Unit 3 · Topic 3.4 Beta

Justifying a Claim Based on a Confidence Interval for a Population Proportion

Interpret an interval with its level, its ends and the parameter in context; interpret the level as the long-run capture rate of the method.

Practice 4: Interpret Results

Part 1 · Hook

Why this matters

A school board says more than 60% of parents want a later start time. A newspaper's random sample of 500 parents finds 66% in favor. Is that enough to back the board? The sample is above 60%, but samples bounce around. A confidence interval answers the question, if you know how to read it and what not to say about it.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A random sample of 500 gives $\hat{p} = 0.66$ . Which is the 95% confidence interval for $p$ ?

$0.66 \pm 1.96\sqrt{\frac{0.66(0.34)}{500}}$
$0.66 \pm 1.645\sqrt{\frac{0.66(0.34)}{500}}$
$0.66 \pm 1.96\frac{0.66(0.34)}{500}$
$0.66 \pm 1.96\sqrt{0.66(0.34)}$

Show the answer

Statistic ± (critical value)(standard error), with $z^* = 1.96$ for 95% and the square root over $\frac{\hat{p}(1 - \hat{p})}{n}$ .

Correct: $0.66 \pm 1.96\sqrt{\frac{0.66(0.34)}{500}}$ :
$0.66 \pm 1.645\sqrt{\frac{0.66(0.34)}{500}}$ :
$0.66 \pm 1.96\frac{0.66(0.34)}{500}$ :
$0.66 \pm 1.96\sqrt{0.66(0.34)}$ :

2. In a 95% confidence interval, what does the critical value 1.96 come from?

The middle 95% of the standard normal distribution lies between $-1.96$ and $1.96$
1.96 is 95% of 2
The sample proportion is 1.96 standard errors from $p$
Large counts requires 1.96 successes

Show the answer

A 95% interval keeps the middle 95% of the standard normal area, leaving 2.5% in each tail.

Correct: The middle 95% of the standard normal distribution lies between $-1.96$ and $1.96$ :
1.96 is 95% of 2:
The sample proportion is 1.96 standard errors from $p$ :
Large counts requires 1.96 successes:

Part 4 · See it

See it first

Twenty-five 95% confidence intervals, one per row, each from a simulated random sample of 400 from a population where the true proportion is p = 0.60, shown as a dashed vertical line. Each interval is drawn as a bar with a dot at its sample proportion. 22 of the 25 intervals cross the line and capture p; 3 miss it and are drawn in orange. The intervals have nearly the same length but are centered at different sample proportions. — Twenty-five 95% intervals from random samples when p = 0.60. Each interval is centered at its own sample proportion; most capture p, a few miss. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Each random sample gives its own $\hat{p}$ each sample gives its own interval, while the true proportion stays fixed
The method captures $p$ in about 95% of sampleswe say we are 95% confident in one interval, because we cannot tell whether it is one of the misses
A claimed value lies inside the intervalit is a plausible value, so the interval gives no convincing evidence against it (and does not prove it)
The whole interval lies above (or below) a claimed valuethat value is not plausible, so the interval gives convincing evidence that the parameter is greater (or less) than it

Part 6 · Key ideas

Key ideas

Interval: "We are 95% confident that the interval from __ to __ captures the true proportion of all __."
Level: "If we took many random samples of this size and built a 95% interval from each, about 95% of the intervals would capture the true proportion."
Values inside the interval are plausible. A claim is supported (convincing evidence) only when the whole interval is on the claimed side.
Never: "95% probability that $p$ is in the interval", "95% of the data", or anything about the sample proportion.
Higher confidence gives a longer interval, so more values stay plausible.

Part 7 · Misconception

A common mistake

The wrong idea: The claimed value 0.70 is inside the interval, so the interval shows the claim is true.

What actually happens: Inside the interval means plausible, not proven: many other values are plausible too. Say the interval does not give convincing evidence against the claim.

