ogives!

WS: #30 and 1.20ab
Ch. 6 #2, 3

standard deviation

Ch. 5 #6, 21, 26, 36, 41

Outlier test

Ch. 5 #12bcd, 14, 15, 23b

boxplots

Ch. 5 #23cd, 35b + boxplot

Stem and leaf!

Ch. 4 #12, 17 & 24
Article #1
statsarticles.blogspot.com

Histograms

Ch. 3 #23
Ch. 4 #8, 18
Article #1 due Friday!

CUSS!

Ch. 4 #3 & 5

Independent?

Ch. 3 #15, 16 & 18
Friday's work: worksheet
All of this is due Tuesday.
And don't forget your die!
And the syllabus quiz!

Pie and bar

Ch. 2 #6
Ch. 3 #6, 8

Chapter 2!!

#2-5, that's it!

First day!

Read the syllabus:  quiz Friday.
Read chapter one:  quiz tomorrow!
Start making your die:  due next Tuesday.

first day of school!

HW:  Read the syllabus and chapter 1.

Test day

My civic duty is complete.  I'll be at school all day.  Feel free to come by before noon to say hi.  I'll come by the MPR before you start.  If you're studying tonight and want to e-mail a question, feel free.

Carne Asada Feast:  May 31!!!

sunday night

well, you're probably studying hard now.  I definitely have to go do my civic duty at jury duty tomorrow.  If there's anything that is stressing you out, feel free to drop me an e-mail!

mr d

O-6

OK, folks. O-6 will challenge you. Here's the answers:

a) Ho: p = 0.5; Ha: p=/ 0.5

b) Conditions for a 1-prop z-test are NOT satisfied, because 8*0.5 = 4 which is a lot less than 10. In other words, the expected number of successes is too small.

c) To work around this problem, we use binomial pdf. We use p = 0.5, n = 8 and fill in the table:

0 0.00391
1 0.03125
2 0.10937
3 0.21875
4 0.27344
5 0.21875
6 0.10937
7 0.03125
8 0.00391

d) Since the test is two-sided, we see this this:

If x = 0 or 8, p = 2*0.0391 which is less than 5%
If x <= 1, then we add up the lower 2 and upper 2 and get 0.07312, which is greater than 5%
So 5% is not possible.

e) For the provided data, x = 2. So the p-value is 2*P(x <= 2) = 2*(0.003906 + 0.03125 + 0.109375) = 28.9% We would fail to reject and fail to find evidence of any difference between the two brands.

f) Increase sample size! It lowers the expected sd and enables us to more clearly see any differences that might exist between the two brands. sqrt(p*1-p/n) gets smaller!

That's a doozy of a problem, but hopefully this helps!

More??

OK folks, here's what I've done:

• Flashcards powerpoint
• Name that green sheet website
• Answers to test O
• Name that green sheet answers for tests I through Q

Keep scrolling down and you'll see all of that.

I'm more than happy to do more, but I will now switch to waiting for e-mail requests. I you want more answers (P or Q) or something else, just send me an e-mail:

mrmathman@(no-spam)gmail.com

Obviously, take out the no spam thing.

I'm more than happy to type up more answers if you're working hard and want them. Just drop me that e-mail and I'll put them up.

Good luck and happy studying to all!

Flashcards

I put the entire flashcard powerpoint here:

http://gallery.mac.com/mrmathman#100034

Enjoy!

web site for inference

This web site loads slowly, but is pretty nifty:

 

http://www.ltcconline.net/greenl/java/Statistics/StatsMatch/StatsMatch.htm

 

Note: the wording on this website is more vague than the AP test. The AP test always asks for "a 95% confidence interval" if it wants an interval (or some other %). Also, the AP problems are longer, and thus have more information that will guide you to deciding what to do.

However, I think that after you 10 or so of these, you'll get used to the wording of this site and it will be good practice. If it is just always frustrating, switch to finding the tests and intervals in your notebook. Every released test (A, B, C,...) has at least one.

Note: this website includes prediction intervals, which you do not need to know.

