Statistics and Modelling Course 2011 Topic: Confidence Intervals Achievement Standard 90642...

Statistics and Modelling Course

2011

Topic: Confidence Intervals

Achievement Standard 90642

Calculate Confidence Intervals for Population Parameters

3 Credits

Externally Assessed

NuLake Pages 63101

LESSON 1 – Sampling

Handout with gaps to fill in – goes with the following slides.

STARTER: Look at the following 2 examples of bad sampling technique & discuss what’s wrong in each case.

1. Discuss how you’d obtain a representative sample from our school roll.

2. Notes on sampling and inference.3. Population and Samples ‘Policemen’ worksheet (from Achieving in

Statistics). Complete for HW.

Describe some faults with each of these sampling methods.

Sampling


(a) A survey on magazine readership is conducted by phoning households between 1 and 4pm.

People who aren’t at home during those times cannot be surveyed.

Some people don’t have a phone

Sampling


(b) A talkback radio station asks listeners to phone in with a quick ‘yes’ or ‘no’ answer to the question “Should NZ have capital punishment?”

Only people who are listening at the time can participate.

Self-selected sample. Only those with a strong opinion will ring in.

Sampling

SamplingYou are asked the question: “How tall are St. Thomas students?”

•You only have time to measure the heights of 35 students.

Q1: How would you choose which 35 students to measure.

Q2: Once you’ve measured your 35 students’ heights, how would

you use this data to answer the question: “How tall are St.

Thomas students?

Purpose of a Sample

POPULATIONSAMPLE

Make an inference

Sampling terminology

Purpose of a Sample

POPULATIONSAMPLE InferencesMake an inference

POPULATION:Target Population: All items under investigation.

We usually just call it the “Population”.


Purpose of a Sample

POPULATIONSAMPLE InferencesMake an inference

SAMPLES:Sample: Subset selected to REPRESENT the population.

Sampling Frame: A list/database of items from which we select our sample. (Should include all items in the Target Population)


We usually just call it the “Population.”




For a sample to be Representative of a given population: The Sampling Frame must match the Target Population.

Sample statistic Population parameter







Number of items







Number of items n: Sample size







Number of items n: Sample size N: Population size








Mean






X



Mean Standard deviation






X




s






X




s

Proportion






X




s

Proportion p






X

A representative sample should have…• Sample-size large enough to allow the results to be

meaningful (rough guide: sample size of n > 30).

• No Bias – Sample selection is said to be “biased” if some items are more likely to be chosen than others. Every item in the target population should be equally-likely to be chosen. Random selection ensures this.

• Minimal Non-response – difficult to control this.




s

Proportion p

X

Example:A home security firm is hoping to sell as many burglar alarms as possible to householders in a certain town.Usually each house only needs one burglar alarm.Before the firm orders the alarms from their supplier, they wish to have an indication of how many alarms they might sell.

1.) What is the target population?A. all the people who live in the town.B. the head of each household.C. the houses in the town.


1.) What is the target population? Answer: C. the houses in the town.

2.) What is the sampling frame?A. the electoral roll for the town.B. a list of all the people who live in the town.C. a list of all the houses in the town.


1.) What is the target population? Answer: C. the houses in the town.

2.) What is the sampling frame? Answer: C. a list of all the houses in the town.

3.) How would you select a representative sample of the houses in the town? (discus s as a class)

Do Population and Samples ‘Policemen’ worksheet. Finish by Monday. Will mark as a class.

EXTRA ON SAMPLING TECHNIQUES IF TIME (schol

students)

Otherwise skip to Lesson 3:

Distribution of Sample Means 1

Extension Lesson:Other sampling techniques

Good sampling techniques:1.Simple Random Sampling2.Systematic Sampling3.Stratified Sampling4.Cluster Sampling

Bad sampling techniques (biased selection):• Convenience sampling.• Self-selected sampling.

Random selection

Q: What does the word “random” actually mean?

Q: How would you select a student at random from this school?

Simple random sampling.21.03

Generate 20 different random numbers between 1 and 100.

42 67 2 12 77 49 60 20 45 15 64 7 8 21 15 64 58 14 29 68 26 90

If a random number has already occurred, generate more as needed.

Calculator formula1 + 100×RAN#

1. Simple Random Sampling1. Obtain a list of all N items in the target population,

numbering them 1 to N (e.g. the school roll: 1-600).

2. Decide how many you will select for your sample (n).

3. Use the random number generator on your calculator to select numbers at random between 1 and N:

On calculator, type: 1 + Population size × RAN#4. Keep pressing ‘equals’ until you have selected n

different items. Discard any repeats.

Advantage of SR sampling: Ensures that every item in the population has an equal chance of being selected – so no chance of bias.

