Task Sets: Estimate population parameters with margin of error - Common Core Math - HS Statistics and Probability

Recall / Warm-Up

A random sample of 80 town workers has a mean commute of $\bar{x} = 27$ minutes.
What is the point estimate of the population mean commute $\mu$ for all town workers?

$27$ minutes

$80$ minutes

It cannot be estimated from a sample.

$\mu$ is exactly $27$ minutes for the whole population.

A random sample of 60 students gives $\hat{p} = 0.65$ for the proportion who would re-enroll.
Which statement is the most honest report of what this tells us about the population proportion $p$ ?

Exactly 65% of all students would re-enroll.

$0.65$ is a point estimate of $p$ ; another random sample would likely give a somewhat different value, so the estimate needs a margin.

The sample is wrong because it did not survey every student.

$p$ must be between $0$ and $0.65$ .

The simulated sampling distribution of $\hat{p}$ for a survey has a standard deviation of about
$0.05$ . The margin of error is about two standard deviations. What is the margin of error?

Fluency Practice

A pollster reports " $52\%$ , with a margin of error of $\pm 3$ percentage points." A student
says, "That means the pollster might have made mistakes recording the answers." What is the
best correction?

The student is right -- margin of error measures recording and measurement mistakes.

The margin of error measures how much the estimate would move across different random samples (sampling variability), even with perfectly collected data.

The margin of error guarantees the true value is exactly $52\%$ .

The margin of error means the survey is unreliable and should be discarded.

A survey of $n = 60$ students gives $\hat{p} = 0.65$ . We do not know the true proportion $p$ ,
so we cannot simulate from it directly. Describe a simulated trial we can run to build the
sampling distribution of $\hat{p}$ , and state what the center and the spread of that
distribution each tell us.

The dot plot below shows about 200 simulated values of $\hat{p}$ for a re-enrollment survey
( $n = 60$ ). The estimates center at $\hat{p} = 0.65$ , and the shaded middle-95% band runs
from $0.53$ to $0.77$ . Using the rule "margin of error = about two standard deviations =
half the width of the middle-95% band," what is the margin of error?

A survey gives $\hat{p} = 0.65$ and the simulated sampling distribution has standard
deviation about $0.06$ . Compute the margin of error and write the interval estimate.
Margin of error $=$ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ . Interval estimate: from ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ to ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ .

margin of error:

lower endpoint:

upper endpoint:

A random sample of town workers gives a mean commute $\bar{x} = 27$ minutes. The simulated
sampling distribution of $\bar{x}$ has standard deviation about $1.5$ minutes. What is the
margin of error, in minutes? (margin of error = about two standard deviations)

Exercises: Estimate a Population Mean or Proportion with a Margin of Error

Recall / Warm-Up

Fluency Practice