8.11

Using Statistics To Make Inferences 8

Summary  

Contingency tables.Goodness of fit test.

Thursday 20 April 2023 06:29 PM

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Goals To assess contingency tables for independence.To perform and interpret a goodness of fit test.

 Practical

Construct and analyse contingency tables.

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Recall

To compare a population and sample variance we employed?χ2Cc

cc

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Today

The probability approach from last week is employed to tell if “observed” data confirms to the pattern “expected” under a given model.

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Categorical Data - Example

Assessed intelligence of athletic and non-athletic schoolboys.

bright stupid Total

athletic

581 567 1148

lazy 209 351 560

Total 790 918 1708K. Pearson “On The Relationship Of Intelligence To Size And Shape Of Head, And To Other Physical And Mental Characters”, Biometrika, 1906, 5, 105-146, data on page 144.

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Procedure1. Formulate a null hypothesis. Typically

the null hypothesis is that there is no association between the factors.

2. Calculate expected frequencies for the cells in the table on the assumption that the null hypothesis is true.

3. Calculate the chi-squared statistic. This is for an r x c table with entries in row i and column j.

r

i

c

j jijijiobserved

1 1

22

,expected,expected,

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Procedure4. Compare the calculated statistic with

tabulated values of the chi-squared distribution with ν degrees of freedom.

ν   =  (rows ‑ 1)(columns ‑ 1) = (r – 1)(c – 1)

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Key Assumptions1. Independence of the observations. The data

found in each cell of the contingency table used in the chi-squared test must be independent observations and non-correlated.

2. Large enough expected cell counts. As described by Yates et al., "No more than 20% of the expected counts are less than 5 and all individual expected counts are 1 or greater" (Yates, Moore & McCabe, 1999, The Practice of Statistics, New York: W.H. Freeman p. 734).

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Key Assumptions3. Randomness of data. The data in the table

should be randomly selected.

4. Sufficient Sample Size. It is also generally assumed that the sample size for the entire contingency table is sufficiently large to prevent falsely accepting the null hypothesis when the null hypothesis is true.

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Example

Assessed intelligence of athletic and non athletic schoolboys.

 Observed

bright stupid Total

athletic

581 567 1148

lazy 209 351 560

Total 790 918 1708

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ProbabilitiesThe probability a random boy is athletic is

6721.017081148

The probability a random boy is bright is

4625.01708790

Assuming independence, the probability a random boy is both athletic and bright is

98.5301708

7901148

3109.01708790

17081148

bright stupid Total

athletic

581 567 1148

lazy 209 351 560

Total 790 918 1708

For 1708 respondents the expected number of athletic bright boys is

CCCCCCCCCCCCCCC

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Expected

bright stupid Total

athletic

530.98 1148

lazy 560

Total 790 918 1708

The expected number of athletic bright boys is

98.5301708

7901148

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Expected

bright stupid Total

athletic

530.98 ? 1148

lazy 560

Total 790 918 1708

The expected number of athletic stupid boys is

8.1414

Expected

bright stupid Total

athletic

530.98 617.02 1148

lazy 560

Total 790 918 1708

The expected number of athletic stupid boys is

1148 – 530.98 = 617.02

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Expected

bright stupid Total

athletic

530.98 617.02 1148

lazy ? 560

Total 790 918 1708

The expected number of lazy bright boys is

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Expected

bright stupid Total

athletic

530.98 617.02 1148

lazy 259.02 ? 560

Total 790 918 1708

The expected number of stupid lazy boys is

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Expected

bright stupid Total

athletic

530.98 617.02 1148

lazy 259.02 300.98 560

Total 790 918 1708

The expected number of stupid lazy boys is

918 – 617.02 = 300.98

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Expected

bright stupid Total

athletic

530.98 617.02 1148

lazy 259.02 300.98 560

Total 790 918 1708

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χ2

73.26

98.30098.300351

02.25902.259209

02.61702.617567

98.53098.530581

22

22

2

calc

Expected

Expected - Observed 2

111 cr

1708918790Total

560351209lazy

1148567581athletic

Totalstupidbright

1708918790Total

560351209lazy

1148567581athletic

Totalstupidbright

Observed Expected

1708918790Total

560300.98259.02lazy

1148617.02530.98athletic

Totalstupidbright

1708918790Total

560300.98259.02lazy

1148617.02530.98athletic

Totalstupidbright

Only one cell is free.

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χ2

As a general rule to employ this statistic,all expected frequencies should exceed 5.

If this is not the case categories are pooled (merged) to achieve this goal. See the Prussian data later.

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Conclusion73.262 calc 1

84.305.2

1

ν p=0.1

p=0.05

p=0.025

p=0.01

p=0.005

p=0.002

1 2.706 3.841 5.024 6.635 7.879 9.550

The result is significant (26.73 > 3.84) at the 5% level. So we reject the hypothesis of independence between athletic prowess and intelligence.

