© Nuffield Foundation 2012 Nuffield Free-Standing Mathematics Activity Successful HE applicants ©...

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© Nuffield Foundation 2012 Nuffield Free-Standing Mathematics Activity Successful HE applicants © Rudolf Stricker

Transcript of © Nuffield Foundation 2012 Nuffield Free-Standing Mathematics Activity Successful HE applicants ©...

Page 1: © Nuffield Foundation 2012 Nuffield Free-Standing Mathematics Activity Successful HE applicants © Rudolf Stricker.

© Nuffield Foundation 2012

Nuffield Free-Standing Mathematics Activity

Successful HE applicants

© Rudolf Stricker

Page 2: © Nuffield Foundation 2012 Nuffield Free-Standing Mathematics Activity Successful HE applicants © Rudolf Stricker.

© Nuffield Foundation 2010

Are 50% of the applications for courses in higher

education from females?

Are the same proportion of male and female applicants

successful?

Carrying out significance tests on proportions, and the difference between proportions, can help to answer such questions.

In this activity you will use data on successful applications for courses in higher education to carry out significance tests of this type.

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Distribution of a sample proportion

p

ps

n

pqp 3

If the sample size, n, is large (>30)the sample proportion, ps, will follow a normal distribution

n

pqand standard deviation

with mean p

n

pqp 3

where q = 1 – p

Think aboutWhy is it important that the sample is large?

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Calculate the test statistic:

npq

pspz

Summary of method for testing a proportionState the null hypothesis:

H0: population proportion, p = value suggested

and the alternative hypothesis:

H1: p ≠ value suggested (2-tail test)

or p < value suggested or p > value suggested (1-tail test)

Think aboutCan you explain this formula?

where ps is the proportion in a sample of size n and q = 1 – p

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If the test statistic is in the critical region (tail of the distribution)

Compare the test statistic with the critical value of z:

Summary of method for testing a proportion

For a 1-tail test

1% level, critical value = 2.33 or –2.33

reject the null hypothesis and accept the alternative.

5% level, critical value = 1.65 or –1.65

For a 2-tail test

5% level, critical values = 1.96

1% level, critical values = 2.58

-1.65 z0

5%

95%

95%

z0 1.96

2.5%

-1.96

2.5%

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Testing a proportion: Example

In 2010 a newspaper article said that the proportion of people accepted on higher education courses over 20 years old was 16%.

Null hypothesis H0: p = 0.16

Alternative hypothesis H1: p < 0.16

Test statistic

= –2.77

npq

pspz

From 2010 data:

q = 1 – p

p = 0.16

= 0.84

415 150.840.160.160.1518

z

n = 15 415

1-tail test

Using 2010 data to test this percentage:

Think aboutWhy is a 1-tail test used?

= 0.1518 415 152340 ps

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Test statistic = –2.77

–2.77

The test statistic, z, is in the critical region.

The result is significant at the 1% level, so reject the null hypothesis.

For a 1-tail 1% significance test:

ConclusionThere is strong evidence that the proportion reported is too high.

npq

pspz

–2.33 z0

1%

99%

Think aboutExplain the reasoning behind this conclusion.

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samples in number Totalattribute with itemsof number Totalp

Calculate the test statistic:

Compare with the critical value of z.

State the null hypothesis: H0: pA = pB

H1: pA ≠ pB

Summary of method for testing the difference between proportions

and alternative hypothesis:

or pA < pB or pA > pB

2-tail test

1-tail test

where

BA

SBSA

nnpq

ppz

11 q = 1 – p

(pA – pB = 0)

Think aboutExplain the formula for the test statistic. If the test statistic is in the critical region

reject the null hypothesis and accept the alternative.

pSA , pSB , nA and nB are values from the samples.

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Using 2010 data to test whether the proportion of males that were accepted is equal to the proportion of females that were accepted.

Test statistic:

H0: pM = pF (pM – pF = 0)

H1: pM ≠ pF 2-tail test

Testing the difference between proportions: Example

FM

SFSM

nnpq

ppz

11

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Testing the difference between proportions: Example

In the 2010 sample: 7062 out of 45 455 males and 8353 out of 54 030 females were successful.

030 54 455 458353 7062

p = 0.154 948

q = 1 – 0.154 948 = 0.845 052

455 457062SMp = 0.155 362

030 548353SFp = 0.154 599

Test statistic:

FM

SFSM

nnpq

ppz

11

030 541

455 451052 0.845948 0.154

599 0.154362 0.155z = 0.331

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z = 0.331

95%

z0 1.96

2.5%

- 1.96

2.5%

0.331

Conclusion There is no significant difference between the proportion of males and the proportion of females accepted.

The test statistic is not in the critical region.

Think aboutExplain the reasoning behind this conclusion.

For a 5% testTest statistic:

Testing the difference between proportions: Example

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At the end of the activity

• What are the mean and standard deviation of the distribution of a sample proportion?

• Describe the steps in a significance test for a sample proportion.

• Describe the steps in a significance test for the difference between sample proportions.

• When should you use a one-tail test and when a two-tail test?

• Would you be more confident in a significant result from a 5% significance test or a 1% significance test? Explain why.

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