Answers for O #1-5

O-1a) 40% of the sales agents had sales of $300,000 or less.

O-1b) 10% (show work!)

O-1c) No agents had sales between $1 million and $1.1 million

O-1d) $800,000 (with explanation!)

O-2a) 2-prop z-interval (name or formula)

success = 188, 180 > 10

failures = 12, 20 > 10

Independent samples

(-0.0156, 0.0956)

We are 96% confident that the difference in the proportions of parts meeting specs for the two shifts is between -0.0156 and 0.0956.

O-2b) No, because 0 is in the interval, zero is a possible value. So we are 96% confident that zero difference in shift specs is possible. We make this decision using alpha = 0.04!!

O-3a) z > 1.14 = 12.71%

O-3b) 1 - (8729)^5

O-3c) invnorm(0.99) = 2.33

algebra: (291.2 - M)/2.8 = 2.33

M =

Show work!!!

O-4): matched pairs t-test

Ho: mu = 0

Ha: mu > 0 (mean diff (after - before) is greater than 0)

Name or formula: t = (x-bar - 0)/(s/sqrt(n))

SRS and graph to check nearly normal

t = 3.75

df = 11

p-value = 0.002

Reject Ho. Is evidence that people who complete program have increased dexterity.

O-5a)

rv: amount of draft

treatments: standard hitch and new hitch

experimental units: 2 large plots of land

O-5b)

Yes, the hitches are randomly assigned to the plots.

O-5c)

No. Each treatment was applied to only one plot. Replication is needed.

O-5d)

Even though 25 measurements were taken, each hitch was used on only one plot only. If a difference in the draft is observed, we will not know whether the difference is due to the hitch or the plot. The treatment (hitches) are confounded with the plots.

Name that green sheet

I-5c:  Chi-square goodness of fit

J-5:  Chi-sq 2-way table or 2-prop z

K-4:  2-mean t-interval

L-4:  1-prop z-test

M-4:  mp's 1-mean t-interval

N-4:  2-prop z interval

O-2:  2-prop z interval

O-4:  mp's 1 mean test

P-4:  mp's 1 mean t test

Q-5:  2-mean t test

Calendar info

Below you see the link for the review calendar. Please note:

**Click on "Week" to view a week at a time and better see the details
of what we did each day.

**Homework and classwork is listed for each day.

I hope this is helpful!

Review Calendar

To see a schedule of what we are going to review when, go to:

http://www.google.com/calendar/embed?src=5dskgriao5era1dj6un4ipgo7c%40group.calendar.google.com&ctz=America/Los_Angeles

Quest on Thursday

Last section of the book, aka, Unit VII Review: #10

2-way tables!!

Ch. 26 #12 and 13 (skip part a)

Goodness of fit

Ch. 26 #2 & 8

#6 is even? OK...

#6
69 and 103 > 10

69/172 +/- 1.96*sqrt(.401*.599/172)

I am 95% confident that the true percentage of Adirondack streams that
have a shale substrate is between 32.8% and 47.4%.

a few answers

#8:
histogram looks very nice and normal! (did you draw it?)

7.963 +/- (t*)(8.79/sqrt(27))
t* is for 26 df

I am 95% confident that the true mean difference between boys and
girls who have been drunk at least twice is between 4.5% and 11.4%

#10a)
1772/2000 = 88.6%

b) 250000/304266 = 82.2%

c) Type I = certify the petition when there are not enough sigs.

d) Type II = valid petition not certified.

e) Ho: p = 82.2% Ha: p > 82.2%

1772 and 228 > 10 check!

z = (.886 - .822)/0.008553

Note: use 0.822 in sd formula

z = 7.48, p-value = 0

We reject and say we found sig evidence that the % is greater than
82.%. Certify!

f) To increase power, collect even more sigs. Although it seems like
we have plenty of power.

#16
SRS
2 independent years
260, 240, 270 and 230 are all > 10

(0.54 - 0.52) +/- 1.96*sqrt( (.54*.46/500) + (.52*.48/500) )

I am 95% confident that the change in proportion of students who
choose to enroll is between -4.2% and 8.2%

16b) Since zero is in the interval, we would think that there is not
evidence of a change.

