17 Day 17

Review

Sampling Distribution

Given any population \(Y \sim N(\mu,\sigma^2)\)

  • Sample \(X \sim N(\mu_X,\sigma^2_X)\)

  • Sample mean \(\bar{x} \sim N(\mu_{\bar{x}},\sigma^2_{\bar{x}})\)

    • Where: \(\mu_{\bar{x}} = \mu\)

    • And: \({\sqrt{\sigma^2_{\bar{x}}}}=\sigma_{\bar{x}} = {\sigma \over \sqrt{n}}\)


Central Limit Theorem

  • Let \(\bar{x}\) be the mean of a large random sample (\(n > 30\)) from any population

    • With mean \(\mu\) and standard deviation \(\sigma\)

  • The distribution of \(\bar{x}\) is approximately normal

    • Mean \(\mu_{\bar{x}} = \mu\)

    • Standard deviation \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\).


If \(n\) is large enough, we have:

\[\bar{x} \sim N(\mu, \frac{\sigma^2}{n})\]

  • Regardless of the original population’s distribution


How large does \(n\) need to be?


Population Proportion

  • Proportions are just percentages of the population


  • Say the percentage of the population who participate in early voting is \(40\%\)

    • \({40\over 100}=0.40\)

    • The proportion of the population who early vote, \(p=0.40\)

  • If we poll a sample of 100 Manhattan residents and find that \(31\%\) early vote:

    • The proportion of our sample who early vote, \(\hat{p}=0.31\)


Just like every other statistic, sample proportions are random variables

  • So their distribution is the sampling distribution of the proportion


All of our previous rules and ideas apply

  • As we take samples from our population we will see they aren’t consistent

  • The more we sample the closer we get to true values


  • Mean of the sample proportion \(\hat{p}\) is:

\[\mu_{\hat{p}} = p \quad \text{(population proportion)}\]

  • Standard deviation of sample proportion \(\hat{p}\) is:

\[\sigma_{\hat{p}} = \sqrt{\frac{p(1 - p)}{n}}\]


Proportion Central Limit Theorem

  • If \(np \geq 10\) and \(n(1 - p) \geq 10\)

  • Distribution of \(\hat{p}\) is approximately normal

    • Mean \(\mu_{\hat{p}} = p\)

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

So:

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


  • Point estimates are a deterministic result

    • Statistics deals with probabilistic results


Confidence Intervals

  • Since: the value of \(\bar{x}\) varies with each sample

    • We need to quantify the uncertainty associated with \(\bar{x}\)


Example:

A random sample of \(120\) students admitted to top business schools yielded an average GPA of \(3.45\)

\[\bar{x} = 3.45\] This is a point estimate of \(\mu\)

  • One number, no additional information provided



A confidence interval (CI) provides a range of values that contains:

  • The population parameter

  • With a certain level of confidence

    • We refer to this as the confidence level


Formula for the CI:

\[\text{Point estimate} \pm \text{Margin of Error}\]

  • The confidence interval for \(\mu\):

\[ \bar{x} \pm \text{Margin of Error} \]


\[ (\bar{x} - \text{Margin of Error}, \bar{x} + \text{Margin of Error}) \]


Margin of error

The farthest distance we believe our estimate \(\bar{x}\) is from \(\mu\)

  • The size of the margin of error is determined by the sampling distribution of \(\bar{x}\) and the confidence level


  • Confidence level is denoted by \(100(1 - \alpha)\%\)

    • Typically \(90\%\), \(95\%\), or \(99\%\)


For a population with unknown \(\mu\) but known \(\sigma\), a \(100(1 - \alpha)\%\) confidence interval for \(\mu\) is computed as:

\[\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\]

Where \(z_{\alpha/2}\) is the z-score with an area of \(\alpha/2\) to its right


When construction a confidence interval for \(\mu\)

  • We have to consider our assumptions


At least one of the following must hold:

  1. The sample size is large (\(n > 30\))

  2. The original population is normally distributed


In most practical cases, \(\sigma\) is unknown, and we must use the sample standard deviation \(s\)

The formula for the confidence interval is:

\[\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]

Where \(t_{\alpha/2}\) is the critical value from the Student’s t-distribution, and \(s\) is the sample standard deviation


Student’s t-Distribution

  • The (Student’s) t-distribution is similar to the standard normal distribution

    • Unimodal

    • Symmetric around \(0\)

  • But it has wider (or heavier) tails than the standard normal

    • Meaning it’s more spread out

  • The t-distribution is distinguished by degrees of freedom (\(df = n - 1\))

    • As \(df\) increase the t-distribution converges to a normal distribution


The critical value \(t_{\alpha/2}\) is a \(t\) value separating an area of \(\alpha/2\) in the right tail of the \(t\) distribution

When using the \(t\) distribution ot construct a confidence interval for \(\mu\):

  • Degrees of freedom (\(df\)) is \(1\) less than the sample size


Example 1: Finding Critical Value

Find the critical value \(t_{\alpha/2}\) for a 95% confidence interval with \(n = 8\)

  • Set \(1 - \alpha = 0.95\), then \(\alpha = 0.05\), and \(\alpha/2 = 0.025\)

  • For \(n = 8 \Rightarrow df = n-1 = 7\)

The critical value is \(t_{\alpha/2} = 2.365\)


What is the \(df\) I’m looking for isn’t in the table?

  • Round down to the nearest value on the table

    • If \(df=59\), round down to \(df=50\)

    • At \(95\%\) confidence, \(t_{\alpha/2}=2.009\)


Summary of CI for Population Mean \(\mu\):

Check your assumptions for construction a CI of \(\mu\):

  • Sample size is large (\(n>30\)) or the population is normal


\(100(1-\alpha)\%\) confidence interval is computed as:

  • Case 1: \(\sigma\) is known, use the z-method:

    \[\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}\]

  • Case 2: \(\sigma\) is unknown, use the t-method:

    \[\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\]


Example 2: Constructing a CI

Given a sample of size \(n = 5\) from a normal population, \(\bar{x} = 4.31\), and \(s = 2.7\), construct a 95% confidence interval for \(\mu\)


  1. Should we use \(z\) method or \(t\) method?
  • \(\sigma\) is unknown


  1. Compute the margin of error for this \(95\%\) confidence interval:
  • With \(df = 4\) and \(t_{\alpha/2} = 2.776\), calculate:

\[\text{Margin of Error} = 2.776 \times \frac{2.7}{\sqrt{5}} \approx 3.352\]


  1. Construct a \(95\%\) confidence interval for \(\mu\) and interpret your result:

\[4.31 \pm 3.352 \quad \text{or} \quad (0.958, 7.662)\]

We are 95% confident that the true population mean lies between 0.958 and 7.662


  1. If the population were not normal, would the confidence interval in (c) be valid?


Interpreting a CI

  • Suppose we take many random samples and construct a \(95\%\) confidence interval from each sample

    • \(95\%\) of those intervals would contain the true population mean, \(\mu\)


In practice:

  • We say that we’re \(95\%\) confident that the true value of \(\mu\) is within our confidence interval