MAS1403 Quantitative Methods for Business Managementnag48/teaching/MAS1403/slides1beam.pdf ·...

MAS1403

Quantitative Methods for

Business Management

Semester 2, 2013–14

Lecturer: Dr. Andy Golightly

Lecturer details

Me: Dr Andrew (Andy) Golightly

Office: Room 2.32 Herschel

Phone: 0191 2087312

Email: andrew.golightly@ncl.ac.uk

www: www.mas.ncl.ac.uk/∼nag48/

My involvement with MAS1403

Lecturer since 2006

First taught MAS1403 in 2009/10,

Module leader in 2010/11, 2012/13, 2013/14

Helped set up

Computer based exam,Revision DVD (!)

New arrangements for 2014

Lecture time has changed! Monday 12 noon (Curtis)

Tutorial arrangements have changed:

– Tuesday 10, Herschel LT2

– Tuesday 2, Herschel TR2 (level 4)

– Thursday 10, Herschel LT2

Check your personal timetable to see your allocated slot!

Tutorial arrangements have changed:

– Tuesday 2, Herschel TR2 (level 4)

Check your personal timetable to see your allocated slot!

A Minitab practical will take the place of the tutorial in week 7

Further arrangements for 2014

There will be three CBAs

There will be one written assignment over the Easter holidays

There will be a computer based exam at the end of Semester 2covering material from the entire year !

You should refer to the week–by–week schedule for this coursefor CBA/assignment deadlines, computer practicals etc. etc.

The semester 2 webpage is available through Blackboard

1 Will the format be the same as semester 1?

1 Will the format be the same as semester 1? Yes!

2 Will semester 2 be harder than semester 1?

2 Will semester 2 be harder than semester 1? Only a little - wewill build on semester 1 material

3 Do I need to be able to derive all formulae etc to pass thecourse?

3 Do I need to be able to derive all formulae etc to pass thecourse? No! Remember, the exam is open book!

4 I only did GCSE maths, is that a problem?

4 I only did GCSE maths, is that a problem? Absolutely not...

5 Do I need to attend tutorials?

4 I only did GCSE maths, is that a problem? Absolutely not...

5 Do I need to attend tutorials? Defo! We also take anattendance register...

Lecture 1

INTERVAL ESTIMATION

Motivation and Aims

Consider a population of interest e.g.

Heights of all males in the UK,IQ,Starting salaries of UK graduates,Blood pressure after a particular drug treatment.

Motivation and Aims

Suppose we are interested in some summary of thepopulation.

Motivation and Aims

We take a random sample from the population and use thissample to say something about the summary of interest.

Motivation and Aims

We take a random sample from the population and use thissample to say something about the summary of interest.

Today: construction of a confidence interval for apopulation mean

Recap and introduction

Recall that data can be summarised in two ways:

1. Graphical summaries

Stem–and-leaf plots;

Bar charts;

Histograms;

Bar charts;

Histograms;

Relative frequency histograms;

Bar charts;

Histograms;

Relative frequency histograms;

Frequency polygons.

2. Numerical summaries

Measures of location

(i) Sample mean;(ii) Sample median;(iii) Sample mode.

2. Numerical summaries

Measures of location

(i) Sample mean;(ii) Sample median;(iii) Sample mode.

Measures of spread

(i) Range;(ii) Variance (and standard deviation);(iii) Interquartile range.

What does our sample tell us about the population?

We can rarely observe the entire population, so thepopulation mean and population variance are hardly everknown exactly ;

These unknown quantities are called parameters;

We use Greek letters to denote them – µ for the mean, andσ2 for the variance (and so σ for the standard deviation);

We hope that the sample mean (x̄) will be quite close to thetrue mean (µ);

But how do we know if it is?

The distribution of the sample mean

Let x1, x2, . . . , xn be a random sample from a N(µ, σ2)distribution. We can calculate the mean from this sample –call this x̄1;

Let x1, x2, . . . , xn be a random sample from another N(µ, σ2)distribution. We can calculate the mean from this sample too– call this x̄2;

We can calculate the means from many samples, and look atthe distribution of the x̄ ’s!

It turns out that, if the populations from which the sampleswere drawn follow normal distributions, then X̄ will also followa normal distribution; in fact,

X̄ ∼ N(µ, σ2/n).

It turns out that, if the populations from which the sampleswere drawn follow normal distributions, then X̄ will also followa normal distribution; in fact,

X̄ ∼ N(µ, σ2/n).

The Central Limit Theorem goes one step further and saysthat, if n is large, then this result will (approximately) hold no

matter what the ‘parent’ population distribution!

Interval estimation

x̄ is a point estimate of the population mean µ. We can improveestimation by constructing an interval estimate.

Interval estimation

To construct such an interval, we first calculate the samplemean x̄ ;

Interval estimation

We then go a little bit to the left of x̄ and a little bit to theright of x̄ to create an interval to (hopefully!) ‘capture’ µ;

Interval estimation

We then go a little bit to the left of x̄ and a little bit to theright of x̄ to create an interval to (hopefully!) ‘capture’ µ;

It’s more likely that µ will fall within this interval than exactly‘on top of’ the point estimate.

