HW Day 5: (Re) Read pp. 43-49
Do Check p.59 17,19,20 Read 55-6, "Organizing...". then: Ahead:
Finish Ch. 2 pp. 49-55(standard
deviation, & using technology). ("Check" problems:
2.21, do 22-24)
Do the 5-number summaries required here by hand (with a calculator if needed for means, finding middles between 2 numbers).
Read Ahead in Ch.3, 67-72 density curves, & ahead Normal Distributions 73-86: There's a lot there, and will need repetition.
Hand in ..
p. 61, 2.36 days of births, Canada
book's question is very open-ended. Answer
instead the questions
just below the HW box,*
p. 53, 2.11 xbar=7.50, s =
the same for both dist's. Don't do the
calculations--just make back to
back or side by side stemplots & compare their
|Read, to discuss
p. 62, 2.42 Play with summary numbers. Use the Applet, One variable statistical calculator; type data in at the Data tab
Questions for 2.36, p. 61 (Days of
births, Canada ):
A. a) Which day had the lowest Median (and about what was that number)?
b) Which day had the highest Median (and about what was that number)?
c) Which day had the highest variability (spread), measured by:
--IQR (about what are the quartiles for this day)?
--Range (about what are min and max for this day)?
d) Tuesday appears to be somewhat skewed. Left, or Right skewed?
B. To compare the Canadian with the American data p. 11 #1.5:
a) Is the general pattern the same in the Canadian and American data? Discuss briefly the common findings.
b) Going deeper: (Following the 4-step method, p. 55-7:)
State the issue: Is the weekend/weekday difference greater in Canada or in the US (or are they similar?)
Plan how to find an appropriate answer: In both countries, Tuesday is highest, Sunday is lowest. Relate the number of Tuesday's births to the number of Sunday's births for each country. Proportion/ percents will show the relationship best, since different types of summary numbers are given for the two countries. We'll find Sunday births as a percent of Tuesday's.
Solve: For Canada,you have (part A) estimated the median number of births for Tuesday and also for Sunday, from the graph. Take the number for Sunday, divide by Tuesday's number, restate as a percent. For U.S., use the numbers on p. 11, dividing Sunday by Tuesday.
Conclude, something like this: " In Canada, on Sunday(s), the number of Sunday births was ___% of the number of births on Tuesday. In US [make the parallel statement.] Therefore the difference is greater(?) in (Canada?US?). This may indicate that proportionately more "planned births" occur in (Canada?US?)." (Remember we decided the most likely reason for the weekday/weekend difference was planned births--induced and Caesarians.)
c) The picture for 2.36 makes the difference between weekdays and weekend days look more extreme than it actually is. Why/how?
d) To make the numbers more comparable, (U.S. Means (per day) of all births in a year of Sundays/Tuesdays, Canada median number per Sunday/Tuesday) it would be better if we had the Canadian Means also. Is it likely to make much difference? To address this, look at the boxplots and tell (using skewness) whether the Canadian mean for Tuesday would be less than the median, about the same, or more than the median. Do the same for Sunday.
Go to Mac 101 Computer Lab Friday
(Feb. 3) for SPSS.
Bring Flash drive.
up today for 10:30 session if you can. Back here Monday.-- At
point we'll be using SPSS heavily for about 3 1/2 weeks, then not
again till the very end of term. SPSS
First hourly exam, a week from Friday: Feb. 10, Day 9 .
Sample exam handed out Friday or Monday. Solutions will be linked from Day pages. Closed book, but bring one sheet of notes (anything you like) and a calculator.
Exam will cover thru what is assigned on this coming Monday, Plus reading SPSS output. (We'll be going to the computer lab to learn SPSS this Friday.) You may be asked to read SPSS output (as we'll see it on the sample exam), but not how to produce it.
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Homework questions? Day 4
I didn't say Wed:
Median: Middle one if n is odd, or average the 2 middle if n is even.
Formula: Count in how far? (n+1)/2 places. ( 14 items--> 7 1/2 places? go halfway =average the 7th and 8th observations. HW: 2.1, wood n = 20; (20+1)/2 = 10.5: halfway between 10th &11th: average the 10th and 11th.)
SPREAD: Quartiles, five number summary, boxplot, IQR. Notes Day 4
4-step process (Day 4, bottom)
SPSS Friday; start
of Middle & Spread continued--"Systems:"
-- (Midrange, Range Very sensitive to outliers--they use only the max and min!)
-- Median, IQR (+ Quartiles Q1, Q3, 5-number summary), based on percentiles ( j'th percentile is > j% of the data)
-- Mean, StandardDeviation "x-bar" (or "y-bar"), "s" (good for symmetric unimodal, no outliers)
deviation (measure of Spread that goes with mean)
Variance s2: (almost) average of squared deviations from the mean.
(Divide by (n-1) "degrees of freedom")
s : Standard deviation is the square root of the variance.
Computation: I will require you to know how to do it by hand for 4 or 5 observations
(see BPS5e p. 49-51 for formula & computation example. )
Demo: 1,1,2,4, mean = 2, sum of squared deviations = 6, variance = 2, s = 1.41 (Using Table to calculate sum of squared deviations below)
1,1,2,4,12, mean = 4, sum of squared deviations = 86,
variance = 21.5, s = 4.64.
(Midcomputation check: Sum of deviations from the mean (before squaring each) always = 0 )
--s is Always > 0 (0 only if all observations are =)
--s units the same as those of the observations (squared and squarerooted).
Physics: angular momemtum (spinning ice skater)
Not so weird: High school geometry?
Remember Pythagorean theorem:
c2 = a2 + b2:
Hypotenuse of right triangle is also square root of a sum of squares.
Very sensitive to outliers (the
outliers contribute much more than their share to the Sum of
Deviations from the Mean) Note contribution
12 is disproportionate.
SPSS, for simple computation: Handout
|8 = Sum. xbar = 8/4=2
||0 = sum (always!)
||6 = sum of squared