Test coming!

Page 510 and following: #5-10, 13, 16 19, 27
NAME them all.
Do three.

Read page 508

Organize

Today's assignment is to make a 4-quadrant graphic organizer that sorts all the information about the 4 different inference procedures.

Tuesday, March 4th

Ch 24 #8, 12, 23, 24

2-t!

Ch. 24 #7 and 9

2 samples, coming at ya!

Friday Article #9 is due!

test ideas

CLT?  Do Ch. 18 #21 and 23
1-mean?  Ch. 23 #21
matched pairs mean? Ch. 25 #14abc is great
Alpha?  Ch. 21 #11
Read?  "What Can Go Wrong?" on p. 415 and pages 444 and 445 (beer!)

t-test Thursday!!

Ch. 25 #14abc

Ch. 18 #24 and 28

matched pairs

Ch. 21 #14

Ch. 23 #24 interval and test

Ch. 25 #7 and 11

t from data

Ch. 23 #23:  test and interval

Class data:  test and interval

t-test

Ch. 23 #8, 9, 12, 22

Mr t, beer and mermaids

Ch. 23 #15

Note:  you'll need to use t* for 60, even though that isn't quite right...

Bonus reminder

Can you calculate sample size?

Here comes a test!

We did a crossword today:  study the vocabulary in chapters 19, 20 and 21.

 

The test:

Multiple Choice (study Ch. 19 #7 & 8 and...?)

A green sheet

Type I and II errors (study Ch. 21 #7 & 8)

Outlier test!!!!

 

Good luck!

alpha

HW:  Ch. 20 #22, Ch. 21 #9 and 10

Good review for Friday's test

HW:  Ch. 19 #26b, Ch. 20 #20, Ch. 21 #6-8

Quest tomorrow!

Ch. 19 #21
Ch. 20 #21

no homework

Quest on Friday
You might want to work on perfecting your homework and getting ready
for your Quest!

got behind, sorry...

Friday:  Ch. 20 #9 and 10

Monday:  Ch. 20 #11 (put a confidence interval and test on one green sheet)

Tuesday:  Ch. 20 #12 (in class) #13 and 14 (just check conditions) #15 & 16 (add a confidence interval on the bottom)

Article #8 + ...

Article #8
statsarticles.blogspot.com
Ch. 19 #4 and 25

More intervals

Ch. 18 #9
Ch. 19 #7, 8, 12, 13cd

Confidence intervals!

HW:  Ch. 19 #13ab (and if you were absent, 14ab also!)

Quiz tomorrow

HW: Ch. 18 #17-19

mean mean mean

Ch. 18 #21-23

Start of Semester 2!!

HW:  Ch. 18 #21ab and 22ab

Extra?

Want a few extra problems to study this weekend?
Here's a few ideas!

Ch. 3 #15
Ch. 5 #23
Ch. 6 #27
Ch. 8 #35
Ch. 11 #11
Ch. 13 #29
Ch. 17 #19, 21a

Unit 3 review answers

4a)  The problem would be that if a group of people who work together all get the same treatment, they might all respond in a similar fashion, because they all might have fun together and then you'll think that that treatment lowered stress when in fact it was the fun they were having together.  Random assignment of treatment will balance out any similarities these people might have with each other and help create homogeneous treatment groups for a more clear view of how the treatment effected the volunteers.

4b)  A control group gives us a baseline for comparison.  Something might happen during the course of the study, say, everyone gets a raise.  Then we could see how the stress level was reduced overall and will be able to tell if the treatment(s) lowered the stress even more.

4c)  No, the study is done on volunteers, not a random sample of the company.  There might be differences from the volunteers compared the average worker.