2. Decide how many you will select for your sample (n).3. Use the random number generator on your calculator

to select numbers at random between 1 and N: On calculator, type: 1 + Population size × RAN#4. Keep pressing ‘equals’ until you have selected n

different items. Discard any repeats.

Advantage of SR sampling: Ensures that every item in the population has an equal chance of being selected – so no chance of bias.

Disadvantage:• Does not ensure that all subgroups of the population

are represented in proportion (e.g. some racial, socio-economic groups could be over/under-represented).

Select a sample of 35 students from the St. Thomas school roll.

HW: Old Sigma Pg. 130 – Ex. 9.1 (all), then Pg. 134 – Ex. 9.2 – just Q1.

Cluster samplingStratified sampling

3 other good sampling techniques

1. Obtain a list of all N items in the target popn (numbered 1N).

2. Pick a random starting point (e.g. item number 7)

3. Sample every kth item after that, where k=N/n until you have selected n items.

Use when the population consists of categories (strata), (e.g. racial groups)

1. Divide sampling frame into the strata (categories).2. Select a separate random sample from each stratum in proportion to the percentage of the population found in each. (Called Proportional Allocation )

Systematic sampling

Use when the population is distributed into naturally-occurring groups or ‘clusters’ (e.g. towns and cities in NZ)

Stage 1: Select the clusters:Select a representative sample of the clusters themselves.

Stage 2: Select a random sample of items within chosen clusters. Must be in proportion to the percentage of the population found in each. (Called Proportional Allocation )

Simple random sampling Systematic sampling

Cluster samplingStratified sampling

21.03 Comparison of samples.

Stratified sampling

3 other good sampling techniques

1. Obtain a list of all N items in the target popn (numbered 1N).

2. Pick a random starting point (e.g. item number 7)


Use when the population consists of categories (strata), (e.g. racial groups)

1. Divide sampling frame into the strata (categories).2. Select a separate random sample from each stratum in proportion to the percentage of the population found in each. (Called Proportional Allocation )

Systematic sampling

Use when the population is distributed into naturally-occurring groups or ‘clusters’ (e.g. towns and cities in NZ)

Stage 1: Select the clusters:Select a representative sample of the clusters themselves.

Stage 2: Select a random sample of items within chosen clusters. Must be in proportion to the percentage of the population found in each (Proportional Allocation).

HW: Do Old Sigma (2nd edition) p137: Ex. 9.3.

Cluster sampling

1. Select a sample of between 30 and 36 students from the school roll using each of these 3 methods.

2. Write down at least one advantage and at least one disadvantage/risk associated with each of these 3 techniques.

21.03

Choose a starting point at random between 1 and 100.

Suppose this gives 5.87352 5.

So start at item number 5.

Then choose every kth item,

where k = N/n.

Using calculator1 + 100×RAN# =

Systematic sampling.

To obtain a systematic sample of size 20 from this data.

= 100/20

= 5. So sample every 5th item.

Systematic Sampling1. Obtain a list of all N items in the target population.2. Decide on your sample size, n .3. Pick a random starting point (e.g. item number 7)


Advantages:• Ensures that sample is selected from throughout the

breadth of the sampling frame.• Convenient and fast – easier to collect info on items that

are in a sequence (every 5th house) than from a random sample where they are scattered all over.


Advantages:• Ensures that sample is selected from throughout the

breadth of the sampling frame.• Convenient and fast – easier to collect info on items that

are in a sequence (every 5th house) than from a random sample where they are scattered all over.

Disadvantage:Be careful that the list itself has no systematic pattern. If

every 2nd house on a street were sampled, all would be on the same side of the street!

21.03

Suppose the avocados are of 3 different varieties.

Hass: 1–40 40%

Hopkins: 71–100 30%

Fuerte: 41–70 30%

The number in each strata of the sample should be proportional to the number in each group in the population.

Hass: 40% x 20

Fuerte: 30% x 20

Hopkins: 30% x 20

Stratified sampling.

= 8

= 6

= 6

21.03

Thus generate random numbers as follows:

Hass: 1–40 8 random nos.33 17 12 25 9 9 33 16 39 8

Fuerte: 41–70 6 random nos.58 59 67 43 53 56

Hopkins: 71–100 6 random nos.98 85 96 99 90 81

Stratified sampling.

Stratified samplingUse when the population consists of categories (strata),

and you wish to represent each ‘stratum’ proportionally (e.g. racial groups, one-story and multi-story homes within a city).

1. Obtain a list of all N items in the target population.2. Decide on your sample size, n .3. Divide list into the strata (categories).4. Select a separate random sample from each

stratum in proportion to the percentage of the population found in each.

Proportional Allocation: Selecting from each stratum in proportion to its percentage of the population.

1. Obtain a list of all N items in the target population.2. Decide on your sample size, n .3. Divide list into the strata (categories).4. Select a separate random sample from each



E.g. If 12% of a city’s citizens are Pacific Islanders, then 12% of the sample size should be selected from among the Pacific Island citizens.

