Notes
Sampling distributions
- A statistic varies from sample to sample (sampling variability). The sampling distribution shows how.
- For a proportion: $\mu_{\hat{p}} = p$ and $\displaystyle \sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}$. For a mean: $\mu_{\bar{x}} = \mu$ and $\displaystyle \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$.
- Larger $n$ means less variability. By the Central Limit Theorem, $\bar{x}$ is approximately normal for large $n$.
Confidence intervals
- Interval $=$ statistic $\pm$ margin of error, with margin of error $= z^* \cdot \text{SE}$. For $95\%$ confidence use $z^* \approx 2$.
- For a proportion, $\displaystyle \text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Conditions: random sample and at least 10 successes and 10 failures.
Interpreting
- "We are $C\%$ confident that the true [parameter in context] is between ... and ...". The confidence level is the long-run success rate of the method.
- Higher confidence gives a wider interval. Larger $n$ gives a narrower interval.
Topics in this unit
Each topic has its own short notes and practice.
Practice
A statistic naturally changes from sample to sample.
$\displaystyle \sqrt{\frac{0.2 \cdot 0.8}{100}} = \sqrt{0.0016} = 0.04$.
$\displaystyle \frac{12}{\sqrt{16}} = \frac{12}{4} = 3$.
$\displaystyle \text{SE} = \sqrt{\frac{0.25}{100}} = 0.05$ and margin $= 2(0.05) = 0.10$.
The interval is a statement about the population mean, not individuals.
Quadrupling $n$ cuts SE, and therefore the margin of error, in half.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A random sample of 400 students from a large school district is asked whether they prefer a later start time. 320 say yes.
(a) Find the sample proportion and verify the sample is large enough to use a normal model (at least 10 successes and 10 failures).
Show the answer and scoring guide
$\displaystyle \hat{p} = \frac{320}{400} = 0.8$. There are 320 successes and 80 failures, both at least 10.
Check off the points you earned:
(b) Find the standard error and a $95\%$ confidence interval using $z^* = 2$.
Show the answer and scoring guide
$\displaystyle \text{SE} = \sqrt{\frac{0.8 \cdot 0.2}{400}} = 0.02$. Margin of error $= 2(0.02) = 0.04$. Interval: $0.76$ to $0.84$.
Check off the points you earned:
(c) Interpret the interval in context. Would a $99\%$ interval from the same data be wider or narrower?
Show the answer and scoring guide
We are $95\%$ confident that the true proportion of students in the district who prefer a later start is between $0.76$ and $0.84$. A $99\%$ interval would be wider, because a higher confidence level needs a larger $z^*$.
Check off the points you earned: