Unformatted text preview: 1/24/2011 TA Office Hours: SOM 321 B
Day
Time TA
Monday 10 a.m.‐12 p.m. Jon
Monday 2‐3 p.m. Justin Monday 3‐5 p.m. Robin Tuesday 11 a.m.‐1 p.m. Ilya
Tuesday 2‐4 p.m. Chelsea
Wednesday 10 a.m.‐12 p.m. Jon Wednesday 12‐2 p.m. Arjun
Wednesday 2‐3 p.m. Justin Thursday 2‐3 p.m. Chelsea Thursday 3‐5 p.m. Will TA Office Hours: Stockbridge 210
Day
Time TA
Tuesday 4‐6 p.m. Chris
Thursday 4‐6 p.m. Chris Friday Discussions: Sections, Times, Rooms, TAs
Section
Time Room
TA
1 9:05 a.m.‐9:55 a.m. SOM 126 Jon
2 9:05 a.m.‐9:55 a.m. SOM 117 Arjun
3 9:05 a.m.‐9:55 a.m. SOM 123 Justin
4 10:10 a.m.‐11:00 a.m. SOM 118 Arjun
5 10:10 a.m.‐11:00 a.m. SOM 123 Justin
6 10:10 a.m.‐11:00 a.m. SOM 129 Will
7 11:15 a.m.‐12:05 p.m. SOM 26 Ilya
8 11:15 a.m.‐12:05 p.m. SOM 118 Robin
9 11:15 a.m.‐12:05 p.m. SOM 127 Will
10 12:20 p.m.‐1:10 p.m. SOM 119 Chelsea
11 12:20 p.m.‐1:10 p.m. SOM 118 Robin
12 12:20 p.m.‐1:10 p.m. SOM 127 Ilya 1 1/24/2011 On the syllabus, I use the phrase
• “Connect Pre‐lecture Assignment”
• I want to get away from this name.
• I will use a new name in place of “Connect Pre‐lecture Assignment.”
• The new name is just plain, old Homework . Lecture 2: January 24, 2011
• Homework 1 opens at 8:00 p.m. this evening:
–
–
–
– 4 questions: page 307, Problems 8.4 and 8.6
Sampling Distributions and The Central Limit Theorem It closes Wednesday, January 26 at 6:00 p.m.
Unlimited tries until it closes! Feedback! • Non‐Connect Practice problems:
– Pages 306‐307: Problems 8.1 and 8.3
– Answers are in the back of the book
– If you need help with these, see a TA during office hours. • Readings:
– Chapter 8, pages 294‐318
– Chapter 9, pages 340‐365 (One‐Sample Hypothesis Tests) Again … Revised Vocabulary
• Old Name:
– Connect Pre‐lecture Assignment • New Name:
– Homework 2 1/24/2011 Next class: Bring Two Handouts
• z‐table
• Steps in Hypothesis Testing What we did at the end of last class:
• We started looking at the behavior of a sample statisticlike .
X
• A statistic is important because we use it to make inferences.
• It takes time, energy, and money to generate a good statistic.
• We certainly hope that the statistic that we generate mimics that way things really are. • How do we know that it will?
• One thing we know for sure:
– A larger sample size will make the statistic of interest a better representative of truth. • We can get a really good idea about the behavior of the single statistic that we generate by visiting the Central Limit Theorem. Three Important Players in the Central Limit Theorem
• Any sample statistic – like the sample mean – has X
three important characteristics:
– Its central value or central tendency
(Its true – but unknown – center is µ) .
– Its spread or variability (based on other s X
calculated using the same sample size n).
(Its true – but unknown standard deviation is
/ n .
– Its shape or probability distribution 3 1/24/2011 We are now going to do an experiment.
• It will be highly unrealistic.
• We will calculate 10,000 sample means.
• In applied work, we never calculate 10,000 means. We calculate only one mean.
• Q: Why are we doing this, then?
• A: To illustrate the behavior of that single statistic that we do calculate. Where do we get the draws upon which each sample mean is based? • Each mean will be based on draws from a really weird population.
• The population will be extremely • Non‐normal . Example of an “anything but normal” parent population:
The exponential:
• Characteristics
• Formula: P(X) = 1/µ ∙ exp(‐X/µ)
• Values of X are greater than zero.
• µ is the population mean.
• The standard deviation σ equals µ. •Let’s take an example:
• Let µ=4=σ.
• P(X) = ¼ ∙ exp(‐X/4)
• Shape: To the right μ=4 X 4 1/24/2011 Take 10,000 repeated samples, each of size n=2 from our exponential distribution.
Calculate a sample mean (an ) for each sample of size n=2, X
so that we have 10,000 X s: X1 , X 2 ,... , X10,000 • •
•
• • •
X μ=4 X μ=4 X
X1 =
2 X2 = X
2 X μ=4 . . . X10,000 = X
2 Q: How do these 10,000 look when they are plotted?
Xs Take 10,000 repeated samples, each of size n=10 from our exponential distribution.
Calculate a sample mean (an ) for each sample of size n=10, X
so that we have 10,000 X s: X1 , X 2 ,... , X10,000
•
• X2 = X X
10 • •
•
••
•
μ=4 •• • μ=4 . . . X10,000 = • X X
10 • μ=4
X1 = • • • • •• • • •
•• • • ••
•
• X
X
10 Xs
Q: How do these 10,000 look when they are plotted? Take 10,000 repeated samples, each of size n=30 from our exponential distribution.
Calculate a sample mean (an ) for each sample of size n=30, X
so that we have 10,000 X s: X1 , X 2 ,... , X10,000
•
• ••
•
•
• • • ••
•
•
• • ••
•• •
•
•
• ••• • •
μ=4
X
X1 =
30 X •
••
•
•• •
•
••
•
•
• ••
•
••
•
• • • ••
•
••
•
•
μ=4 • •
• • •• •• •
• •• • •
•
• ••
••
• •• •
••
••
•
•
• • X X
X2 =
30 μ=4 . . . X10,000 = X X
30 Xs
Q: How do these 10,000 look when they are plotted? 5 1/24/2011 Behavior of when n=2:
X μX = 4 σX = σ
4
4
=
=
= 2.828
n
2 1.4142 X
• Probability distribution of should be skewed to the right.
X
Behavior of when n=10: μX = 4 σX = σ
4
4
=
=
= 1.2649
n
10 3.1622 X
• Probability distribution of should be less skewed to the right. Behavior of when n=30:
X μX = 4 σX = σ
=
n 4
4
=
= 0.73
30 5.4772 X
• Probability distribution of will be approximately normal. Let’s confirm these numbers with Minitab
• Check out the handout that I asked you to download, titled Minitab Experiment –
Central Limit Theorem.
• Let’s also take a peek at the summarized output of the experiment itself. This is in the handout titled Minitab Experiment Output. 6 ...
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 Spring '08
 KOUZEHKANANI
 Central Limit Theorem, TA, Tier One, Scaled Composites

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