But how much do we go to the left and right? This depends on:

(i) The size of our sample;

(ii) How ‘confident’ we want to be that our interval captures µ,and

(iii) What (if anything) we know about the population.

But how much do we go to the left and right? This depends on:

(i) The size of our sample;

(ii) How ‘confident’ we want to be that our interval captures µ,and

(iii) What (if anything) we know about the population.

Regarding point (iii), we will begin by assuming that thepopulation variance σ2 is known

Construction of a confidence interval

We know from the Central Limit Theorem that

X̄ ∼ N

µ,σ2

X̄ ∼ N

µ,σ2

We can ‘standardise’ X̄ , using “slide–squash”, i.e.

Z =X̄ − µ√

σ2/n,

X̄ ∼ N

µ,σ2

Z =X̄ − µ√

σ2/n,

where Z ∼

X̄ ∼ N

µ,σ2

Z =X̄ − µ√

σ2/n,

where Z ∼ N(0,1).

Construction of a 95% confidence interval

We know that (from tables)

Pr(–1.96 < Z < 1.96) = 0.95;

We can think about this graphically:

Pr(–1.96 < Z < 1.96) = 0.95;

–1.96 <X̄ − µ√

σ2/n< 1.96

= 0.95;

Rearranging the LHS, we get

X̄ − 1.96×√

σ2/n < µ < X̄ + 1.96×√

= 0.95

Rearranging the LHS, we get

X̄ − 1.96×√

σ2/n < µ < X̄ + 1.96×√

= 0.95

If we want a 99% confidence interval, the only thing that willchange is the value 1.96.

Case 1: Known variance σ2 (bottom, page 4)

If we know the population variance σ2, we can just bung ournumbers into the formula on the previous slide! Remember, the(95%) confidence interval is

x̄ ± 1.96×√

σ2/n,

x̄ is the sample mean;

σ2 is the population variance, and

n is the sample size.

Example (page 5)

A coffee machine fills cups with hot water; the variance of thefilling process is known to be σ2 = 10 (ml)2.

A sample of 100 filled cups gives a sample mean and we havecalculated a sample mean of x̄ = 40ml.

What is the 95% confidence interval of the population mean µ?

Example (page 5)

We already have a formula for the 95% confidence interval:

x̄ ± 1.96√

σ2/n.

Example (page 5)

x̄ ± 1.96√

σ2/n.

So, inputting our values, we get

Example (page 5)

x̄ ± 1.96√

σ2/n.

40 ± 1.96√

10/100, i.e.

Example (page 5)

x̄ ± 1.96√

σ2/n.

40 ± 1.96√

10/100, i.e.

40 ± 0.61.

Example (page 5)

x̄ ± 1.96√

σ2/n.

40 ± 1.96√

10/100, i.e.

40 ± 0.61.

Hence, the 95% confidence interval for the population mean µ is(39.39, 40.61).

Example (page 5)

What would happen if the sample size increased to 200 andeverything else remained the same?

Example (page 5)

What would happen if the sample size increased to 200 andeverything else remained the same? We’d get

Example (page 5)

40 ± 1.96√

10/200, i.e.

Example (page 5)

40 ± 1.96√

10/200, i.e.

40 ± 0.44.

Example (page 5)

40 ± 1.96√

10/200, i.e.

40 ± 0.44.

Example (page 5)

40 ± 1.96√

10/200, i.e.

40 ± 0.44.

This should be intuitive, since as the sample size increases we arebecoming more sure of our estimate for the population value.

Example (page 5)

What would be the 99% confidence interval in this case? Fromtables for the standard normal distribution, we can find that

Pr(−2.576 < Z < 2.576) = 0.99;

Example (page 5)

Pr(−2.576 < Z < 2.576) = 0.99;

hence, the 99% confidence interval is given by

x̄ ± 2.576√

σ2/n,

Example (page 5)

Pr(−2.576 < Z < 2.576) = 0.99;

x̄ ± 2.576√

σ2/n,

in this case giving

40 ± 2.576√

10/200, i.e.

Example (page 5)

Pr(−2.576 < Z < 2.576) = 0.99;

x̄ ± 2.576√

σ2/n,

in this case giving

40 ± 2.576√

10/200, i.e.

40 ± 0.58.

Example (page 5)

Pr(−2.576 < Z < 2.576) = 0.99;

x̄ ± 2.576√

σ2/n,

in this case giving

40 ± 2.576√

10/200, i.e.

40 ± 0.58.

Confidence Intervals (Known σ2) – Summary

(i) Calculate the sample mean x̄ from the data;

(ii) Calculate your interval! For example,

for a 90% confidence interval, use the formula

x̄ ± 1.645×√

σ2/n;

x̄ ± 1.96×√

σ2/n;

x̄ ± 2.576×√

σ2/n.

Looking ahead to next week...

Typically the population variance σ2 will be unknown

In reality we have to estimate σ2

A 95% confidence interval for the population mean µ musttake this additional estimate into account...

MAS1403 Quantitative Methods for Business Managementnag48/teaching/MAS1403/slides1beam.pdf ·...

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