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SPSSRaw data

Note v1 are the row labelsv2 are the column labelsv3 is the frequency

for each cell

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SPSSData > Weight Cases

Since frequency data has been input, necessary to weight.This is essential, do not use percentages.

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SPSSAnalyze > Descriptive Statistics > Crosstabs

Set row and column variables.

Frequencies already set.

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SPSS

Select chi-square

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SPSS

SelectObserved – input dataExpected – output data,

under the model

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SPSS

V1 * V2 Crosstabulation

581 567 1148

531.0 617.0 1148.0

209 351 560

259.0 301.0 560.0

790 918 1708

790.0 918.0 1708.0

Count

Expected Count

Count

Expected Count

Count

Expected Count

athletic

lazy

V1

Total

bright stupid

V2

Total

Expected cell frequencies

Expected under the model.

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SPSS

Chi-Square Tests

26.736b 1 .000

26.204 1 .000

26.973 1 .000

.000 .000

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Pearson Chi-Square

Continuity Correctiona

Likelihood Ratio

Fisher's Exact Test

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)Exact Sig.(2-sided)

Exact Sig.(1-sided)

Computed only for a 2x2 tablea.

0 cells (.0%) have expected count less than 5. The minimum expected count is 259.02.

b.

Pearson Chi Square is the required statistic

Do not report p = .000, rather p < .001

Note Fisher’s exact test, only available in SPSS for 2x2 tables (see next slide).

ff

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What If We Have Small Cell Counts?

Fisher's exact test

The Fisher's exact test is used when you want to conduct a chi-square test but one or more of your cells has an expected frequency of five or less. Remember that the chi-square test assumes that each cell has an expected frequency of five or more, but the Fisher's exact test has no such assumption and can be used regardless of how small the expected frequency is. In SPSS, unless you have the SPSS Exact Test Module, you can only perform a Fisher's exact test on a 2x2 table, and these results are presented by default.

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AsideTwo dials were compared. A subject was asked to read each dial many times, and the experimenter recorded his errors. Altogether 7 subjects were tested. The data shows how many errors each subject produced. Do the two conditions differ at the 0.05 significance level (give the appropriate p value)?

Observed data1 2 3 4 5 6 736 31 31 29 32 25 2629 35 34 35 34 35 30

What key word describes this data?

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Aside

What tests are available for paired data?

One sample t test

Sign test

Wilcoxon Signed Ranks Test

CCCCCCCCCc

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Aside

What tests are available for paired data? What assumptions are made?One sample t test

Sign test

Wilcoxon Signed Ranks Test

normality

Resembles the SignTest in scope, but it is much more sensitive. In fact, for large numbers it is almost as sensitive as the Student t-test

No assumption of normality

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Aside

What tests are available for paired data? One sample t test

Sign test

Wilcoxon Signed Ranks Test

Sign test answers the question How Often?, whereas other tests answer the question How Much?

One sample t test – meanWilcoxon Signed Ranks Test - median

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Example

The table is based on case-records of women employees in Royal Ordnance factories during 1943-6. The same test being carried out on the left eye (columns) and right eye (rows).

Stuart “The estimation and comparison of strengths of association in contingency tables”, Biometrika, 1953, 40, 105-110.

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ObservedHighes

tSecon

dThird Lowes

tTotal

Highest

1520 266 124 66 1976

Second

234 1512 432 78 2256

Third 117 362 1772 205 2456

Lowest

36 82 179 492 789

Total 1907 2222 2507 841 7477

Is there any obvious structure?

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Expected

In general to find the expected frequency in a particular cell the equation is

Row total x Column total / Grand total

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Expected

7477841250722221907Total

7894921798236Lowest

24562051772362117Third

2256784321512234Second

1976661242661520Highest

TotalLowestThirdSecondHighest

7477841250722221907Total

7894921798236Lowest

24562051772362117Third

2256784321512234Second

1976661242661520Highest

TotalLowestThirdSecondHighestIn general to find the expected frequency in a particular cell the equation is

