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Lecture

# UASTAT141Ch23.pdf

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School
University of Alberta
Department
Statistics
Course
STAT141
Professor
Paul Cartledge
Semester
Winter

Description
Ch. 23 – CI for a Population Mean - common situation is that σ is unknown, so the sample data must estimate it. Recall Z = Y − µ . We now have Z ≠ Y − µ = t , where t is a diff. standardized variable. σ / n s/ n The value of s may not be all that close to σ, especially when n is small. Consequently, there is extra variability and the distribution of t is more spread out than the z curve. t-distributions: As with normal curves, there exists a family of t-curves. The normal distribution has 2 parameters: µ and σ; the t-distribution has a single parameter: degrees of freedom (df). Range of t: similar to range of z; range of df: 1, 2, 3, …, ∞ Properties of the t-distribution: 1. The t-curve, with any fixed df, is bell-shaped and centered at 0 (just like z-curve). 2. Each t-curve is more spread out than the z-curve.α/2> zα/2 3. As df increases, the spread of corresponding t-curve decreases. 4. As df increases, the corresponding sequence of t-curves approaches the z-curve. Let y1, 2 , …, n constitute a random sample from a normal population distribution. Then the probability distribution of the standardized variable Y − µ t = ~ tdf (df = n – 1) s/ n One-Sample t Confidence Interval: Assumptions: 1. Sample is random. 2. The population distribution is normal OR the sample size is large (n ≥ 30). 3. σ, the population standard deviation, is unknown. ⎛ s ⎞ y ± (α/2, n – 1⎜ ⎟ ⎝ n⎠ Table T gives critical values appropriate for the most common confidence levels. Sample size (n) n ≥ 30 n < 30 (approx.) normal OK OK Normality NOT normal OK DON’T use t! Ex23.1) A naïve astrophysicist wants to determine the mean radius of all of Jupiter’s numerous moons. Only taking the 4 largest moons, he discovers a sample mean of 2103.75 km with a standard deviation of 495 km. Construct and interpret a 90% confidence interval for the population mean radius of Jupiter’s moons. (Note: changes were made in class to this example to make the assumptions hold.) - Parameter = µ - Sample size: n = 4 - Estimate: y = 2103.75 - Sample standard deviation: s = 495 - Standard Error = s/ n = 495/ 4 = 247.50 - Confidence Level is 90%, so critical value3t 2.353 - Interval is 2103.75 ± (2.353)(247.50) = 2103.75 ± 582.37 Æ (1521.238,6.12) With 90% confidence, the population mean radius of Jupiter’s
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