Recall our procedure for hypothesis testing:

1. Collect data: for each of 8 trials, was the prediction correct?
2. Calculate a sample statistic (called the test statistic):
   • \(x =\) total number correct (8 in our case)
3. Obtain the sampling distribution of the test statistic, assuming a null hypothesis of no effect (in this case, assuming Paul is just guessing)
4. Calculate the p-value: probability of getting a test statistic "at least as extreme" as what we observed in step 2
5. If the p-value is low, reject the null hypothesis and conclude that Paul is psychic!

1. Simulation:
   • Repeatedly simulate 8 trials with probability of success = 0.5. In each simulation, count the number of successes.
   • As the number of simulations increases, we get a more accurate approximation to the sampling distribution.
2. Probability:
   • Calculate probabilties from the sampling distribution exactly using a \(\text{Binomial}(8, 0.5)\) model

• So, we now have 2 ways to get the sampling distribution for the total number of successes in \(n\) trials!
• Let's discuss sampling distributions for two other common sample statistics:
  • The proportion of successes in \(n\) trials
  • The sample mean of a quantitative variable
• For today (and possibly the rest of this class), we'll just focus on the approach using probability

• If \(Y_1, Y_2, \ldots, Y_n\) are independent observations from a population having mean \(\mu\) and finite standard deviation \(\sigma\), then the sampling distribution of \(\bar{Y} = \frac{1}{n}\sum_{i=1}^n Y_i\) is approximately Normal(\(\mu\), \(\sigma/\sqrt{n}\)) for large enough \(n\).

(On the board: derive the mean and standard deviation of \(\bar{Y}\))

• You will explore the part about an approximately normal distribution in a lab.

• For quantitative variables, there's no rule for how big \(n\) must be – depends on how skewed the distribution is (see lab)

• But remember that we don't want to calculate means anyways if the distribution is very skewed!

• Suppose we want to estimate the proportion \(p\) of US households who own the home they live in.

• We take a sample of size \(n\) and count the number of households in our sample who own their home:

\[X \sim \text{Binomial}(n, p)\]

• How can we estimate \(p\) using \(X\)?

• We will estimate the probability of success using \[\hat{p} = \frac{X}{n}\]

• Remember that we can write \(X\) as a sum of independent Bernoulli Random Variables: \(X = X_1 + X_2 + \cdots + X_n\)

• So \(\hat{p} = \frac{X}{n} = \frac{1}{n} \sum_i X_i\) is a sample mean of independent Bernoulli random variables

• Since \(\hat{p} = \frac{1}{n} \sum_i X_i\), the Central Limit Theorem tells us the approximate sampling distribution of \(\hat{p}\), for large enough \(n\).

• For a single Bernoulli random variable,
  • \(E(X_i) = p\)
  • \(SD(X_i) = \sqrt{p(1 - p)}\)

• The CLT says that for large enough \(n\), the sampling distribution of \(\hat{p}\) is approximately \[\hat{p} \sim \text{Normal}(p, \sqrt{p(1 - p)/n})\]

• For estimating a proportion/probability \(p\), we say \(n\) is large enough if the success/failures condition is satisfied:
  • \(np \geq 10\) and \(n(1 - p) \geq 10\)

• See lab for more