Part 8 · Check yourself

Check yourself

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

Graph

Twenty-five intervals from one population

A computer takes 25 random samples of 400 people from a population in which the true proportion with a certain trait is $p = 0.60$ . From each sample it builds a 95% confidence interval for $p$ . Each bar is one interval, with a dot at that sample's $\hat{p}$ ; the dashed line is $p = 0.60$ .

1. How many of the 25 intervals capture the true proportion, and is that surprising?

22, about what 95% confidence predicts for 25 intervals
25, because a 95% interval captures $p$ in each sample
22, which shows the method gives far less than 95% confidence
3, the intervals drawn in a different color

Show the answer

22 intervals cross the dashed line at 0.60 and 3 miss it. In the long run about 95% capture $p$ , which is about 23.75 of 25. Getting 22 in one set of 25 is ordinary chance variation.

Correct: 22, about what 95% confidence predicts for 25 intervals: Right: 22 of 25 is close to 95%, given the chance variation in 25 tries.
25, because a 95% interval captures $p$ in each sample: About 5% of 95% intervals miss $p$ ; here some bars stop short of the dashed line.
22, which shows the method gives far less than 95% confidence: The 95% is a long-run rate. 22 out of 25 is well within normal chance variation from 95%.
3, the intervals drawn in a different color: The orange intervals are the ones that miss the line. Count the ones that cross it.

2. What does "95% confidence" mean in this picture?

About 95% of the population has the trait
Each interval has a 95% chance of containing its own $\hat{p}$
In the long run, about 95% of such intervals capture $p$
About 95% of each sample of 400 falls inside its interval

Show the answer

The confidence level describes the method: if we kept taking random samples and building intervals this way, about 95% of the intervals would capture $p$ .

About 95% of the population has the trait: The population proportion is 0.60. The 95% describes how often the method captures it.
Each interval has a 95% chance of containing its own $\hat{p}$ : Each interval is centered at its own $\hat{p}$ , so it contains $\hat{p}$ every time.
Correct: In the long run, about 95% of such intervals capture $p$ : Right: the level is a success rate of the method over many samples.
About 95% of each sample of 400 falls inside its interval: An interval for a proportion is about $p$ , not about where individual people fall.

3. Look at the interval in the eighth row, which misses $p$ . A student says, "There was a 95% chance that this interval would contain 0.60." Which statement is correct?

It describes the method; this finished interval does not contain 0.60
It is correct: this interval still has a 95% chance of containing 0.60
It is correct, because $p$ moves around from sample to sample
It is wrong: this interval had a 5% chance of being built

Show the answer

Before the sample was taken, the method had a 95% chance of producing an interval that captures $p$ . Once this interval is built, it either contains 0.60 or it does not; this one does not.

Correct: It describes the method; this finished interval does not contain 0.60: Right: probability belongs to the method, not to a finished interval.
It is correct: this interval still has a 95% chance of containing 0.60: The interval is fixed and so is $p$ . It misses 0.60, and no probability changes that.
It is correct, because $p$ moves around from sample to sample: The parameter is fixed at 0.60. The intervals are what change from sample to sample.
It is wrong: this interval had a 5% chance of being built: The 5% is the long-run share of intervals that miss, not the chance of building one particular interval.

Table

A later school start time

A school board says that more than 60% of the district's parents support starting high school later. A local paper selects a random sample of 500 of the district's 9,000 parents, and 330 of them ( $\hat{p} = 0.66$ ) support the change. The conditions for inference are met.

Confidence intervals for the proportion of all the district's parents who support a later start
Confidence level	Interval
95%	from 0.618 to 0.702
99%	from 0.605 to 0.715

4. A teacher claims that 70% of the district's parents support the later start. What does the 95% interval say about this claim?

