Lecture Material Stat PPT

Types of data

Gender male/female

continuousAge in years

ContinuousTime is on a continuous scaleWeight nearest Kg

continuousWeekly expenditure

Continuous

Again a continuous scale but you might argue that it is discreteNumber of siblings

discreteParty Labour, Cons, Lib Dem, other

categoricalTerm accommodation Halls, Home, Private rented, other

categoricalAssignment grade A, B, C, D, E

Ranked because we put grade A as better than grade B better than C etc

Tabulation of data

MalesFemalesTotalHalls75100175Home303868Private rented21930Other437Total130150280

X 360

Pie chart angle formula

Halls 75/130 x360 =208Home 30/130x360= 83Private 21/130x360=58Other 4/130x360= 11

Must add up to 360Pie chart angle calculations

Pie Chart

Chart3

75

30

21

4

males

Sheet1

malesfemales

Halls75100

Home3038

Private rented219

other43

Sheet1

males

Sheet2

females

Sheet3

Presentation of informationA simple bar chart has non touching bars with the height of each bar proportional to the frequency. In Excel a bar chart with vertical bars is called a Column chartAnd with horizontal bars is called a Bar chartMultiple Bar Chart: the bars are split into several to show another variable. In Excel such a bar chart is known as a clustered column with vertical bars and a clustered bar with horizontal bars.Component Bar Chart:components are stacked together to show another variable. In Excel such a bar chart is known as a stacked column with vertical bars and a stacked bar chart with horizontal bars.Percentage Component Bar charts convey the information in percentage form rather than the actual frequency values and thus highlight differences in proportions of one variable.

Frequency table

TimeFrequency BoundariesClass sizeClass midpoint0-

Open ended classesUnder 5 becomes 0 to under 5

40 and over becomes 40 to under 45

Histogramof delivery times

Chart1

0

3

3

7

7

10

10

12

12

8

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2

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2

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0

Delivery times (days)

Frequency

Histogram of delivery times

histogram1

0

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0 0 0 0 0 0 0 0 0

time (days)

frequency

histogram of delivery times

histogram blank grid

00

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histogram blank grid

0

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blank grids

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blank grids

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Sheet1

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histograms2

Time (days)Frequency50

5053

103103

157107

2010157

25121510

3082010

3562012

4022512

452258

308

306

356

352

402

402

452

450

histograms2


Frequency


cumultive frequencytable


0

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cumultive f graph

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cumultive f graph measures

uppper boundarycumulative frequency

delivery timefrequencyless than 550

5 -< 103less than 10103

10 -

Frequency polygon of delivery times

Chart1

0

3

7

10

12

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6

2

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0

frequency


frequency

frequency polygon

histogram equal

50

53Histogram equal classes

103

107

157

1510

2010

2012

2512

258

308

306

356

352

402

402

452

450

histogram equal


Frequency


histogram1

0

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0 0 0 0 0 0 0 0 0

time (days)

frequency


unequal classes

50

100

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500

01.5

51.5

101.5

107

157

1510

2010

2012

2512

258

308

303.3

353.3

403.3

453.3

450

00.3

50.3

100.3

101.4

151.4

152

202

202.4

252.4

251.6

301.6

300.7

350.7

400.7

450.7

450

unequal classes

0

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histograms2

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1.5

1.5

1.5

7

7

10

10

12

12

8

8

3.3

3.3

3.3

3.3

0

delivery times (days)

frequency per class size 5

histogram using standard class size 5

cumultive f graph

0.3

0.3

0.3

1.4

1.4

2

2

2.4

2.4

1.6

1.6

0.7

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0.7

0


frequency density

histogram using frequency density

frequecny polygon

frequecny polygon

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5 -< 103less than 10103

10 -

Unequal class sizes

Delivery timeNumber of deliveriesFrequency densityAdjusted height using standard class width 50-

Blank grids

Chart2

0

0

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histogram equal

50


103

107

157

1510

2010

2012

2512

258

308

306

356

352

402

402

452

450

histogram equal


Frequency


histogram1

0

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time (days)

frequency


blank grids

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blank grids

histograms2



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cumultive f graph

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5 -< 103less than 10103

10 -

Histogram standard class size 5

Chart4

1.5

1.5

1.5

7

7

10

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12

12

8

8

3.3

3.3

3.3

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0




histogram equal

50


103

107

157

1510

2010

2012

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356

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452

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histogram equal


Frequency


histogram1

0

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0 0 0 0 0 0 0 0 0

time (days)

frequency


blank grids

50

100

150

200

250

300

350

400

450

500

01.5

51.5

101.5

107

157

1510

2010

2012

2512

258

308

303.3

353.3

403.3

453.3

450

blank grids

histograms2





cumultive f graph

cumultive f graph

0

0

0

0

0

0

0

0

0

0


0

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5 -< 103less than 10103

10 -

Histogram using Frequency density

Chart6

0.3

0.3

0.3

1.4

1.4

2

2

2.4

2.4

1.6

1.6

0.7

0.7

0.7

0.7

0


frequency density


histogram equal

50


103

107

157

1510

2010

2012

2512

258

308

306

356

352

402

402

452

450

histogram equal


Frequency


histogram1

0

0

0

0

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0 0 0 0 0 0 0 0 0

time (days)

frequency


blank grids

50

100

150

200

250

300

350

400

450

500

01.5

51.5

101.5

107

157

1510

2010

2012

2512

258

308

303.3

353.3

403.3

453.3

450

00.3

50.3

100.3

101.4

151.4

152

202

202.4

252.4

251.6

301.6

300.7

350.7

400.7

450.7

450

blank grids

histograms2





cumultive f graph


frequency density




0

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5 -< 103less than 10103

10 -

Guidelines for grouping dataSo, in the example,Largest observation = 41 Smallest = 5Require say 8 classesclass width =