Unit 2 review answers

1a.  Yes, the residual plot has no pattern.
1b.  233.517 = ABOUT 233.5 more aircraft per year.
1c.  89.9% of the change in aircraft is explained by the regression on year.
1d.  2939.93 + 233.517*(2) = 3406.964 aircraft
1e.  3406.964 + 40 = 3446.964 = 3447 aircraft
1f.  sqrt(0.899) = 0.948 = r = strong, positive, linear relationship between year and # of aircraft
1g. s = 33.43 = sd of the residuals = the LSRL missed the data by an average of 33.43 aircraft
1h.  aircraft-hat = 2939.93 + 233.517*year
1i.  Yes!  In 1990 (year zero) we PREDICT that there were 2939.93 aircraft.

Change!

For Ch. 17 #11 I recommend you ONLY do (a)!

end of the semester

Due Wed:  Unit I review + Ch. 17 #1, 11, 16
Due Thur:  Unit III review

Unit I review answers

1.  The center is at 2.1 hurricanes.  The spread is an IQR of 3 ( 4 - 1), with a range of 7 (7 - 0).  The shape is skewed right.

2a.  increase;  b.  same;  c.  inc.;  d.  same;  e.  inc.

3a.  Since the data is skewed right (draw a little box-plot using the 5 # summary to see this clearly), the mean will be greater than the median.  It is pulled up by the skewness.

3b.  IQR = 3.3 - 2.8 = 0.5
1.5*IQR = 1.5*0.5 = 0.75
2.8 - 0.75 = 2.05 is the lower outlier fence
3.3 + 0.75 = 4.05 is the upper outlier fence

SO:  4.2 is a high outlier and the other two fish are not outliers.

Chapter 6 #25 is odd!  Check is carefully!  Here's a little work:
25b)  z = 0.5;  0.5 is the lower bound and 999 is upper
25c)  lower bound = z = -1.583 and upper bound = z = -0.75
25d)  invnorm(0.25) = - 0.674 = z;  invnorm(0.75) = z = 0.674;  then do algebra to find Q1 and Q3;  IQR = Q3 - Q1

1.  Who = 200 adults
What:  education level and smoking habits
When:  ??
Where:  mall
How:  ??
Why: ??

2.  two categorical variables:  education level and smoking status

3a.  32/200 = 16%;  b.  32/93 = 34.4%;  c.  32/50 = 64%

4.  2 bars:  one for smokers and one for non.  The bars should both add to 100%.  The HS part of smoker bar should be 64%, whereas the 4+ non should be the biggest (48%).

5.  These data provide evidence of an association between smoking and education level.  64% of smokers had only a hs diploma, whereas 40.7% of non-smokers had only a hs diploma.

6.  We have no idea if the behavior changes over time.  This data was only taken at one point in time.

Binomial

Ch. 17 #7, 14
Review worksheet, Unit 2

Answer to basketball simulation

If I set up my simulation like this:

01-72 = makes shot
73-00 = misses

here's some digits:

5730 3485 3246 75 4563
2pts 1pt 2pts 0pts 2pts.

For the 5 runs above my average would be: (2 + 1 + 2 + 0 + 2)/5 = 1.4

That's the simulation.

Now for the expected value

P(0) = 0.28 (misses first shot and is done)
P(1) = (.72)*(.28) (makes the first, misses the second)
P(2) = (.72)^2 (makes both)

Now just run the expected value formula with 0, 1, 2 and the 3 probabilities.

Answers, comments and hints to the review

3a) Note that the problem says that the tickets are bought at the same time for the same price. This makes the formula 3C + 5F.

If the each ticket was bought separately, that would make the formula C + C + C + F + F + ... The mean is the same for either formula. But if the second formula was used, you'd have to do the Pythagorean formula thing.

3b) First do 3C = 450 and 5F = 500; then do sqrt(450^2 + 500^2)

3c) This is only about ONE of each tix: C - F. You can take it from there...

6) Make a 2-way table: 51% in the upper left corner. The other two numbers go on the outside.

a) only 3% is left in the lower right corner
b) P(left|right) = 51/82 = 62.2%. Since P(left) = 66%, this is fairly different, so they are not independent.

