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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