• The \(t\) distribution is similar to the Normal\((0, 1)\)
• The \(t\) has more probability in the tails
• As the degrees of freedom increases, the \(t\) becomes more like a Normal\((0, 1)\)

• The degrees of freedom for the \(t\) distribution is \(n - 1\) because we use a denominator of \(n-1\) in the sample standard deviation/variance: \[s^2 = \frac{\sum_i (x_i - \bar{x})^2}{n - 1} \qquad \qquad s = \sqrt{\frac{\sum_i (x_i - \bar{x})^2}{n - 1}}\]
• We use \(n -1\) so that the average value of \(s^2\) is \(\sigma^2\)
• For example, suppose \(\sigma^2 = 1\):

The U.S. Bureau of Transportation Statistics reported the percentage of flights that were delayed each month from 1994 through October of 2013 (238 months in total). Treat these as a representative sample of all months. Here's a histogram:

• Would it be appropriate to use these data to calculate a confidence interval for the mean percent of flights that are delayed per month?
  • Symmetric? Skewed a little to the right, but not too badly
  • Unimodal? Yes
  • Sample Size? The sample size is fairly large, so a normal model for the sample mean is appropriate

• Calculate a 95% confidence interval for the mean percent of flights that are delayed per month

t.test(delays$delayed_pct, conf.level = 0.95)

## 
##  One Sample t-test
## 
## data:  delays$delayed_pct
## t = 71.31, df = 237, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  19.20733 20.29872
## sample estimates:
## mean of x 
##  19.75303

• Calculate a 95% confidence interval for the mean percent of flights that are delayed per month

n <- nrow(delays) # 238 observations
sample_mean <- mean(delays$delayed_pct) # sample mean = 19.75
sample_sd <- sd(delays$delayed_pct) # sample standard deviation = 4.27
mean_se <- sample_sd / sqrt(n) # standard error of sample mean = 0.28
t_critical <- qt(0.975, df = n - 1) # critical value: use .975 for a 95% CI!

sample_mean - t_critical * mean_se # lower CI bound

## [1] 19.20733

sample_mean + t_critical * mean_se # upper CI bound

## [1] 20.29872

• Interpret the 95% confidence interval in context

We are 95% confident that the mean percent of flights that are delayed per month is between 19.2% and 20.3%. If we took a lot of different samples of months and computed different confidence intervals from each, we would expect about 95% of the resulting intervals to contain the mean percent of flights that are delayed per month.