7a) 1 - (89/90)^10
7b) 1 - (9/10)^10
7c) 1 - (89/90)^5*(9/10)^5

25d) (0.93)^4*(0.07)

28a) mean = 4

28b) sd = 3.2

28c) Think and read carefully! If the first is bigger than the second then:

(first - second) > zero

So we want to use the mean and sd from above to compare to zero:
z = (0 - 4)/3.2 = -1.249
P(z>-1.249) = (using normalcdf) = 89.4%

42a) 1/100 = 0.01

42b) (.99)(.99)(.01) = 0.009801

42c) (.99)^100 = 0.366

42d) You want to be first!


42e) It doesn't matter! Everyone has a 1% chance!

Now if you're thinking carefully about (e), you might be thinking: "Hey, don't the probabilities change?" Watch this!

Prob(3rd person wins) = (99/100)*(98/99)*(1/98) = 1% (notice how all the fractions reduce!)
Pretty cool, huh?

I will check e-mail at about 10-ish tonight. If you are feeling frustrated, drop a line to:

mrmathman @ gmail.com
I will reply tonight.

Good luck!

Review Answers

It is 8:20pm and I'm starting to put up some answers. Here they come...

THE LAST ASSIGNMENT OF 07!!!

Unit IV Review #3, 6, 7, 23, 25, 28, 42
Ch. 11 #15:  Run 10 times and find the expected value of this problem in theory.

07 is finishing...

Friday:  Ch. 8 #25, 35;  Ch. 10 #1
Monday:  Ch. 11 #2, 5, 19

Review

Wednesday:  we did a 2-page green worksheet that is an excellent probability review.
Thursday:  Unit I review; Page 106 #8, 15, 16, 30

Rules!

Ch. 16 #24, 28

Answers:
24a)  60, 12
b) 6, 1.5
c) 92, 12.37
d) 68, 12.37
e) X + X = 160, 16.97

28a) 10
b) 2.68
c) The seed packets need to be independent of each other.  If they all came from the same batch, they might not be independent.

The end of Unit 07

No homework tonight.

Review

Unit IV Review #1, 5a, 15ab, 16ac, 35-37

Even Answers:
16a) 38.9%
16c) 79.3%

36a) 59.29%
36b) 40.71%
36c) 5.29%
36d) 0.25%

Thursday

Article #7

More Expected Value

Ch. 16 #3, 5, 21
Scramble #2

Expected Value

Ch. 16 #16, 17

Quiz day

Scramble #1 for homework

Answers to Friday's work

Chapter 14 #12a)
1. 0.45
2. 0.87

Chapter 14 #12b)
1. (0.55)^2 = 0.3025
2. (0.45)^2 = 0.2025
3. 1 - (0.87)^2 = 0.2431

Chapter 15 #8
a) P(male | cat) = 6/18 = 0.333
b) P(cat | female) = 12/28 = 0.429
c) P(female | dog) = 16/24

Chapter 15 #10
a) 0.62
b) 0.26/0.30 = 0.867
c) 0.12/0.62 = 0.194
d) 0.66

Chapter 15 #24
No! P(death penalty) = 62%, but P(dp | rep) = 26/30 = 86.7% and P(dp
| dem) = 12/36 = 33%, so party is clearly NOT independent of party!

Quiz on Monday!!!

Answers to Thursday's work

Ch. 15 #4:
a) 0.14
b) 0.23
c) 0.77

Ch. 15 #20
a) P(Canada | Mexico) = 0.04/0.09 = 0.444

b) No, 4% have been to both

c) No, P(Canada) = 18%, which does not equal part (a), above, so not
independent.

Friday HW

Ch. 14 #12, 21
Ch. 15 #8, 10, 15, 24

Start of 15

Ch. 15 #2, 4, 20

Chapter 15

We started chapter 15, but no homework yet.
Get those app's done tonight!!!

Tuesday HW

Ch. 3 #14
Ch. 14 #19, 20, 24, 25

Back from break!

Page 110 #24
Ch. 14 #11 and 14

The last 4 days before Thanksgiving

Tuesday:  Unit 3 Review, #23, 24, 29;  Article #6

Wednesday:  Crossword and flipbook

Thursday:  Test on Unit 3 (1 page multiple choice + normal, 2 experiments, 1 survey)

Friday:  Bias project: poster and presentation

down the river

We started a stratifying activity today. Make sure you find all four
means. We'll wrap it up tomorrow.

