Interval for Two Means — AP® Statistics Lesson

Part 1 · Hook

Why this matters

Twenty-eight volunteers learned new words, half in short sessions spread over a week and half in one long cram. A week later the spaced group recalled 5.7 more words on average. How big is the real advantage? A two-sample t interval turns the two groups' means and spreads into a range of plausible values for the true difference.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For independent samples, the standard deviation of $\bar{x}_1 - \bar{x}_2$ is

$\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$
$\frac{\sigma_1}{\sqrt{n_1}} + \frac{\sigma_2}{\sqrt{n_2}}$
$\sqrt{\frac{\sigma_1^2}{n_1} - \frac{\sigma_2^2}{n_2}}$
$\frac{\sigma_1 - \sigma_2}{n_1 + n_2}$

Show the answer

Variances add, then take the square root.

Correct: $\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ :
$\frac{\sigma_1}{\sqrt{n_1}} + \frac{\sigma_2}{\sqrt{n_2}}$ :
$\sqrt{\frac{\sigma_1^2}{n_1} - \frac{\sigma_2^2}{n_2}}$ :
$\frac{\sigma_1 - \sigma_2}{n_1 + n_2}$ :

2. A one-sample t interval for a mean is

$\bar{x} \pm t^*\frac{s}{\sqrt{n}}$ with $n - 1$ df
$\bar{x} \pm z^*\frac{s}{n}$
$\bar{x} \pm t^* s$
$\mu \pm t^*\frac{s}{\sqrt{n}}$

Show the answer

Statistic plus or minus $t^*$ times the standard error $\frac{s}{\sqrt{n}}$ .

Correct: $\bar{x} \pm t^*\frac{s}{\sqrt{n}}$ with $n - 1$ df:
$\bar{x} \pm z^*\frac{s}{n}$ :
$\bar{x} \pm t^* s$ :
$\mu \pm t^*\frac{s}{\sqrt{n}}$ :

Part 4 · See it

See it first

Top: side-by-side boxplots of words recalled. Spaced practice: min 25, Q1 29, median 35, Q3 39, max 44. Massed practice: min 16, Q1 25, median 29, Q3 34, max 39. Neither has outliers. Bottom: the 95% confidence interval for the true difference in mean recall, spaced minus massed, from 0.61 to 10.81 words, centered at 5.71, entirely to the right of a dashed line at 0. — Top: boxplots of words recalled by the two randomly assigned groups. Bottom: the 95% two-sample t interval for $\mu_S - \mu_M$ , from 0.61 to 10.81 words, centered at the observed difference 5.71. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The parameter is $\mu_1 - \mu_2$ , the difference in true meansits point estimate is $\bar{x}_1 - \bar{x}_2$
Replace $\sigma_1$ and $\sigma_2$ with $s_1$ and $s_2$ the standard error is $\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$
Estimating two standard deviations adds uncertaintyuse $t^*$ , with technology df (Welch) or the conservative df, the smaller of $n_1 - 1$ and $n_2 - 1$
Check random, 10% and normal/large sample for each groupbuild $(\bar{x}_1 - \bar{x}_2)$ $\pm t^*\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ and interpret it in context, order of subtraction stated

Part 6 · Key ideas

Key ideas

Two-sample t interval: $(\bar{x}_1 - \bar{x}_2)$ $\pm t^*\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$ .
Degrees of freedom: technology (Welch) gives a non-integer df; the conservative df is the smaller of $n_1 - 1$ and $n_2 - 1$ , which gives a slightly wider interval. Both are accepted.
Conditions for each group: random (two independent random samples or random assignment), 10% (when sampling), normal/large sample.
Two independent groups use this interval; two linked measurements on each unit are paired and use a one-sample t interval on the differences.

Part 7 · Misconception

A common mistake

The wrong idea: Build a t interval for each group; if they overlap, there is no difference.

What actually happens: Two separate intervals answer a different question and can overlap even when the difference is clear. Estimate the difference directly with one two-sample t interval for $\mu_1 - \mu_2$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Spaced versus massed practice

Twenty-eight student volunteers learned 50 new vocabulary words. They were randomly assigned, 14 to each group, to study in four short sessions spread over a week (spaced) or in one long session (massed). A week later each took the same recall test (invented data).

Words recalled (out of 50)
Group	n	Mean	SD
Spaced	14	34.21	6.13
Massed	14	28.50	6.97

1. Which procedure estimates how much better spaced practice is, on average?

A paired t interval for the mean difference
A two-sample z interval for $p_S - p_M$
Two one-sample t intervals, one per group
A two-sample t interval for $\mu_S - \mu_M$

Show the answer

Parameter: $\mu_S - \mu_M$ , the true mean recall with spaced practice minus with massed practice, for volunteers like these. Procedure: a two-sample t interval.

