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A few years ago, 120 researchers and 400 undergraduates were presented with six false interpretations of a confidence interval. On average, the group endorsed more than three of the six falsehoods, with researchers worse than undergrads. If PhD researchers struggle with inference, what hope is there for senior students studying Applications to identify an association or for those studying Methods and Specialist to interpret a confidence interval? Intro Inference Page 1

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A few years ago, 120 researchers and 400 undergraduates were presented with six false interpretations of a confidence interval.

On average, the group endorsed more than three of the six falsehoods, with researchers worse than undergrads.

If PhD researchers struggle with inference, what hope is there for senior students studying Applications to identify an association or for those studying Methods and Specialist to interpret a confidence interval?

Intro

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Methods Year 11Conditional probability and independenceCan we infer from data that events are independent?

Applications Year 12Identifying and describing associations between two categorical variablesCan we infer from differences across categories that an association exists?

Identifying and describing associations between two numerical variablesCan we infer from the correlation coefficient that a linear association exists between two variables?Methods Year 12Random samplingWhy sample? Why might our sample not be random? Does it matter?

Confidence interval for a proportionCan we infer a population proportion from a sample proportion?

Specialist Year 12Confidence interval for a meanCan we infer a population mean from a sample mean?

Statistical Inference - drawing conclusions about a population from a sample.

Australian Senior Curriculum :: Inference Topics

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Conditional probability and independence: use relative frequencies obtained from data as point estimates of conditional probabilities and as indications of possible independence of events.

Observed Year 11 Year 12 Total

Science 91 75 166

Non-science 10 21 31

Total 101 96 197

Expected Year 11 Year 12 Total

Science 166

Non-science 31

Total 101 96 197

(𝑖𝑣) 𝑃(𝑆𝑐|𝑌𝑟11) = 91 ÷ 101 ≈ 0.90

(ii) 𝑃(𝑆𝑐) = 166 ÷ 197 ≈ 0.84

Recall the chi-square test of independence ?

χ =(𝑂 − 𝐸)

𝐸⎯⎯⎯⎯⎯⎯⎯⎯

 

 

How close for independence?

𝑛(𝑆𝑐 ∩ 𝑌𝑟11) =101 × 166

197⎯⎯⎯⎯⎯⎯⎯⎯⎯≈ 85

NB If 𝐴 ∗ 𝐵 denotes 𝐴 is independent of 𝐵 then

it follows that 𝐵 ∗ 𝐴, 𝐴 ∗ 𝐵, �̅� ∗ 𝐵

and so on.

Methods Unit 1 - Independence of events

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If 'survey' a census then we would have to answer NOT independent as 𝑃(𝑆𝑐) ≠ 𝑃(𝑆𝑐|𝑌𝑟11).

But if a sample, then we have a case for independence as our test indicates such a difference would occur by chance about 2% of the time.

Col % Year 11 Year 12

Science 90% 78%

Non-science 10% 22%

Total 100% 100%

Row % Year 11 Year 12 Total

Science 55% 45% 100%

Non-science 32% 68% 100%

ClassPad: Stats, Calc, Test, 𝜒 Test.Matrix A

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Do we need an explanatory and response variable?

Can we have an explanatory and response variable?

Example: 2002 General Social Survey about gun ownership and gun laws

Counts are difficult to interpret, especially with unequal numbers of observations in the rows and columns. Percentages are more useful than counts for describing how two categorical variables are related.

Conditional Percentages as evidence of an associationDefinition : Two categorical variables are associated in a sample if at least two rows (columns) noticeably differ in the pattern of row (column) percentages.

For gun ownership, the two columns clearly have different sets of columns percentages. Gun ownership and opinion about gun laws are associated.

Applications U3 Categorical Variables

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18

41

41

TAS GEN 2018 Q1

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Categorical data has morphed into numerical data, ready for bivariate analysis!

Apps 2017 Q5

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So many handy guides to 'strength' of correlation.

But what about sample size?

𝑛 = 2 and 𝑟 = ±1?

Valid?!

Strong, negative, non-linear assn.

Assn Numerical Variables - Apps Unit 3

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Excel simulation

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(a)

(b)

Apps 2017 Q9 (parts of)

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Altitude v max temp

Apps 2019 Q16

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the 95% confidence interval for the mean ranges from 0.1 to 0.4

Statements 1-4 assign probabilities to parameters or hypotheses.Statements 4-6 include specific boundaries.

A correct statement is

“If we were to repeat the experiment over and over, then 95 % of the time the confidence intervals contain the true mean.”

Once an interval is computed for a particular sample, it either

Professor Bumbledorf reports that

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Once an interval is computed for a particular sample, it either contains the true mean or it does not; there is no longer anything random about it.

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What does the level of confidence mean?

Take 100 samples from the same population•Calculate sample statistic e.g. proportion•Use the 100 sample proportions to construct 100 CI's•95 of these 100 CI's are expected to contain the true population parameter

•

Confidence is in the method - not any particular interval.

Assumptions made when constructing CI's (using z-scores)Random - data comes from random sample of populationNormal - sampling distribution is normalIndependent - individual observations are independent

Level of Confidence

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Independent - individual observations are independent(i.e. sample size < 10% of population)

Normal assumptionFor proportion, commonly check that and .(Note both courses require approximate CI's based on normality)

For mean, if population is normal then OK, otherwise central limit theorem says better approximation as increases… …

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Methods Unit 4 - CI for proportion

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�̂� = 56 ÷ 250 ≈ 0.224

𝐸 = 0.0517

𝑠 = 0.0264, 𝑧 . = 1.96

(0.1723, 0.2757)

No. The interval in (a)(ii) doesn't contain the true proportion, but that is to be expected in 5% of such samples. The CI is valid because the sample was random, it was sufficiently large to assume normality and we can assume independence of households selected.

Methods 2019 Q8

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VIC METH 2018 P2 Q4

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𝑋~𝐵(500, 0.6) and 𝑃(�̂� < 0.58) = 𝑃(𝑋 < 0.58 × 500) = 𝑃(𝑋 ≤ 289) = 0.1688

WA METHODS ATAR EXAM 2018 - Q17

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Spec U4 CI Mean

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58% average score on this question compared to exam average of 63%.

𝑋 = 40.62 − (1.08 ÷ 2) = 40.08

0.54 = 2.5758 ×𝑠

√400⎯⎯⎯ ⎯⎯⎯⎯⎯⇒ 𝑠 = 4.19

(39.54, 40.62)

True - interval is wider

False - intervals are constructed from random samples and so may or may not overlap

No way to tell. We do not know 𝜇 and intervals are constructed from random samples.

WA SPEC ATAR EXAM 2018 - Q16

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