Class Notes (808,131)
Canada (493,086)
Statistics (237)
STAT141 (21)


3 Pages
Unlock Document

University of Alberta
Paul Cartledge

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
More Less

Related notes for STAT141

Log In


Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.