Row total x Column total / Grand total

So for highest right and bottom left the equation becomes

1976 x 1907 / 7477 = 503.98

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ExpectedHighest Secon

dThird Lowes

tTotal

Highest

503.98 ? 1976

Second

? 2256

Third ? 2456

Lowest

? ? ? ? 789

Total 1907 2222 2507 841 7477

Row total x Column total / Grand total1976 x 1907 / 7477 = 503.98

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ExpectedHighest Secon

dThird Lowes

tTotal

Highest

503.98 587.22 662.54 ? 1976

Second

575.39 670.43 756.43 ? 2256

Third 626.40 729.87 823.48 ? 2456

Lowest

? ? ? ? 789

Total 1907 2222 2507 841 7477

Row total x Column total / Grand total

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ExpectedHighest Secon

dThird Lowes

tTotal

Highest

503.98 587.22 662.54 ? 1976

Second

575.39 670.43 756.43 ? 2256

Third 626.40 729.87 823.48 ? 2456

Lowest

? ? ? ? 789

Total 1907 2222 2507 841 7477

The missing values are simply found by subtraction

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ExpectedHighest Secon

dThird Lowes

tTotal

Highest

503.98 587.22 662.54 ? 1976

Second

575.39 670.43 756.43 2256

Third 626.40 729.87 823.48 2456

Lowest

789

Total 1907 2222 2507 841 7477

1976 – 503.98 – 587.22 – 662.54 = 222.26

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Expected

Highest Second

Third Lowest

Total

Highest

503.98 587.22 662.54 222.26 1976

Second

575.39 670.43 756.43 2256

Third 626.40 729.87 823.48 2456

Lowest

789

Total 1907 2222 2507 841 7477

1976 – 503.98 – 587.22 – 662.54 = 222.26

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ExpectedHighest Secon

dThird Lowes

tTotal

Highest

503.98 587.22 662.54 222.26 1976

Second

575.39 670.43 756.43 ? 2256

Third 626.40 729.87 823.48 ? 2456

Lowest

? ? ? ? 789

Total 1907 2222 2507 841 7477

Similarly for the remaining cells

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ExpectedHighest Secon

dThird Lowes

tTotal

Highest

503.98 587.22 662.54 222.26 1976

Second

575.39 670.43 756.43 253.75 2256

Third 626.40 729.87 823.48 276.25 2456

Lowest

201.23 234.47 264.55 88.75 789

Total 1907 2222 2507 841 7477

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Short Cut

Contributions to the χ2 statistic,

for the top left cell the contribution is

expected

expectedobserved 2

32.2048

98.50398.5031520 2

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Conclusion32.20482 calc 911 cr

ν p=0.1 p=0.05 p=0.025

p=0.01 p=0.005

p=0.002

9 14.684 16.919 19.023 21.666 23.589 26.056

92.1605.2

9

The above statistic makes it very clear that there is some relationship between the quality of the right and left eyes.

For the top left cell only.

Nine cells are free.

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Highest

Second Third Lowest Total

Highest

2048.32

175.72 437.75 109.86

Second 202.55 1056.38

139.14 121.73

Third 414.25 185.41 1092.53

18.38

Lowest 135.67 99.15 27.66 1832.37

Total 8097

Total χ2

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Conclusion87.80962 calc 911 cr

ν p=0.1 p=0.05 p=0.025

p=0.01 p=0.005

p=0.002

9 14.684 16.919 19.023 21.666 23.589 26.056

92.1605.2

9

The above statistic makes it very clear that there is some relationship between the quality of the right and left eyes.

For all cells.

Nine cells are free.

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SPSS

Raw data

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SPSSExpected cell frequencies

V1 * V2 Crosstabulation

1520 36 234 117 1907

504.0 201.2 575.4 626.4 1907.0

66 492 78 205 841

222.3 88.7 253.8 276.2 841.0

266 82 1512 362 2222

587.2 234.5 670.4 729.9 2222.0

124 179 432 1772 2507

662.5 264.5 756.4 823.5 2507.0

1976 789 2256 2456 7477

1976.0 789.0 2256.0 2456.0 7477.0

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

Count

Expected Count

Highest

Lowest

Second

Third

V1

Total

Highest Lowest Second Third

V2

Total

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SPSS

Pearson Chi Square is the required statistic

Chi-Square Tests

8096.877a 9 .000

6671.512 9 .000

7477

Pearson Chi-Square

Likelihood Ratio

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

0 cells (.0%) have expected count less than 5. Theminimum expected count is 88.75.

a.

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Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume.

Typical applications are to queues/arrivals. The number of phone calls received per day.The occurrence of accidents/industrial injuries.More exotically, birth defects and the number of genetic mutations. The occurrence of rare diseases.

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Poisson Distribution

1 discrete events which are independent.

2 events occur at a fixed rate λ per unit continuum.(λ lambda)

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Poisson Distribution x successes

!

;Probxe

xx

e is approximately equal to 2.718

λ is the rate per unit continuum

the mean is λ the variance is λ

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Casio 83ES

exp or “e”

exp(1) = 2.7182818

exp(2) = 7.389056

Its inverse, on the same key is ln, so

ln(2.7182818) = 1

ln(7.389056) = 2

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Alternate applications

A similar approach may be employed to test if simple models are plausible.