HW: Ch. 13 #9-15, 32, 34, 36; Ch. 12 #7, 9--Due Friday!

More design

HW: Ch. 13 #3-5, 28-30

More experiments

Read about observational studies: pages 246-247
Do: Ch. 13 #1, 2, 6, 26

See you Monday!

Work for Thursday night and Friday:
Ch. 13 #19-24

Article #5

Article #5 is due Thursday

Monday Homework

Ch. 12 #15, 16, 23, 24

Here are some of the class notes:

Calling on students for chance cards.

Simple Random Sample:
Use the Random Integer command on my calculator to randomly pick who to call on.
OR:
Put everyone's name in a hat and draw 5 names.

Stratified Random Sample:
I might stratify by class grades. Divide the class into five groups
(A, B, C, D, F) and randomly pick some students to participate from
each group. I think students with high grades are more likely to
participate and vice versa, so this will give me a good
representation.

Cluster sample:
Randomly pick one of the 8 table groups and call on everyone in that group.

Systematic Random Sample:
Use the roll sheet and pick every 5th person.
OR:
Pick every 3rd person as you arrive to class.

Population: Rancho Students
Question: Is RCHS a quality school?
Possible Strata?

Stratifying by GPA:
I would divide the student body into 3 groups: high, med and low GPA.
I would then randomly choose some students to survey from each group.
I think that each of these groups will have very different opinions
about RCHS and I want to make sure that each group is represented.

More chapter 12...

#8, 10, 12, 14

HOW do you collect that data?

Ch. 12 #2, 4, 6

Test tomorrow!

Test on regression tomorrow:

Study chapters 7 through 10 and normal problems
Study your worksheet from Chapter 10
Study your old regression test on temp/crawling
Study Ch. 9 #1
Study extrapolation, outliers and influential points and the standard
deviation of the residuals
Chapter 10 #1 and 2

End of regression, really!

Page 204 #2, 4 and for #28 please interpret s in context.

2 days

Thursday we did a worksheet and Article #4
Friday: Page 207 #17, 37, 40

Onward in 10

We did Chapter 9 #1 in class, and so should you.
HW: Ch. 10 #1, 2, 7

Starting Chapter 10

1978—63,042
1979—226,260
1980—907,075
1981—2,826,095
Year vs. acres devastated by the gypsy moth.
Predict for 1982, please!

Also, MC packet #6

Influential?

Wandering point worksheet
Ch. 9 #11-16
Article #4 due Friday--just write a half-page summary

Review Answers

Well, you lazy bums. I suppose you're too busy on MySpace to e-mail me any answers, so here's a few:

17c) First take the squareroot of 0.924. I forget exactly what this is, but it is 0.9something. This tells us there is a strong, positive, linear relationship between tar and nicotine.

17d) For every 1 more mg of tar, we predict about 0.065 more mg of nicotine.

17e) Even with no tar, we still predict about 0.154 mg of nicotine.

1d) 92.3% of the variation in age can be explained by the regression on age.

1f) She is shorter than predicted, for her age.

1g) Wait until next week.

See you in the morning!
Now go to bed! :o)

Before the test

Finish the crossword and finish your flipbook.

Practice Test

Ch. 8 #10, 32, 36
Add to #10 the interpretation of the y-intercept
Parts of this assignment will graded tomorrow
and this grade will be part of your test grade.

End of chapter 8

HW: Ch. 8 #8abc, 21, 23, 25d

Thursdays HW

Ch. 8:
#25ef, 29c
#30: slope, y-intercept, r, R2, and prediction for 2002
Article #3!!!

No homework for Friday

R^2

HW Ch. 8:
#25bg, 29abd
For #29 find the residual for (250, 95).
Is this data point above or below the LSRL?

Chapter 8

HW: 
Ch. 8 #25ac + find the residual (19, 410)
Ch. 8 #28acd