Cumulative frequency table

Delivery timeFrequency CumulativeFrequencyLess than 505-

Cumulative frequency graph

Finding measures from the cumulative frequency graph

Measures for this examplemedian look at cumulative frequency of 25 on graph22 daysupper quartile -cumulative frequency of 37.5 on graph28 dayslower quartile - cumulative frequency of 12.5 on graph16 daysinter-quartile range is UQ LQ 28-16=12 days20th percentile look at cumulative frequency of 10 on graph 15 dayslook at 30 days horizontal axis to give 40 deliveries so 50-40 =10 deliveries are more than 30 days90% of deliveries take less 34 days

Measures of locationThe mode : Most frequently occurring item 35

The median: Middle number. The mean

28 28 35 35 35 36 39 44 44 50

Using frequency formula

XFFX2825635310536136391394428850150

Measure from a grouped frequency table

TimeFrequencyFMidpointXFXCumulative frequency0-

MeasuresMean =

Mode is estimated to be 22.5, the middle of the modal class

Median

Which measure is best3 33 45 710 10 10 25 40Mean= 10.9~11Mode= 3, 10Median = 7

quartilesLower quartileUpper quartile

Measures of spread

ArmstrongBarrett3 6 3 4 4 6 4 2 4 5 3 5 4 4 3 5Ordered2 3 3 4 4 4 6 6 3 3 4 4 4 4 5 5Mean4 weeks4Mode44Median44ConcludeLittle differenceRange6-2 = 45-3=2Inter-quartile rangeStandard deviationCoefficient of variation

Standard deviation

X23344466-2-1-1000224110004414

1.32

Coefficient of variationThe higher the ratio, the greater the spread around the mean. Lengths mean=55standard deviation = 28.7coefficient of variation=52%Weightsmean=5.5standard deviation = 2.8.7coefficient of variation = 52%

Mean and Standard Deviation for Armstrong

XFFXFX20.51473.51.51522.533.752.51845112.53.516561964.51567.5303.755.51160.5332.756.51171.5464.75totals3301447

Standard deviation

Probabilty examplesExamplesThrow a coin. The probability of a head = 0.5

There are three counters in a bag, one red, one green and one blue. One counter is pulled out.The probability that the counter is red =

The counter is then replaced and a second pulled out.List all the outcomes: RR, RG, RB, GR, GG, GB, BR, BG, BB

the probability that both the first and the second were red =

exampleOver the last month (November) a machine has broken down on three days. What is the probability the machine breaks down?

exampleIn a sample of adults these probabilities were found: P(male) = 0.5P(Married)=0.6P(full time job)=0.9A person is selected at random. What is the probability that the person ismarried and male

ii) male and in a full time job

iii) female 0.5

exampleP(male)=0.7P(aged 40 to 59)=0.4P(aged 60 to 69)=0.15P(aged 70 or more)=0.1

Female1-0.7 = 0.3

Probability(40 to 59) or (60 to 69) years old0.4+0.15=0.55

Probability (female) and (40 to 59) years old0.30.4 = 0.12

male or aged 40 to 59? CANNOT SAY NOT 0.7 + 0.4 = 1.1

exampleThe probability that firm A makes a profit has been assessed to be 0.6. The probability that the firm breaks even is 0.3.What is the probability that the firm makes a loss?1- 0.6 - 0.3 = 0.1

example

Firm ProfitBreak evenlossA0.60.30.1B0.70.10.2

exampleboth firms make a profit 0.60.7=0.42firm A does not make a profit 1-0.6=0.4firm A makes a profit or breaks even0.6+0.3 = 0.9firm B does not make a profit 0.3neither firms make a profit 0.40.3=0.12at least one firm makes a profit1-0.12=0.88only one firm makes a profit 1- 0.42 0.12 = 0.46

Throw a dieExpected score is

expectationExpected number of minutes late00.4 + 30.3 + 50.15 + 100.1 + 150.05 = 3.4 minutes

03510150.40.300.150.10.05

Spread of valueswith =3000 hrs and =200 hrsapproximately 68% of the bulbs will last between 2800 hours and 3200 hours,

approximately 95% of the bulbs will last between 2600 hours and 3400 hours,

approximately 99.75% of the bulbs will last between 2400 hours and 3600 hours.

Using normal tablesiP(Z1.3) Read directly from the table =0.0968

iiiP(Z

Find Kfind k such that P(Z>K) = 0.15

also means P(Z

Solution to example = 2000 hours and = 250 hours(a) less than 1750 hours

from tables area to the left of -1 = 0.1587

more than 2350 hours,

from tables area to the left of -1.4 = 0.0808

Example (c)between 1800 hours and 2400 hours?

area to the left of -0.8= 0.2119

area to left of 1.6 = 1- 0.0548 = 0.9452

area in between = 0.9452-0.2119 = 0.7333

Example(d)4% fail

P(Z

Example (e) best 6%P(Z>k) = 0.06 means also that P(Z

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Lecture Material Stat PPT