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χ2 Goodness of Fit Test

ii

iicalc E

EO 2

2

The degrees of freedom are ν = m – n – 1, where there are m frequencies left in the problem, after pooling, and n parameters have been fitted from the raw data.

For example…

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Example

The number of Prussian army corps in which soldiers died from the kicks of a horse in a year.

 Typical “industrial injury” data

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Which distribution is appropriate?

Is the data discrete or continuous?

Discrete, since a simple countccccccccccccccccccccccc

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Check list of distributionsDiscrete Continuous

Binomial Normal

Poisson Exponential

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Check list of distribution parameters

Discrete Continuous

Binomial Normal

Poisson Exponential

n p μ σ2

λ

cccccccccccccccccccccccccc

Discrete, no “n” implies Poissonccccccc

λcccccccccccccccccccccccccc

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Observed Data

Number deaths in a corps

Observed frequency (Oi)

0 144

1 91

2 32

3 11

4 2

5 or more 0

Total 280

We need to estimate the Poisson parameter λ. Which is the mean of the distribution.

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Observed Data

Number deaths in a corps

Observed frequency (Oi)

0 144

1 91

2 32

3 11

4 2

5 or more 0

Total 280

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Mean

7.02113291144

241133229111440

ccccccccccccccccccccc280Total

05 or more24

113

322911

1440

Observed frequency (Oi)

Number deaths in a corps

280Total

05 or more24

113

322911

1440

Observed frequency (Oi)

Number deaths in a corps

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Expected

Number deaths in a

corps

Poisson model

Expected probability

0 0.4966

1 0.3476

2 0.1217

3 0.0284

4 0.0050

5 or more By subtraction

?

Total 1 1

e e

!2/2 e!3/3 e

!4/4 e

λ = 0.7 and “e” is a constant on your calculator

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Expected

Number deaths in a

corps

Poisson model

Expected probability

0 0.4966

1 0.3476

2 0.1217

3 0.0284

4 0.0050

5 or more By subtraction

0.0008

Total 1 1

e e

!2/2 e!3/3 e

!4/4 e

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Expected FrequencyExpected frequency for no deaths 280 x 0.4966 =

139.04

Number deaths in a

corps

Expected probabilit

y

Expected frequency (Ei)

0 0.4966 139.04

1 0.3476

2 0.1217

3 0.0284

4 0.0050

5 or more 0.0008

Total 1

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Expected FrequencyExpected frequency for remaining rows

280 × probability = frequency

Number deaths in a corps

Expected probability

Expected frequency (Ei)

0 0.4966 139.04

1 0.3476 97.33

2 0.1217 34.07

3 0.0284 7.95

4 0.0050 1.39

5 or more 0.0008 0.22

Total 1 280

Note the two expected frequencies less than 5!

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χ2 Calculation

Number deaths

in a corps

Observed

frequency (Oi)

Expected

frequency (Ei)

0 144 139.04 0.18

1 91 97.33 0.41

2 32 34.07 0.13

3 or more

13 9.56 1.24

Total 280 280 1.95

i

ii

EEO 2

Pool to ensure all expected frequencies exceed 5

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ConclusionHere m (frequencies) = 4, n (fitted parameters) = 1 then ν = m – n – 1 = 4 – 1 – 1 = 2

ν p=0.1 p=0.05 p=0.025

p=0.01 p=0.005

p=0.002

2 4.605 5.991 7.378 9.210 10.597 12.429

991.505.2

2 95.12 calc

The hypothesis, that the data comes from a Poisson distribution would be accepted (5.991 > 1.95).

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Next Week

Bring your calculators next week

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Read

Read Howitt and Cramer pages 134-152

Read Howitt and Cramer (e-text) pages 125-134

Read Russo (e-text) pages 100-119

Read Davis and Smith pages 434-448

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Practical 8

This material is available from the module web page.

http://www.staff.ncl.ac.uk/mike.cox

Module Web Page

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Practical 8

This material for the practical is available.

Instructions for the practical

Practical 8

Material for the practicalPractical 8

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Assignment 2

You will find submission details on the module web site

Note the dialers lower down the page give access to your individual assignment. It is necessary to enter your student number exactly as it appears on your smart card.

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Assignment 2As a general rule make sure you can perform the calculations manually.

It does no harm to check your calculations using a software package.

Some software employ non-standard definitions and should be used with caution.

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Assignment 2

All submissions must be typed.

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Whoops!

Researchers at Cardiff University School of Social Science claim errors made by the Hawk-Eye line - calling technology can be greater than 3.6mm - the average error quoted by the manufacturers.

Teletext, p388

12 June 2008

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Whoops!

Kate Middleton 'marries Prince Harry' on souvenir mugThe Telegraph - Thursday 17 March 2011

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Whoops!

Poldark - BBC - 8 March 2015