This section is a revisit of what we learned in Chapter 5.

• Parameter \(p\) and point estimate \(\hat{p}\)

• Central Limit Theorem for sample proportions

• Parameter \(p\) and point estimate \(\hat{p}\)

• Standard Error

• Margin of Error (ME)

• Confidence Interval (CI)

• Hypothesis Testing (HT)

• Sample Size

• Parameter of interest: Proportion of all Americans who have good intuition about experimental design.

  \[\mathbf{p} \ \text{(a population proportion)}\]

• Point estimate: Proportion of sampled Americans who have good intuition about experimental design.

  \[\hat{\mathbf{p}} \ \text{(a sample proportion)}\]

• We can estimate a population proportion using a sample and \(\color{blue}{confidence \hspace{0.2cm}interval}\), which we know is always of the form

\[{\textbf{point estimate} \mathbf{\pm} \textbf{ME}}\]

• The margin of error is

  \(\color{blue}{ME = critical\hspace{0.2cm} value \times standard \hspace{0.2cm} error}\) of the point estimate

• The point estimate for proportions is \(\color{blue}{\hat{p}}\)

• The critical value depends on the confidence level

• Standard error of a sample proportion \(\color{blue}{SE_{\hat{p}}= \sqrt{\frac{p(1-p)}{n}} \sim \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}\)

Given: \(n = 670, \hat{p} = 0.85\). First, check conditions.

• Independence: The sample is random, and 670 < 10% of all Americans, therefore we can assume that one respondent’s response is independent of another.

• Success-Failure: 571 people answered correctly (successes) and 99 answered incorrectly (failures), both are greater than 10.

We are given that \(n = 670, \hspace{0.2cm} \hat{p}= 0.85\), the standard error of the sample proportion is \(SE = \sqrt{\frac{p(1-p)}{n}} \sim \sqrt{\frac{\hat{p}(1-\hat{p)}}{n}}\). Which of the below is the correct calculation of the 95% confidence interval?

1. \({0.85} \pm 1.96 \times \sqrt{\frac{0.85 \times 0.15}{670}} \rightarrow \color{red}{(0.82, 0.88)}\)

2. \(0.85 \pm 1.65 \times \sqrt{\frac{0.85 \times 0.15}{670}}\)

3. \(0.85 \pm 1.96 \times \frac{0.85 \times 0.15}{\sqrt{670}}\)

4. \(571 \pm 1.96 \times \sqrt{\frac{571 \times 99}{670}}\)

How many people should you sample in order to cut the margin of error of a 95% confidence interval down to 1%?

\[ME = z_{\frac{\alpha}{2}} \times SE\] \[ \begin{align*} 0.01 &\geq 1.96 \times \sqrt{\frac{0.85 \times 0.15}{n}}\color{red}{\rightarrow \text{Use } \hat{p} \text{ from previous study}}\\ 0.01^2 &\geq 1.96^2 \times \frac{0.85 \times 0.15}{n}\\ n &\geq \frac{1.96^2 \times 0.85 \times 0.15}{0.01^2}\\ n &\geq 4898.04 \color{red}{\rightarrow \text{n should be at least 4,899}} \end{align*} \]

Use \[M.E = z_{\frac{\alpha}{2}} \sqrt{\frac{p(1-p)}{n}} \rightarrow \frac{M.E}{z_{\frac{\alpha}{2}}} = \sqrt{\frac{p(1-p)}{n}}\]

\[ n \ge {\frac{p(1-p)z^2_{{\alpha}/2}}{(M.E)^2}}\] (i.e., take the ceiling integer)

• Success-Failure conditions:

  • CI: At least 10 observed successes and failures.
  • HT: At least 10 expected successes and failures, calculated using the null value \(p_0\)

• Standard error:

  • CI: Calculate using observed sample proportion: \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\).

  • HT: Calculate using the null value: \(SE = \sqrt{\frac{p_0(1-p_0)}{n}}\)

Since the p-value is low, we reject \(H_0\). The data provide convincing evidence that more than 80% of Americans have a good intuition on experimental design.

• Popular parameter: \(p\). point estimate: \(\hat{p}\)

• Conditions:

  • Independence - Random sample and 10% condition
  • At least 10 successes and failures

• Standard error: \(SE = \sqrt{\frac{p(1-p)}{n}}\)

  • for CI: use \(\hat{p}\)
  • for HT: use \(p_0\)