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Lecture

# UASTAT141Ch23.pdf

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University of Alberta

Statistics

STAT141

Paul Cartledge

Winter

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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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