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MATH 155 MODERN STATISTICS Prerequisites: High School Algebra II and Plane Geometry Textbook 1: A First Course in Statistics , 9 th ed by McClave and Sincich

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math 155 modern statistics

MATH 155MODERN STATISTICS

Prerequisites: High School Algebra II

and Plane Geometry

Textbook 1: A First Course in Statistics, 9thed

by McClave and Sincich

Textbook 2: Statistics by Power, 6thed

by Philip E. Luft

ASK AT THE INFO DESK

THIS IS NON-RETURNABLE

slide4

IF YOU ARE NEW TODAY, Be Sure to Read All the Power Point Slides for

LESSONS 1 AND 2

Now We Look at

Part of LESSON 3

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 19

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Sometimes the data are given in a sentence or two, without a data set.

EXAMPLE C1: A college of 1000 students has 700 day students (and the rest night students). Also 600 of the students are men (and the rest are women). If exactly 400 students are both men and day students, how many are women night students?

SOLUTION C1: To analyze this problem, make a list of the sets mentioned:

M = men

W = women

D = day students

N = night students

Next, we draw a two-way table.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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EXAMPLE C1: A college of 1000 students has 700 day students (and the rest night students). Also 600 of the students are men (and the rest are women). If exactly 400 students are both men day students, how many are women night students?

SOLUTION C1: Next, we draw a two-way table, showing sex in the rows and day/night in the columns.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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EXAMPLE C1: A college of 1000 students has 700 day students (and the rest night students). Also 600 of the students are men (and the rest are women). If exactly 400 students are both men day students, how many are women night students?

SOLUTION C1: Next, enter the numbers for marginal sets.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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EXAMPLE C1: A college of 1000 students has 700 day students (and the rest night students). Also 600 of the students are men (and the rest are women). If exactly 400 students are both men day students, how many are women night students?

SOLUTION C1: Next, enter the numbers for intersections of marginal events.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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SOLUTION C1: Finally, enter the rest of the numbers by subtraction.

Thus we conclude 100 students are women night students.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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1.4 Intersection and Union

The intersection AB of two sets A, B consists of all subjects which belong to both A and B. To find the intersection of M and FT, list all the elements 3,5,6,7,10 of the set FE and cross out those which do not belong to PT, leaving FEPT. Thus we represent the students who are female and part-time as

FEPT = { 3, 5, 6, 7, 10 } = { 3, 7, 10 }

Similarly, we represent the students who are male and full-time as

MFT = { 1, 2, 4, 8, 9 } = { 1, 4, 8, 9 }

Such intersections can be shown in a two-way table.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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The union AB of two sets A, B consists of all subjects which belong to either A or B (or both). This common type of or is called an inclusiveor. Thus we represent the students who are female or full time as

FEFT = { 3, 5, 6, 7, 10, 1, 4, 5, 6, 8, 9 }

= { 3, 5, 6, 7, 10, 1, 4, 8, 9 }

where we have crossed out repeated mention of subjects 5, 6 because they belong to both M and FT. The union is shaded in the above table.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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This information is usually reduced to a table of counts, showing how many subjects fall within each cell. (A cell represents the intersection of a row category with a column category.) Then the row and column totals tell how many subjects belong to a row category or a column category.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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Since the variables Sex and Load are in the students data set, we can use MINITAB to form the table. Choose

Stat > Tables > Cross-tabulation, double-click on the variable names, and choose Counts.

Rows: Sex Columns: Load

full part All

female 2 3 5

male 4 1 5

All 6 4 10

Cell Contents --

Count

Notice that MINITAB omits the grid lines, so the table is harder to read. You might want to draw them in.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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Sometimes the data are given in a sentence or two, without a data set.

EXAMPLE C2: A college of 1000 students has 700 day students (and the rest night students). Also 600 of the students are men (and the rest are women). If exactly 400 students are both men and day students, how many are women and night students? men or day students?

