What is the difference between z* and t*? When to use z* and when to you t*?

To be more precise - if you know the standard deviation of the ENTIRE population, then use z*. Otherwise, calculate the standard deviation of the sample and use 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

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.

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?

I hope this helps, It states that they are wearing both watches at the same time, not that they are running in a group together at the same time. Since the runners were randomly selected, and the runners according to the study were not running together, they weren't able to reasonably influence each other's results.

You made a typo. You wrote "here are the data" instead of "here is the date." It's at Step 3.

Data is a plural word and thus the correct phrase would be "here are the data" instead of "here is the data". The singular form of "data" is "datum".

Main content

Making a t interval for paired data

Google Classroom

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

[Wait, how do you find the critical value for 95% confidence?]

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?

Want to join the conversation?

Sort by:

Makara.p
Posted 6 years ago. Direct link to Makara.p's post “What is the difference be...”
What is the difference between z* and t*? When to use z* and when to you t*?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Angels Angels
  Posted 6 years ago. Direct link to Angels Angels's post “To be more precise - if y...”
  To be more precise - if you know the standard deviation of the ENTIRE population, then use z*. Otherwise, calculate the standard deviation of the sample and use t*.
  Comment on Angels Angels's post “To be more precise - if y...”
  (17 votes)
ahmedmetiaz
Posted 4 years ago. Direct link to ahmedmetiaz's post “"Since there was no contr...”
"Since there was no control group for comparison we shouldn't say there's a causal relationship here" I didn't understand this explanation
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- MFogleman
  Posted 4 years ago. Direct link to MFogleman's post “The only way to tell if s...”
  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.
  Button navigates to signup page
  (15 votes)
Hieu Le
Posted 5 years ago. Direct link to Hieu Le's post “How did you find the stan...”
How did you find the standard deviation in the table in step 2?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- deka
  Posted a year ago. Direct link to deka's post “using our familiar genera...”
  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
  Comment on deka's post “using our familiar genera...”
  (1 vote)
phillicia makanatleng
Posted 3 years ago. Direct link to phillicia makanatleng's post “how was the standard devi...”
how was the standard deviations in example 1 calculated?
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(4 votes)
Answer
pfarheen
Posted 4 years ago. Direct link to pfarheen's post “You made a typo. You wrot...”
You made a typo. You wrote "here are the data" instead of "here is the date." It's at Step 3.
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(0 votes)
Answer
- Saivishnu Tulugu
  Posted 4 years ago. Direct link to Saivishnu Tulugu's post “Data is a plural word and...”
  Data is a plural word and thus the correct phrase would be "here are the data" instead of "here is the data". The singular form of "data" is "datum".
  Button navigates to signup page
  (11 votes)
Mike Papadakis
Posted 3 years ago. Direct link to Mike Papadakis's post “Is it okay to use the dif...”
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)
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(2 votes)
Answer
jk3109
Posted 5 years ago. Direct link to jk3109's post “In my opinion, Step 2 of ...”
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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Jesse Johnson
  Posted 5 years ago. Direct link to Jesse Johnson's post “I hope this helps, It s...”
  I hope this helps,
  
  It states that they are wearing both watches at the same time, not that they are running in a group together at the same time. Since the runners were randomly selected, and the runners according to the study were not running together, they weren't able to reasonably influence each other's results.
  Comment on Jesse Johnson's post “I hope this helps, It s...”
  (3 votes)
B H
Posted 3 years ago. Direct link to B H's post “How did you calculate the...”
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
5:55
pm CDT: Math AP®︎ Statistics Confidence intervals Confidence intervals for means, example 2.
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(1 vote)
Answer
shadiaroberts.srr
Posted 2 years ago. Direct link to shadiaroberts.srr's post “how did he get the standa...”
how did he get the standard deviation if it wasnt given?
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(1 vote)
Answer
kartikeya.gupta4
Posted 5 years ago. Direct link to kartikeya.gupta4's post “For PROBLEM B (EXAMPLE 2)...”
For PROBLEM B (EXAMPLE 2), should the correct answer be- 'Yes, since, we are only 95% confident of the interval' ?
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(0 votes)
Answer
- NB
  Posted 4 years ago. Direct link to NB's post “Yes, but we can't attribu...”
  Yes, but we can't attribute the gain to practice on the website. We need the scores of those who don't practice to compare with. An experimental design with a control group would ensure that the gain is indeed caused by practice on the website
  So, the confidence interval tells us that the true value for mean gain for those who practice lies within that interval, but not that it is caused by practice on the website. I hope this helps.
  Button navigates to signup page
  (1 vote)

AP®︎/College Statistics

Course: AP®︎/College Statistics > Unit 11