Confidence Intervals for a Mean
- Construct a one-sample t-interval for a population mean
- Verify the conditions for t procedures, including the Normal/large-sample condition
- Interpret the interval in context
The one-sample t-interval
A confidence interval for a mean has the familiar form *x-bar ± t\ · (s/sqrt(n))*, where t\ is the critical value from the t-distribution with n − 1 degrees of freedom. The quantity s/sqrt(n) is the standard error of the mean, and t\*·(s/sqrt(n)) is the margin of error.
Conditions for t procedures
Three conditions: Random (data from an SRS or randomized experiment), 10% (n ≤ 0.10N when sampling without replacement), and Normal/Large Sample. The last is met if the population is Normal, or n ≥ 30 (CLT), or — for small samples — a graph of the data shows no strong skew or outliers. Always address the shape condition, especially when n is small.
A random sample of 25 batteries has x-bar = 40 hours and s = 5 hours. Construct a 95% confidence interval for the mean battery life (t\* = 2.064 for df = 24).
- 1.Conditions: random sample; assume batteries < 10% of all produced; n = 25, so check that a graph shows no strong skew/outliers (assume roughly symmetric).
- 2.Standard error: s/sqrt(n) = 5/sqrt(25) = 5/5 = 1.
- 3.Margin of error: t\*·SE = 2.064 × 1 = 2.064.
- 4.Interval: 40 ± 2.064 = (37.94, 42.06).
For a one-sample t-interval based on n = 20 observations, how many degrees of freedom are used to find t\*?
For small samples, you must address the shape condition explicitly: state that a dotplot/boxplot/stemplot of the data shows no strong skew or outliers, so it is plausible the population is approximately Normal. Skipping this on free response loses the conditions point.
Holding everything else fixed, how does a 99% t-interval compare to a 95% t-interval for the same data?
Answer the 2 checkpoints as you read.
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