‍	$0.1$ ‍	$- 0.2$ ‍	$- 0.1$ ‍

Here are the summary statistics:

	Mean	Standard deviation
Watch A	${\bar{x}}_{A} = 10.00$ ‍	$s_{A} \approx 0.19$ ‍
Watch B	${\bar{x}}_{B} = 10.06$ ‍	$s_{B} \approx 0.11$ ‍
Difference $(B - A)$ ‍	${\bar{x}}_{Diff} = 0.06$ ‍	$s_{Diff} \approx 0.21$ ‍

Since we want to construct a confidence interval for the mean difference, we only need the summary statistics for the differences.

We'll use the formula for a one-sample

t

interval for a mean:

\begin{aligned} (statistic) & \pm (\binom{critical}{value}) (\binom{standard deviation}{of statistic}) \\ {\bar{x}}_{Diff} & \pm t^{*} \cdot \frac{s_{Diff}}{\sqrt{n}} \end{aligned}

Components of formula:

Our statistic is the sample mean

{\bar{x}}_{Diff} = 0.06 km

Our sample size is

n = 5

runners.

Our sample standard deviation is

s_{Diff} = 0.21 km

Our degrees of freedom is

df = 5 - 1 = 4

, so for

95 %

confidence our critical value is

t^{*} = 2.776

Using a calculator:

We can use the inverse

t

function to find the value corresponding to a tail area of

0.025

\begin{aligned} invT \\ area: 0.025 \\ df : 4 \\ = - 2.77645 \end{aligned}

Using a $t$ ‍ table:

We can use a

t

table to find the value corresponding to an area of

0.025

with

df = 4

t^{*} = 2.776

Computations:

\begin{aligned} {\bar{x}}_{Diff} & \pm t^{*} \cdot \frac{s_{Diff}}{\sqrt{n}} \\ 0.06 & \pm 2.776 \cdot \frac{0.21}{\sqrt{5}} \\ 0.06 & \pm (2.776) (0.094) \\ 0.06 & \pm 0.261 \\ 0.06 & - 0.261 = - 0.201 \\ 0.06 & + 0.261 = 0.321 \end{aligned}

Interval

\approx (- 0.20, 0.32)

Step 4: Interpret the interval

Does the interval suggest that there is a difference between the two watches?

We're

95 %

confident that the interval

(- 0.20, 0.32)

captures the mean difference between the distances (in kilometers) reported by the watches on this sort of run. Notice that the interval contains

0 km

—which represents no difference—so it's plausible that there is no difference between the distances reported by Watch A and Watch B.

If the entire interval had been above

0

(all positive values), or if it had been entirely below

0

(all negative values), then it would have suggested a difference between the two watches.

Example 2—Try it!

An educational website offers a practice program for the Law School Admissions Test (LSAT). Users of the program take a pretest and posttest. Here are the scores and gains for a random sample of

6

users:

User	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍	$5$ ‍	$6$ ‍
Pre	$140$ ‍	$152$ ‍	$153$ ‍	$159$ ‍	$150$ ‍	$146$ ‍
Post	$150$ ‍	$159$ ‍	$170$ ‍	$164$ ‍	$148$ ‍	$166$ ‍
Gain $(post - pre)$ ‍	$10$ ‍	$7$ ‍	$17$ ‍	$5$ ‍	$- 2$ ‍	$20$ ‍

Here are summary statistics:

	Mean	Standard deviation
Pre	${\bar{x}}_{pre} = 150$ ‍	$s_{pre} \approx 6.48$ ‍
Post	${\bar{x}}_{post} = 159.5$ ‍	$s_{post} \approx 8.89$ ‍
Gain $(post - pre)$ ‍	${\bar{x}}_{gain} = 9.5$ ‍	$s_{gain} \approx 8.07$ ‍

Problem A (Example 2)

Based on this sample, which is a $95 %$ ‍ confidence interval for the mean gain for users of this program?

Problem B (Example 2)

Is it plausible that users of this program have no mean gain?

The makers of the website say that this interval provides strong evidence that using their program will cause an increase in a user's LSAT score.

Problem C (Example 2)

Is this a valid conclusion?

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