Math151 Day31

Math 151 , Day 31, Monday, April 14, 2008 hit reload.....

HW Day31 (Re)Read Chapter 11 (pp. 286-291 optional). First pp. 271-77 Check p. 294: 11.17, 18 (parameter/statistic, sampling dist.). 11.19, 20 (behavior of xbars, mean & s.d.) 11.22, 23, 24 (behavior of xbars, more) Next: Skip Ch. 12, 13. Do Ch.14 on.
Memorize the 3 yellow-headed boxes on p. 278, 281 (mean and s.d. of sampling dist. of X-bar, Normality & Central Limit Th.)


*Hand in* all. Also complete your HW from Day 30. Try now to do 11.39 and 38 ("backward" problems), but don't lose sleep over them. Be sure you can do the "forward" types (36, 37). These problems use the Central Limit theorem (p. 281) p. 185, 11.10 What does the CLTh say? p. 286 , 11. 12 SAT scores, n = 1 and 70 p. 286, 11.13, insurance (Hint: find P(Xbar> $275)) p. 298, 11.41 auto accidents p. 298, 11.42 airplane overloads (Hint: to do the problem you have to assume all the seats are taken. Maybe not a reasonable assumption, but if there are empty seats, there's likely not a problem with overweight.) A. (preliminary for Ch. 14) Get 4 slips from the Birkenstock box (if you didn't Friday). Record them, return them. HW: Find their mean xbar. Now xbar is your "point estimate" of the unknown mean of the numbers in the box. Calculate xbar - .841, xbar +.841. This is your "point estimate" plus or minus a "margin of error" of .841. (xbar - .841, xbar +.841) is your "interval estimate" for the unknown mean of the box. Be ready to add these to the list on WED, if you haven't.	Read, to discuss	Optional

Exams returned last time: Solutions

Comments: Reading carefully. Knowing what words mean precisely. Writing clearly. Sampling vocabulary:
especially sampling frame: Actual list from which you draw your sample. Ideally would be whole population.
2Bb--I was trying to asking what variable might be confounded with wearing/not wearing orange, so that just making it the law might not make that much difference to the death rate. I was thinking that the careful well trained hunters all probably already wearing orange, and even if some of the others are forced to wear orange, they won't become generally more careful.   (In fact, the proportion of NY hunters wearing orange is about the same as that in states where it's the law. Pataki vetoed the bill requiring it, on the grounds that if it was the law that people had to, people who didn't would be more likely to get shot because hunters wouldn't watch out for them.)

If you haven't: Get 4 slips from the Birkenstock box (outside my door if you missed class). Record them, return them. HW: Find their mean xbar. Now xbar is your "point estimate" of the unknown mean of the numbers in the box.
Calculate xbar - .841, xbar +.841. This is your "point estimate" plus or minus a "margin of error" of .841.
     (xbar - .841, xbar +.841) is your "interval estimate" for the unknown mean of the box.

Closed book Quiz Wednesday: Like this: The population has mean 125 and standard deviation 18.
You take a simple random sample of size 9. The distribution of all possible sample means from such samples has
mean _____ and standard deviation______
Answers: Mean is 125,
standard deviation is 18 divided by the square root of 9.   Square root of 9 is 3, so standard deviation is 18/3 = 6.
that's all.
<>~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
RECAP: What is the distribution of the random variable Xbar, when the experiment is to take a simple random sample of size n? This is the distribution of means of all possible SRS's of size n.
We'll call it the "sampling distribution of the (sample) mean" (p. 275-7, then details 278-86)

Whatever the population distribution of X, that we draw the sample from, (see p. 278)
(as long as the population is large compared to the sample (at least 10 to 20 times sample)
The mean of the x-bars = the mean of the population
The standard deviation of the x-bars =
the s. d. of the population divided by the square root of n.
IF the population is Normal, the sampling distribution of Xbar is Normal.
"The Central Limit Theorem"
In any case, for "large" n, the sampling distribution of Xbar is Approximately Normal.

We talked about the above Friday. HW questions Day 30?
B) "Normal" body temperature 98.6 deg. on average. (Assume this is true.)
Assume normal distribution, & s.d.among many people is 0.6.
   Probability that one (random) healthy individual's normal temperature is above 98.8?
   Probability that the mean of a sample of n is above 98.8?
All of these are P(Xbar>98.8) for different sample sizes n. (Normal table A)

Sample size n	s.d. of Xbars = (pop.s.d.)/sqrt(n)	z = (raw-mean)/s.d.	P(Xbar>98.8)= P(Z>z)
1	.6/1 = .6	(98.8-98.6)/.6 = .2/.6 = .33	P(Z>.33) = .3707
4	.6/2 = .3	(98.8-98.6)/.3 = .2/.3 = .67	P(Z>.67) = .2514
36	.6/6 = .1	(98.8-98.6)/.1 = .2/.1 = 2	P(Z>2) = .0228
100	.6/10 = .06	(98.8-98.6)/.06 = .2/.06 = 3.33	P(Z>3.33) = .0004

Excel pictures for these Xbar distributions. Normal X vs Xbar

Today, the Central Limit Theorem Day 29, details

Central Limit Theorem...
How large is "large"? How approximate is "approximate"?
If the population was close to normal, n doesn't need to be very large.
If the population is not badly skewed or bimodal, n=25 already gives a pretty good approximation to normal.

What if the population is not 10 to 20 times the sample size? The real s.d. of the x-bars will be narrower than the rule above. You may not "get to" normal as a shape. Sample of 4 grades from a population of 10: last semester All possible samples. Will show these Wed.


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