0.70 is plausible, so there is not convincing evidence against the claim
0.70 is in the interval, so the interval shows that 70% of parents support the change
0.70 is far from 0.66, so the interval gives convincing evidence against the claim
Nothing, because the interval was built for the board's claim of 60%

Show the answer

0.70 is inside the interval from 0.618 to 0.702, so it is a plausible value. The interval does not give convincing evidence that the true proportion is different from 0.70; it does not prove that it equals 0.70 either.

Correct: 0.70 is plausible, so there is not convincing evidence against the claim: Right: a value inside the interval cannot be ruled out.
0.70 is in the interval, so the interval shows that 70% of parents support the change: Many values in the interval are plausible, 0.62 and 0.66 among them. Being plausible is not being proven.
0.70 is far from 0.66, so the interval gives convincing evidence against the claim: Distance from $\hat{p}$ is not the test; 0.70 is inside the interval, so it is plausible.
Nothing, because the interval was built for the board's claim of 60%: An interval gives plausible values for $p$ ; any claimed value can be compared with it.

5. Which is the correct interpretation of the 95% interval?

We are 95% confident that the interval from 0.618 to 0.702 captures the true proportion of all the district's parents who support the change
There is a 95% probability that the proportion of the district's parents who support the change is between 0.618 and 0.702
We are 95% confident that the interval from 0.618 to 0.702 captures the proportion of the 500 parents in the sample who support the change
About 95% of the district's parents support the change at a level between 0.618 and 0.702, as the interval shows

Show the answer

Name the level, the interval and the parameter in context. The interval estimates the proportion for all the district's parents.

Correct: We are 95% confident that the interval from 0.618 to 0.702 captures the true proportion of all the district's parents who support the change: Right: level, interval and parameter in context.
There is a 95% probability that the proportion of the district's parents who support the change is between 0.618 and 0.702: The true proportion is fixed; it is in this interval or not. Confidence describes the method, so say "95% confident".
We are 95% confident that the interval from 0.618 to 0.702 captures the proportion of the 500 parents in the sample who support the change: The sample proportion is known exactly, 0.66. The interval is about the whole district.
About 95% of the district's parents support the change at a level between 0.618 and 0.702, as the interval shows: Each parent either supports the change or not. The interval is about one proportion, not individual parents.

6. Build the interpretation of the 95% confidence level in this context.

Write the conclusion in context.

Show the answer

The level describes what happens over many samples of the same size: the long-run share of intervals that capture the true proportion.

A full-credit answer: If we took many random samples of 500 parents and built a 95% interval from each, about 95% of the intervals would capture the true proportion of all the district's parents who support a later start time.

7. A news report says, "In a random sample of likely voters, 52% favor Measure 4, with a margin of error of 3 percentage points (95% confidence)." Which conclusion is justified?

A majority of likely voters favor Measure 4, because 52% is more than half of them
There is not convincing evidence of a majority, since 0.50 is in the interval
The poll shows that exactly 52% of likely voters favor Measure 4
95% of likely voters favor Measure 4 by 3 percentage points

Show the answer

The interval is $0.52 \pm 0.03$ , from 0.49 to 0.55. It includes 0.50 and values below it, so the poll does not give convincing evidence that a majority of all likely voters favor the measure.

A majority of likely voters favor Measure 4, because 52% is more than half of them: The interval reaches down to 0.49, so a proportion of one half or less is still plausible.
Correct: There is not convincing evidence of a majority, since 0.50 is in the interval: Right: 0.50 is a plausible value.
The poll shows that exactly 52% of likely voters favor Measure 4: The 52% describes the sample; the margin of error says the population value may differ.
95% of likely voters favor Measure 4 by 3 percentage points: That mixes up the confidence level with the result. The 95% describes the method.

Part 9 · Summary

Summary

Interpret an interval with its level, its ends and the parameter in context; interpret the level as the long-run capture rate of the method. Values inside the interval are plausible. An interval gives convincing evidence for a claim like "greater than 0.60" only when every value in it is greater than 0.60; a claimed value inside the interval is plausible, which is not the same as proven.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic