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# 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 $95\mathrm{%}$ 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 $\text{difference}=\text{B}-\text{A}$ 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 $\left(\text{B}-\text{A}\right)$$0.3$$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 $\left(\text{B}-\text{A}\right)$$0.3$$0.2$$0.1$$-0.2$$-0.1$
Here are the summary statistics:
MeanStandard deviation
Watch A${\overline{x}}_{\text{A}}=10.00$${s}_{\text{A}}\approx 0.19$
Watch B${\overline{x}}_{\text{B}}=10.06$${s}_{\text{B}}\approx 0.11$
Difference $\left(\text{B}-\text{A}\right)$${\overline{x}}_{\text{Diff}}=0.06$${s}_{\text{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:
Components of formula:
Our statistic is the sample mean .
Our sample size is $n=5$ runners.
Our sample standard deviation is .
Our degrees of freedom is $\text{df}=5-1=4$, so for $95\mathrm{%}$ confidence our critical value is ${t}^{\ast }=2.776$.
Computations:
Interval $\approx \left(-0.20,0.32\right)$

### Step 4: Interpret the interval

Does the interval suggest that there is a difference between the two watches?
We're $95\mathrm{%}$ confident that the interval $\left(-0.20,0.32\right)$ captures the mean difference between the distances (in kilometers) reported by the watches on this sort of run. Notice that the interval contains —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$$166$
Gain $\left(\text{post}-\text{pre}\right)$$10$$7$$17$$5$$-2$$20$
Here are summary statistics:
MeanStandard deviation
Pre${\overline{x}}_{\text{pre}}=150$${s}_{\text{pre}}\approx 6.48$
Post${\overline{x}}_{\text{post}}=159.5$${s}_{\text{post}}\approx 8.89$
Gain $\left(\text{post}-\text{pre}\right)$${\overline{x}}_{\text{gain}}=9.5$${s}_{\text{gain}}\approx 8.07$
Problem A (Example 2)
Based on this sample, which is a $95\mathrm{%}$ 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?

## Want to join the conversation?

• What is the difference between z* and t*? When to use z* and when to you t*? • "Since there was no control group for comparison we shouldn't say there's a causal relationship here" I didn't understand this explanation
(1 vote) • The only way to tell if something causes something else (a causal relationship) is with a controlled experiment, where you can make sure that only one variable changes. In this example, we did an observational study instead of an experiment, so we don't know for sure if the program causes better performance. There could be a lurking variable in the mix, so we would need to assign treatments and perform an experiment to know more.

A control group would be a group that doesn't use the program at all between the pretest and post test, so you can compare the performance between both groups and know that the only change was whether someone used the program or not. You would also need to take some more steps to make sure that this experiment is ironclad, which would let you do more with the results. I would recommend checking out the videos on experimental design, observational studies, and surveys, just to get a bit more depth.
Hope this helps.
• How did you find the standard deviation in the table in step 2? • using our familiar general formula for std,
std_sample = sqrt{sum_on_all_data[(data_i-data_mean)^2]/n-1}
this measures and tells you how each datapoint varies from the mean of them in general

by the way, you can't use the concept and formula for std of the difference between two random variables that sqrt(std_var1^2 + std_var2^2) here. cause the two datasets (pre and post -tests) are not independent here. they are paired each other. thus the std of their difference may be less than that of those from two independent datasets using the formula above. and it is indeed

please, hand on the numbers yourself. it will help you to grasp the concept of variance, standard deviation, and variables more clearly
(1 vote)
• how was the standard deviations in example 1 calculated? • You made a typo. You wrote "here are the data" instead of "here is the date." It's at Step 3. • Is it okay to use the difference of watch a - watch b? this results in negative numbers but if one keeps in mind we talk about the difference in time intervals, no problem should come up. Isn't that right?( example 1) • In my opinion, Step 2 of Example 1 shouldn't be considered independent because I don't believe they "shouldn't influence each other's results". They are running at the same time so isn't it likely they are pressured to keep up at the group pace and thus influenced by each other? If this type of question were on the AP exam, should I state this opinion or assume they aren't going to be influenced and go through with the calculations?
(1 vote) • How did you calculate the standard deviation of the difference between means? You show 8.07; I get 4.49 = [6.48/SQRT(6) + 8.89/SQRT(6)]. Thanks.

At pm CDT: Math AP®︎ Statistics Confidence intervals Confidence intervals for means, example 2.
(1 vote) • how did he get the standard deviation if it wasnt given?
(1 vote) • For PROBLEM B (EXAMPLE 2), should the correct answer be- 'Yes, since, we are only 95% confident of the interval' ? 