Calculate the mean and standard deviation of ˆp, and plot its sampling distribution I n = 100 I n = 400 I n = 1600

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Calculate the mean and standard deviation of ˆp, and plot its sampling
distribution
I n = 100
I n = 400
I n = 1600

- Mean of \( \hat{p} = p \)

- Standard deviation of \( \hat{p} = \sqrt{\frac{p(1-p)}{n}} \)

- \( \hat{p} \sim N \left( p, \sqrt{\frac{(1-p)p}{n}} \right) \)

- Back to our example:
  - Suppose \( p = 0.40 \) prefer Biden, and \( n = 25 \)
  - Mean of \( \hat{p} = p = 0.4 \)
  - Standard deviation of \( \hat{p} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4(1-0.4)}{25}} = 0.098 \)

- Let's plot the sampling distribution of \( \hat{p} \):
Transcribed Image Text:- Mean of \( \hat{p} = p \) - Standard deviation of \( \hat{p} = \sqrt{\frac{p(1-p)}{n}} \) - \( \hat{p} \sim N \left( p, \sqrt{\frac{(1-p)p}{n}} \right) \) - Back to our example: - Suppose \( p = 0.40 \) prefer Biden, and \( n = 25 \) - Mean of \( \hat{p} = p = 0.4 \) - Standard deviation of \( \hat{p} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.4(1-0.4)}{25}} = 0.098 \) - Let's plot the sampling distribution of \( \hat{p} \):
Suppose the share of the U.S. population that favors Biden is \( p \).

But the population parameter \( p \) is unknown, so we draw a random sample of size \( n \) from the population to estimate \( p \).

Each draw has a probability \( p \) of picking a voter who favors Biden, and a \( (1 - p) \) chance of picking a voter who favors Trump.

Our random sample of \( n \) can be considered a sequence of \( n \) independent observations with probability \( p \) of "success."

- If we have \( x \) successes—if \( x \) out of \( n \) in our sample supports Biden—the estimated fraction of Biden supporters is \( \hat{p} = x/n \).
- Note that \( \hat{p} \) is just the sample mean—the average number of successes, since \( X = 1 \) if they vote for Biden, 0 otherwise.

Share of intention of voters who favor Biden in a sample: \( \hat{p} \).

Share of intention of voters who favor Biden in the population \( p \).
Transcribed Image Text:Suppose the share of the U.S. population that favors Biden is \( p \). But the population parameter \( p \) is unknown, so we draw a random sample of size \( n \) from the population to estimate \( p \). Each draw has a probability \( p \) of picking a voter who favors Biden, and a \( (1 - p) \) chance of picking a voter who favors Trump. Our random sample of \( n \) can be considered a sequence of \( n \) independent observations with probability \( p \) of "success." - If we have \( x \) successes—if \( x \) out of \( n \) in our sample supports Biden—the estimated fraction of Biden supporters is \( \hat{p} = x/n \). - Note that \( \hat{p} \) is just the sample mean—the average number of successes, since \( X = 1 \) if they vote for Biden, 0 otherwise. Share of intention of voters who favor Biden in a sample: \( \hat{p} \). Share of intention of voters who favor Biden in the population \( p \).
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