Lec 20

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 a useful way to interpret information about a population and sample without losing very much nuance at all:


  • Proportions are just percentages of the population


Say the percentage of the population who have experienced flu-like symptoms this month is \(40\%\)


  • \({40\over 100}=0.40\)


  • The proportion of the population who have flu-like symptoms, \(p=0.40\)



If we poll a sample of 100 Manhattan residents and find that \(31\%\) have experienced flu-like symptoms in the past month:


  • The proportion of our sample who experienced flu-like symptoms in our pre-defined time window, \(\hat{p}=0.31\)


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



Thus:


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



Point estimates are a deterministic result


  • Statistics deals with probabilistic results


It would be more informative to provide a range of values



Questions?



Goals for today:

  1. Compute confidence intervals for point estimates arising from Normal data

  2. Discuss interpretations of confidence intervals



Uncertainty I


Confidence Intervals


Since the value of \(\bar{x}\) varies with each sample, we need to quantify the uncertainty associated with \(\bar{x}\)



A random sample of \(120\) students admitted to top veterinary schools yielded an average GPA of \(3.85\)


\[\bar{x} = 3.85\]


This is a point estimate of \(\mu\)


  • One number, no additional information provided



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


  1. The population parameter


  1. A certain level of confidence, called 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 (\(n > 30\))


  1. 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




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 if 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, \(t_{\alpha/2}=2.009\)



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:


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



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



Attendance QOTD



Go away