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
Sunday, April 06, 2008
Wednesday, March 19, 2008
Tuesday, March 18, 2008
Monday, March 17, 2008
Wednesday, March 12, 2008
#6 is even? OK...
#6
69 and 103 > 10
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?)
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.
Tuesday, March 11, 2008
Monday, March 10, 2008
Organize
Today's assignment is to make a 4-quadrant graphic organizer that sorts all the information about the 4 different inference procedures.
Thursday, March 06, 2008
Monday, March 03, 2008
Thursday, February 28, 2008
Wednesday, February 27, 2008
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!)
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!)
Tuesday, February 26, 2008
Monday, February 25, 2008
Thursday, February 21, 2008
Wednesday, February 20, 2008
Tuesday, February 19, 2008
Mr t, beer and mermaids
Ch. 23 #15
Note: you'll need to use t* for 60, even though that isn't quite right...
Thursday, February 14, 2008
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!
Wednesday, February 13, 2008
Tuesday, February 12, 2008
Thursday, February 07, 2008
Wednesday, February 06, 2008
no homework
Quest on Friday
You might want to work on perfecting your homework and getting ready
for your Quest!
You might want to work on perfecting your homework and getting ready
for your Quest!
Tuesday, February 05, 2008
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)
Thursday, January 31, 2008
Wednesday, January 30, 2008
Monday, January 28, 2008
Thursday, January 24, 2008
Wednesday, January 23, 2008
Tuesday, January 22, 2008
Thursday, January 10, 2008
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
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
Wednesday, January 09, 2008
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.
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.
Tuesday, January 08, 2008
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.
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.
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.
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.
Monday, January 07, 2008
Wednesday, December 19, 2007
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.
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!
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!
Tuesday, December 18, 2007
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.
Ch. 11 #15: Run 10 times and find the expected value of this problem in theory.
Monday, December 17, 2007
Thursday, December 13, 2007
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
Thursday: Unit I review; Page 106 #8, 15, 16, 30
Tuesday, December 11, 2007
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.
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.
Monday, December 10, 2007
Friday, December 07, 2007
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%
Even Answers:
16a) 38.9%
16c) 79.3%
36a) 59.29%
36b) 40.71%
36c) 5.29%
36d) 0.25%
Thursday, December 06, 2007
Wednesday, December 05, 2007
Tuesday, December 04, 2007
Monday, December 03, 2007
Friday, November 30, 2007
Answers to Friday's work
Chapter 14 #12a)
1. 0.45
2. 0.87
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
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.
Thursday, November 29, 2007
Wednesday, November 28, 2007
Monday, November 26, 2007
Tuesday, November 13, 2007
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
Wednesday, November 07, 2007
down the river
We started a stratifying activity today. Make sure you find all four
means. We'll wrap it up tomorrow.
means. We'll wrap it up tomorrow.
HW: Ch. 13 #9-15, 32, 34, 36; Ch. 12 #7, 9--Due Friday!
Tuesday, November 06, 2007
Monday, November 05, 2007
Thursday, November 01, 2007
Tuesday, October 30, 2007
Monday, October 29, 2007
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.
Friday, October 26, 2007
Thursday, October 25, 2007
Tuesday, October 23, 2007
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
Monday, October 22, 2007
Friday, October 19, 2007
Wednesday, October 17, 2007
Tuesday, October 16, 2007
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!
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
Monday, October 15, 2007
Influential?
Wandering point worksheet
Ch. 9 #11-16
Article #4 due Friday--just write a half-page summary
Ch. 9 #11-16
Article #4 due Friday--just write a half-page summary
Thursday, October 11, 2007
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)
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)
Wednesday, October 10, 2007
Tuesday, October 09, 2007
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.
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.
Monday, October 08, 2007
Friday, October 05, 2007
Thursdays HW
Ch. 8:
#25ef, 29c
#30: slope, y-intercept, r, R2, and prediction for 2002
Article #3!!!
#25ef, 29c
#30: slope, y-intercept, r, R2, and prediction for 2002
Article #3!!!
No homework for Friday
Wednesday, October 03, 2007
R^2
HW Ch. 8:
#25bg, 29abd
For #29 find the residual for (250, 95).
Is this data point above or below the LSRL?
#25bg, 29abd
For #29 find the residual for (250, 95).
Is this data point above or below the LSRL?
Tuesday, October 02, 2007
Thursday, September 27, 2007
Scatterplots
HW: Ch. 7 #4, 6, 8
If you didn't get c, d and e right on the test, you need to also do:
Ch. 6 #28
If you didn't get c, d and e right on the test, you need to also do:
Ch. 6 #28
Tuesday, September 25, 2007
Test tomorrow!
Here is a nice answer from the practice test from chapter 3 about smoking and education level:
The data suggest that a higher percentage of non-smokers continue their education beyond High School than smokers.
Non-smokers have a higher percentage of 4-year college attendees (48%) than smokers (26%). Less of non-smokers have only High School educations (40%) than smokers (64%). Therefore, smoking habits are associated with education level and these variables are NOT independent.
Here is my answer from the CUSS'ing quiz about Halloween candy:
Note to self: since one data set is skewed, I will use the median and IQR:
The boys collected more candy than girls with a median of 62 compared to 45.5.
The boys were skewed left whereas the girls are fairly symmetrical.
(note: no comparison word really needed)
The boys have slightly more spread than the girls. The boys have an
IQR of 31 and a range of 65, while the girls have an IQR of 25 and a
range of 60.
