How can we numerically summarize a single continuous variable?
When is the mean
an appropriate summary measure to calculate?
What assumptions need to be true in order to use a mean to represent your single continuous variable?
Application Exercise
Today we are going to work with letter beads.
We have two sets of letter beads, which is a better deal in terms of common letter frequency?
Each set has over 1,000 beads, instead of counting them all, you will each select a sample of 20.
How can I calculate the average frequency score in my sample?
\[ \Large\bar{y} =\sum_{i=1}^n \frac{y_i}{n} \]
How can I calculate the average frequency score in my sample?
\[ \Large\bar{y} =\sum_{i=1}^{20} \frac{y_i}{20} \]
What if I want to know the average frequency score for the whole population of beads?
How can we quantify how much we’d expect the mean to differ from one random sample to another?
How can we quantify how much we’d expect the mean to differ from one random sample to another?
\[ \Large\frac{s}{\sqrt{n}}\] . . .
where \(s\) is the sample standard deviation of \(y\).
How can we quantify how much we’d expect the mean to differ from one random sample to another?
\[ \Large\frac{s}{\sqrt{20}}\]
where \(s\) is the sample standard deviation of \(y\).
\[ \Large s = \sqrt{\frac{\sum_{i=1}^n (y_i - \bar{y})^2}{n-1}} \]
\[ \Large s = \sqrt{\frac{\sum_{i=1}^n (y_i - \bar{y})^2}{20-1}} \]
If we use the same sampling method to select different samples and computed an interval estimate for each sample, we would expect the true population parameter (the average frequency score) to fall within the interval estimates 95% of the time.
\[\bar{y} \pm t^∗ \times SE_{\bar{y}}\]
Why 0.025?
Why lower.tail = FALSE
?
If we use the same sampling method to select different samples and computed an interval estimate for each sample, we would expect the true population parameter (the mean) to fall within the interval estimates 95% of the time.
Application Exercise
Make sure to use the standard error in your confidence interval, this is the standard deviation divided \(\sqrt{20}\)
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