| Hand in: Prep
for 6.1: You took 4 Numbers (random sample) from the
Birkenstock box in class and wrote them down. (Box is outside my
door if you missed this) For HW, Find mean xbar.
Find xbar + .841. This
is your interval estimate of the unknown mean of the box's
population.
("margin of error" is .841) (Return your numbers afterward.) Next class, you will add your values to the circulating list, and graph your interval on the transparency circulating. If xbar = 8.0 7.159|_____________8.0_____________|8.841 Sec. 5.2 5.29 axle diameter (mean, s.d.) 5.31 abc axle diameter (cf. X and Xbar) 5.35 mean number of moths in a trap 5.41 airplane weight (They suggest changing it into a "mean" problem, but you can do it as a "sum" problem using the algebra of means and variances and noticing that if Xbar is normal, so is the sum of the Xi's) 5.37, 5.39 Sheila's gestational diabetes 5.40 carpet flaws [5.39 and 5.40 both involve the computations needed for "significance testing". The remaining step is this: If our results are very unlikely, they call our assumption (the value of the mean) into question.] Postpone the rest: Linear combinations of independent normal r.v.'s (as p.365-6) 5.51 X+Y couples. 5.49 sum of 4 blocks 5.44 hole and shaft (Y-X) 5.64 screw-on caps break 5.46 difference of means--trustworthiness 5.47 difference of means--general |
Read, discuss
5.65 p. 377 dye pH (Read for intro to "control limits", compare
with 5.29 and 5.31. This is "Quality Control") |
Optional Sec.5.2 (more practice) 5.32 lightning strikes. (Use algebra of means and variances for part a) 5.61 p. 377 car passengers. Chaper 17, on your disk or downloadable, is all about issues like # 5.29, 31, 65--"Quality Control") 5.48 market gain: if results build on one another, usual mean may not be appropriate. |
Homework questions? Day 27
Prep. for 6.1: Take 4 Numbers
(random sample) from the Birkenstock box: Write them down.
For HW, Find mean xbar. Find xbar + .841.
This is your interval estimate of the unknown mean of the box's population.
("margin of error" is .841) (Return your numbers afterward.)
Next class, you will add your values to the circulating list, and graph
your interval on the transparency circulating.
If xbar = 8.0 7.159|_____________8.0_____________|8.841
Mean and variance quizzes back: Mostly quite good. Main problems:
Pesky added constant. Another way to remember what the variance
of a +bX is.
Let Y = a, with probability 1 (no matter what happens, a
is the result).
y | a
Mean of Y = a. Variance of Y = 0
p | 1 Then
sigma2(a + bX) = sigma2(Y + bX) =
sigma2(Y) + sigma2(bX) = 0 +
sigma2(bX) = 0+ b2sigma2(X)
Variance of (aX +b Y) = a2Variance X + b2Variance
Y
Variance of X - Y = Variance of (X +(-1)Y) = Variance of (X + Y), because
(-1)2 = 1.
When do you need independence? Only for variance of sum of 2 random
variables! NOT for mean of sum, NOT for variance of a+bX.
Not a problem but you should notice:
Mean and s.d. of a distribution (population) are
actually completely parallel to our computations for data, viewed correctly:
X
1 2 4
p
.2 .2 .6
# out of 10 2 2 6 Consider
a population of 10, with these proportions.
Mean: (1+1+2+2+4+4+4+4+4+4)/10 = (1·2 + 2·2
+ 4·6)/10 = 1·.2 + 2·.2 + 4·.6 = 3
Population variance (divide by n) [Sample
variance (divide by n-1)--the one we learned in Chapter 1]
= [(1-3)2+(1-3)2+ (2-3)2+(2-3)2
+(4-3)2+(4-3)2+(4-3)2+(4-3)2+(4-3)2+(4-3)2+(4-3)2]/10
= [(1-3)2·2 + (2-3)2·2 + (4-3)2·6]/10
= [(1-3)2·.2 + (2-3)2·.2 + (4-3)2·.6]
- - - - - - - - - - - - - - - -
- - - - - - - - - - - -
Sec. 5.2 (Central Limit Th. and friends) Day
27
Got through most but not all of Central Lim.
Th.stuff on Day 27. Pick up there Wed.
"If X is normal then Xbar is normal" is a special
case of
A linear combination of independent normal random variables
is normal, (p.365-6) with the mean and s.d. you can calculate using
the algebra of means and variances.
If X and Y are independent normal random
variables, then aX + bY is normal also. Use "algebra" to find the
mean and standard deviation.
Example: Ann runs the mile in
X minutes, where X is N(8, 2)
Betty runs the mile in Y minutes, where Y is N(9, 3)
Ann and Betty compete against each other a lot. What
proportion of the time does Betty beat Ann? (Assume independence.)
Want Prob. that Ann's time is longer than
Betty's. P(X > Y) = P(X - Y > 0)
Need distribution of X-Y. It's Normal. Mean = µX
- µY = 8 - 9 = -1
Variance = sigma2X +
(-1)2sigma2Y = 22 + 32
= 4+9 = 13
S.d. = sqrt( 13) = 3.6
Back to P (X - Y > 0). Standardize by subtracting the
mean (-1) and dividing by the s.d. (3.6)
P(X - Y > 0) = P(Z > [(0-(-1))/3.6 ] ) = P(Z
> .28 ) = 1 - .6103 = .3897, almost 40% of the time.
Suppose each records her time for 4 days. What's the probability that Betty's
4-day average is faster than Ann's ?
Want P( Xbar >Ybar ) = P( Xbar - Ybar > 0).
Need distribution of Xbar -Ybar.
The s.d. of Xbar is 2/2 =1 (since sqrt(4) = 2). So
Xbar is N(8, 1)
The s.d. of Ybar is 3/2 =1.5 (since sqrt(4) = 2). So Ybar is N(9,
1.5)
Xbar -Ybar
is Normal. Mean = 8 - 9 = -1.
Variance = 12 + (-1)2(1.5)2 = 1+2.25 = 3.25
S.d = sqrt(3.25) = 1.8
P( Xbar - Ybar > 0) = P(Z > [(0-(-1))/1.8
] ) = P(Z > .56 ) = 1 - .7123 = .2877, almost 30% of the
time.
Statistics application: Experiment:
Two treatments; compare the mean results--what difference do we see? If
there really is an underlying difference (Ann is faster than Betty on average), are we
sure of seeing it? If there's no underlying difference, what's
the chance of seeing something that looks like a difference?
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