The data suggest that a higher percentage of non-smokers continue their education beyond High School than smokers.
Non-smokers have a higher percentage of 4-year college attendees (48%) than smokers (26%). Less of non-smokers have only High School educations (40%) than smokers (64%). Therefore, smoking habits are associated with education level and these variables are NOT independent.
Here is my answer from the CUSS'ing quiz about Halloween candy:
Note to self: since one data set is skewed, I will use the median and IQR:
The boys collected more candy than girls with a median of 62 compared to 45.5.
The boys were skewed left whereas the girls are fairly symmetrical.
(note: no comparison word really needed)
The boys have slightly more spread than the girls. The boys have an
IQR of 31 and a range of 65, while the girls have an IQR of 25 and a
range of 60.
Friday, September 21, 2007
Homecoming
Today in class we made a Flip-chart.
If you finish the review by Monday and give me a chance card.
Thursday, September 20, 2007
Review assignment starting on page 105
UNIT I Review problems
# 5, 21(outlier test on c!),
#25-27, 29(a-g + make a box-plot),
#31, 32
Double stamped:
once Friday (for starting)
once Tuesday (when you should be DONE)
Or by Monday for a Chance Card!
ALSO: Article #2 due Friday
http://www.mrderksen.com/art2.htm
# 5, 21(outlier test on c!),
#25-27, 29(a-g + make a box-plot),
#31, 32
Double stamped:
once Friday (for starting)
once Tuesday (when you should be DONE)
Or by Monday for a Chance Card!
ALSO: Article #2 due Friday
http://www.mrderksen.com/art2.htm
Wednesday, September 19, 2007
Tuesday, September 18, 2007
Monday, September 17, 2007
A nice answer from the quiz
Note to self: since one data set is skewed, I will use the median and IQR:
The boys collected more candy than girls with a median of 62 compared to 45.5.
The boys were skewed left whereas the girls are fairly symmetrical.
(note: no comparison word really needed)
The boys have slightly more spread than the girls. The boys have an
IQR of 31 and a range of 65, while the girls have an IQR of 25 and a
range of 60.
Starting Chapter 6
We started chapter 6.
We will finish it on Wednesday, so hang on tight.
Test on chapters 1-6 next Wednesday.
Don't forget Article #2 is due Friday.
We will finish it on Wednesday, so hang on tight.
Test on chapters 1-6 next Wednesday.
Don't forget Article #2 is due Friday.
Classwork: Ch. 6 #5, 6, 11
HW:
Ch. 6 #7 and 15
Ch. 5 #30 (you have the ogive on your handout to write on)
Friday, September 14, 2007
Ogives!
Today we did a worksheet on ogives. You can/should do #29 in Chapter
5 to see what you missed.
5 to see what you missed.
It looks like our schedule for next week is this:
Mon-Wed: Work very hard and focused and do Chapter 6
Thur-Fri: Review
Mon: Review
Tues or Wed: Test on Unit 1: Chapters 1-6
Thursday, September 13, 2007
Last homework for the week
Ch. 5 #5b, 6, 22
Fun web site o' the day:
http://www.stat.tamu.edu/~west/ph/stddev.html
Wednesday, September 12, 2007
Put that calculator to work!
HW: Ch. 5 #5a, 14, 21, 26, 36, 41
Website from today:
http://www.mathsnet.net/asa2/modules/shodor/plop/
Tuesday, September 11, 2007
Monday, September 10, 2007
Here comes chapter 5
HW: Ch. 2 #8; Ch. 5 #23cd, #35b + boxplot
Tutoring this week: Wed and Thur, 2:30 - 3:15
Last Friday we discussed Ch. 4 #14. We COMPARED the shape, centers
and spreads of the two distributions.
Thursday, September 06, 2007
Most Important Stuff in Chapter 4
In class we made a histogram and a calculator for #18
HW: Ch. 4 #8, 12, 17, 24 + Article #1 summary from the web
HW: Ch. 4 #8, 12, 17, 24 + Article #1 summary from the web
Wednesday, September 05, 2007
No Homework
No HW.
In class we learned how use Stat Edit (page 11) and Stat Plot (page
43) and we did this with the hair length data from each class.
In class we learned how use Stat Edit (page 11) and Stat Plot (page
43) and we did this with the hair length data from each class.
Tuesday, September 04, 2007
Week 2!
In class we took notes from the Workshop Stats Book.
HW: Ch. 4 #3,5; Read your syllabus and Chapter 1
Quiz on Tuesday on syllabus and chapter 1
Thursday, August 30, 2007
Chapter 3, Day 2
Today in class we discussed independence. We analyzed some of our
class data and we did #21.
class data and we did #21.
Homework: Ch. 3 #15, 16, 18
Die is due on Tuesday
Syllabus and chapter 1 quiz on Wednesday
Wednesday, August 29, 2007
Tuesday, August 28, 2007
Day 2!
Today we read the information on Page 8 and did problems 2 and 3 in Chapter 2.
HW: Chapter 2 #4-6
Monday, August 27, 2007
Monday, August 27th
First Day of School:
Due tomorrow: One dot-plot of the M&M data (pick any color or the
total) and 2 sentences describing your dotplot.
Due the Tuesday after Labor Day: 1 homemade die. The cube must be
homemade. 4 pts. for durability, 4 pts. for artistic/creativity, 4
pts. for punctuality.
Stuff to consider: Bring in Kleenex, start bringing your calculator,
read your syllabus.
Thanks for a great first day!
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