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STAT 120

Bastola

CLT: when n is big enough, means and proportions behave like a normal distribution.

- Today we will compute SE using formulas derived from probability theory
- The inference methods in ch. 6+ are “classical” methods that could be done just with pen and paper.

- Resampling methods are intuitive and don’t require lots of statistical theory/background.
- But in your research fields you will likely only see classical methods used

- In the “olden days”, classical methods were the only thing taught in stats methods classes.
- More advanced methods usually do rely on classical theory due to their complexity.

The Central Limit Theorem applies to the distribution of the

- statistic
- parameter
- null value
- data
- standard error

The standard error for \(\hat{p}\) is \[\begin{align*} S E_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \end{align*}\]

The larger the sample size (n), the smaller the SE

For a sufficiently large sample size, the distribution of sample statistics for a mean or a proportion is normal

- One sample proportion] The sampling distribution for a sample proportion is approximately normally distributed:

\[\begin{align*} \hat{p} \approx N\left(p, \sqrt{\frac{p(1-p)}{n}}\right) \end{align*}\]

Need n large enough so np ≥ 10 and n(1 – p) ≥ 10

President Biden won 52.4% of the popular vote in Minnesota in the 2020 election.

- If we had sampled 100 likely voters just prior to the election, what would be the SE for the sample proportion of voters for Biden?

\[\begin{align*} S E=\sqrt{\frac{0.524 \times 0.476}{100}} \approx 0.05 \end{align*}\]

For a single proportion, what is the margin of error (ME)?

\[\begin{align*} \hat{p} \pm z^* \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \end{align*}\]

- \(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
- \(z^* \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
- \(2 \times z^* \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)

\[\begin{align*} M E=z^* \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \end{align*}\]

You can choose your sample size in advance, depending on your desired margin of error!

Given the formula for margin of error, solve for n.

Neither \(p\) nor \(\hat{p}\) is known in advance. To be conservative, use \(p=0.5\). For a \(95 \%\) confidence interval, \(z^* \approx 2\)

\[\begin{align*} n=\left(\frac{z^*}{M E}\right)^2 \hat{p}(1-\hat{p}) \qquad \Longleftrightarrow \qquad n \approx \frac{1}{M E^2} \end{align*}\]

**Maximized at p = 0.5**

When our sample suggests an even split,

there’s more room for variability,

leading to a larger ME

Suppose we want to estimate a proportion with a margin of error of 0.03 with 95% confidence. How large a sample size do we need?

- About 100
- About 500
- About 1000
- About 5000

What should n be to get a margin of error of 3%?

\[\begin{gathered} 0.03=2 \times \text { SE } \\ 0.015=S E=\sqrt{\frac{0.482 \times 0.518}{n}} \\ n=\frac{0.524 \times 0.476}{0.015^2} \approx 1109 \end{gathered}\]\[\begin{align*} \mathrm{H}_0: & \quad p=p_0\\ \mathrm{H}_A: & \quad p\neq p_0 \end{align*}\]

\[\begin{align*} z&=\frac{\hat{p} -p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}} \end{align*}\]

If \(np_0 ≥ 10\) and \(n(1 – p_0) ≥ 10\), then the p-value can be computed as the area in the tail(s) of a standard normal beyond z.

A survey on 2,251 randomly selected individuals conducted in October 2010 found that 1328 answered “Yes” to the question. Do a majority of Americans believe in global warming?

\[\begin{aligned} & H_0: p=0.50 \\ & H_A: p>0.50 \end{aligned}\]\[p=\text { proportion of all Americans who believe in global warming }\]

“Is there solid evidence of global warming?”

Source: “Wide Partisan Divide Over Global Warming”, Pew Research Center, 10/27/10.s

Sample proportion: \[\begin{align*} \hat{p}=\frac{1328}{2251}=0.590 \end{align*}\]

Standardized test stat: \[\begin{align*} z=\frac{0.590-0.50}{\sqrt{\frac{0.50(0.50)}{2251}}}=\frac{0.09}{0.0105}=8.54 \end{align*}\]

P-value: proportion above z=8.54 on a N(0,1) curve. Yes, there is statistically discernible evidence that the percentage of Americans that believe in global warming is greater than 50% (z=8.51, \(p \approx 0\)).

We are 95% confident that between 57% and 61% of Americans believe in global warming.

** We are 95% sure that the true percentage of all Americans that believe there is solid evidence of global warming is between 57.0% and 61.0%.**

Standard error for a sample proportion: Central Limit Theorem for a proportion: If counts for each category are at least 10 (meaning \(n p \geq 10\) and \(n(1-p) \geq 10)\), then

- For a \(\mathrm{CI}\), use \(p\)-hat in place of \(p\)
- For a Hypothesis Test, use \(p_0\) in place of \(p\) when calculating the standardized statistic

- Please download the Class-Activity-17 template from moodle and go to class helper web page

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