SOLUTION C2: In Example C1, we drew this two-way table.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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The general requirement for two sets A and B to appear both in rows (or both in columns) is that A and B be mutually exclusive: they have no common elements, so that AB=, where  is the empty set. If there are three rows representing sets A, B, C then the sets must be pairwise mutually exclusive: AB=, AC=, BC=.

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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1.5 Fraction and Probability

1.5.1 Fraction

From #A we derive the fraction or portion or proportion P(A) of subjects in A as the ratio of the number subjects in the set to the number of subjects in T :

(3)

Dividing (1) by #T gives

P(A) + P(A' ) = 1 (4)

so that P(A' ) = 1 - P(A) (5)

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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Returning to the students data set in §0.1, we write

P(T) = 1 P(T' ) = 1 - 1 = 0

P(M) = .5 = 50% P(M' ) = 1 - .5 = .5 = 50%

P(FE) = .5 = 50% P(FE') = 1 - .5 = .5 = 50%

P(FT) = .6 = 60% P(FT' ) = 1 - .6 = .4 = 40%

P(PT) = .4 = 40% P(PT' ) = 1 - .4 = .6 = 60%

Notice the expression of a fraction as a percent; we interpret % as .01, so that 40%=40(.01)=.4 . However, .4 expresses a percentage 40, a percent without the % sign. This is important, because MINITAB prints percentages (percents without the % sign).

M155 L2: Populations, Samples, Scales, Sets, Pie Charts, Bar Graphs -- Slide 20

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This information is often expressed by a table of fractions, showing the fraction of subjects falling within each cell. (A cell represents the intersection of a row category with a column category.) Then the row and column totals tell the fraction of subjects belonging to a row category or a column category.

EXAMPLE B3: Here we show both the count and the fraction (of all 10 subjects) in each cell.

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We can present the same information using percents:

MINITAB would present this information in percentages (percents without the % sign).

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Since the variables Sex and Load are in the students data set, we can use MINITAB to form the table. Choose Stat > Tables > Cross-tabulation, double-click on the variable names, and choose both Counts and Total percents (not to be confused with Row percentsnorColumn percents).

Rows: Sex Columns: Load

full part All

female 2 3 5

20.00 30.00 50.00

male 4 1 5

40.00 10.00 50.00

All 6 4 10

60.00 40.00 100.00

Cell Contents --

Count

% of Tbl

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1.5.2 Probability

EXAMPLE C3: A college of 1000 students has 700 day students (and the rest night students). Also 600 of the students are men (and the rest women). Also exactly 400 students are both men and day students. If we choose one student at random, what is the probability that the student chosen will be a woman and a night student? What is the probability that the student chosen will be a man or a day student?

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SOLUTION C3: In Example C1 we developed the following table of counts:

The probability of a randomly chosen student being a woman and a night student is 100/1000 = 0.100 .

The probability that the student chosen will be a man or a day student is (300 + 400 + 200)/1000 = 0.900.

slide27

An alternative approach is to divide all the counts in the table above by the table total 1000, giving a table of fractions or probabilities.

Now the table of fractions tells us that a randomly chosen student has probability 0.100 of being a woman and a night student.

And adding the numbers in the shaded region tells us that a randomly chosen student has probability 0.300 + .400 + 0.200 = 0.900

of being a man or day student.

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EXAMPLE D1: Often we are given fractions or probabilities without counts, and we can draw only the table of probabilities. Suppose 30% of the freshman class smoke, 25% drink, and 10% both smoke and drink. What is the probability that a freshman selected at random is either a smoker or a drinker (or both)? In answering such a question, don't try to work it in your head.

SOLUTION D1: First find the events mentioned in the problem, and then draw a table of probabilities, without any of the given numbers.

slide29

Having drawn the table in a CORRECT way, we can insert the fractions or probabilities in their places, and complete the table by subtraction

the probability that a randomly chosen student is a smoker or a drinker:

P(SD) = 0.15 + 0.10 + 0.20 = 0.45

The probability that a freshman chosen at random had exactly one of these health-threatening behaviors (meaning not both) is:

0.15 + 0.20 = 0.35 .