A paired t interval for the mean difference: The volunteers were not paired: each was in one group only, so there are no natural pairs to subtract.
A two-sample z interval for $p_S - p_M$ : Words recalled is quantitative, so the parameter is a difference of means.
Two one-sample t intervals, one per group: Two separate intervals do not estimate the difference, and overlapping intervals are hard to judge.
Correct: A two-sample t interval for $\mu_S - \mu_M$ : Two separate groups from random assignment, a quantitative response and unknown population standard deviations: a two-sample t interval for the difference in true mean recall.

2. Each group has only 14 students. How is the normal/large sample condition met?

The two groups together have 28 students
Neither boxplot shows strong skew or outliers
The volunteers were randomly assigned to the two groups
14 is less than 10% of the students at their school

Show the answer

Conditions: random assignment (random); no 10% check for an experiment; normal/large sample from the two boxplots, each without strong skew or outliers.

The two groups together have 28 students: Each group is checked on its own, and 14 is less than 30.
Correct: Neither boxplot shows strong skew or outliers: With $n < 30$ in both groups, each group's graph must show no strong skew and no outliers. Both boxplots are roughly symmetric with no outliers.
The volunteers were randomly assigned to the two groups: That is the random condition. Shape is a separate check.
14 is less than 10% of the students at their school: The 10% condition does not apply to random assignment, and it is about independence, not shape.

3. How many degrees of freedom does technology give for the two-sample t interval? (The conservative choice is also accepted.)

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Technology uses the Welch formula and gives $\text{df} = 25.58$ . The conservative choice is the smaller $n - 1$ : $14 - 1 = 13$ . Either is accepted on the exam. Reading Table B, use the row at or below the technology df (25), never a rounded-up row.

Answer: 25.58

4. Construct the 95% confidence interval for $\mu_S - \mu_M$ , in words recalled.

Type the interval as "from a to b".

Show the answer

$(\bar{x}_S - \bar{x}_M)$ $\pm t^*\sqrt{\frac{s_S^2}{n_S} + \frac{s_M^2}{n_M}}$ $= (34.21 - 28.50)$ $\pm 2.057\sqrt{\frac{6.13^2}{14} + \frac{6.97^2}{14}}$ $= 5.71$ $\pm 5.10$ , from 0.61 to 10.81 words ( $t^* = 2.057$ at df 25.6; the Table B row for df 25 gives 2.060 and nearly the same interval). On the exam the conservative df 13 is also accepted: $t^* = 2.160$ gives 0.35 to 11.07. Using $z^* = 1.96$ (0.85 to 10.57) is wrong, because $s_S$ and $s_M$ estimate the population standard deviations.

Answer: from 0.61 to 10.81

5. Interpret the 95% interval from 0.61 to 10.81 words in context.

Write the conclusion in context.

Show the answer

Name the level, the interval, the parameter (a difference of true means, in the order spaced minus massed) and the context.

A full-credit answer: We are 95% confident that the interval from 0.61 to 10.81 words captures the true difference in mean recall (spaced minus massed) for volunteers like these.

6. A two-sample t interval uses samples of $n_1 = 18$ and $n_2 = 25$ . What are the conservative degrees of freedom?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Conservative degrees of freedom: the smaller of $n_1 - 1 = 17$ and $n_2 - 1 = 24$ , so 17. Technology would give a larger, non-integer df.

Answer: 17

7. Which study calls for a two-sample t interval rather than a paired t interval?

Battery life of 30 phones of brand A and 30 phones of brand B, chosen at random
Blood pressure of 30 patients measured before and after taking a new medicine for a month
Tread wear on the left and right front tires of 30 cars
Scores of 30 sets of twins, one twin per teaching method

Show the answer

Two-sample procedures need two independent groups. Paired procedures are for two linked measurements on each unit.

Correct: Battery life of 30 phones of brand A and 30 phones of brand B, chosen at random: Two separate random samples with no link between a phone of brand A and a phone of brand B.
Blood pressure of 30 patients measured before and after taking a new medicine for a month: Each patient gives two linked values: paired.
Tread wear on the left and right front tires of 30 cars: Left and right tires on the same car are a natural pair.
Scores of 30 sets of twins, one twin per teaching method: Twins are matched pairs.

Part 9 · Summary

Summary

To estimate

\mu_1 - \mu_2

from two independent samples or randomized groups, use

(\bar{x}_1 - \bar{x}_2)

\pm t^*\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}

with technology df or the conservative df. Check random, 10% and normal/large sample for each group, and interpret the interval as capturing the true difference in means, in the stated order.

Part 10 · Up next

What comes next

Part 11 · Connections