3. Divide list into the strata (categories).4. Select a separate random sample from each



E.g. If 12% of a city’s citizens are Pacific Islanders, then 12% of the sample size should be selected from among the Pacific Island citizens.

Advantage: Guaranteed to be representative of each stratum.

Disadvantage: Time-consuming and expensive because you must collect information about the strata-sizes in advance.

Cluster samplingUse when the population is distributed into naturally-

occurring groups or ‘clusters’ (e.g. towns and cities in a country).

1. Select a representative sample of the clusters themselves (usually a lot so we can’t sample from all).

2. Select a random sample of items from within each chosen cluster.

3. Again, use Proportional Allocation (like with stratified samples). Weight the number selected from each cluster according to the cluster size.

E.g. Selecting samples of New Zealanders by selecting a sample of towns/cities from throughout the country, then a proportional random sample from within each.

1. Select a representative sample of the clusters themselves (usually a lot so we can’t sample from all).

2. Select a random sample of items from within each chosen cluster.

3. Again, use Proportional Allocation (like with stratified samples). Weight the number selected from each cluster according to the cluster size.


Advantage:• Cheaper and faster when sampling from a geographically

large area (data can be collected in groups within chosen clusters rather than being spread out).


Advantage:• Cheaper and faster when sampling from a geographically

large area (data can be collected in groups within chosen clusters rather than being spread out).

Disadvantages:• Items don’t have an equal chance of selection.

– Small clusters are unlikely to be sampled from.– Items that are not in clusters are excluded altogether.E.g. farmers or people in small rural communities may have no chance

of being selected.

• Requires prior knowledge of cluster sizes.

HW: Memorise the 4 types of sampling techniques and the advantages & disadvantages of each.

21.03

Cluster sampling.

Choose four clusters, each of 5 avocados, by selecting four numbers at random from the data, and taking them as the middle item of a ‘cross’.

62 22 2 68 56

If clusters overlap or run outside the boundaries, choose another.

Note: Depending how a cluster is defined, it can exclude some items or make other items more likely to be chosen than under other sampling methods

Here is one way of obtaining a cluster sample of size 20.

Spreadsheet formula99×RAN# + 1 =

LESSON 2 – Distribution of Sample Means

The points of today:

The point of today: Get confident at calculating probabilities involving the distribution of sample means.– Mark HW: “Achieving in Statistics”: pages 30.– Handout to fill in (goes with following slides)

• Then do Achieving in Statistics: pages 31 & 32.

The Distribution of Sample Means

The Distribution of Sample MeansSTARTER ACTIVITY: Each class member has 5 dice.Toss your 5 dice and record the number facing upward for each. Add up to get the total for your 5.My total value from 5 tosses = _____My mean score for each die roll = ________

Your group of 5 dice tosses represents a sample of size n=5.Between us, as a class, we tossed 5 dice ________ times.We got means of: ________________________________

This illustrates the fact that ________________________.

The Distribution of Sample MeansSTARTER ACTIVITY: Each class member has 5 dice.Toss your 5 dice and record the number facing upward for each. Add up to get the total for your 5.My total value from 5 tosses = _____My mean score for each die roll = ________

Your group of 5 dice tosses represents a sample of size n=5.Between us, as a class, we tossed 5 dice ________ times.We got means of: ________________________________

This illustrates the fact that sample means vary.

The Distribution of Sample MeansYour group of 5 dice tosses represents a sample of size n=5.Between us, as a class, we tossed 5 dice ________ times.We got means of: ________________________________


A random sample can be thought of as a collection of n items (n=5 dice-tosses in the experiment we did last time),

The Distribution of Sample MeansYour group of 5 dice tosses represents a sample of size n=5.Between us, as a class, we tossed 5 dice ________ times.We got means of: ________________________________


A random sample can be thought of as a collection of n items (n=5 dice-tosses in the experiment we did last time), each of which has a value that we measure (the number facing upward when a die lands in this case).

The Distribution of Sample MeansThis illustrates the fact that sample means vary.

A random sample can be thought of as a collection of n items (n=5 dice-tosses in the experiment we did last time), each of which has a value that we measure (the number facing upward when a die lands in this case).

When you select items at random from any population, the value of each item, X is a random variable (e.g. height, weight, volume of drink in soft drink bottles etc.). Select a random sample of size n from any population: The sample mean,

X = n

XXXX n ...321

The Distribution of Sample MeansWhen you select items at random from any population, the value of each item, X is a random variable (e.g. height, weight, volume of drink in soft drink bottles etc.). Select a random sample of size n from any population: The sample mean,

Different samples will produce different mean values , just like we got different mean values from tossing our dice.

X = n

XXXX n ...321

The Distribution of Sample MeansSelect a random sample of size n from any population: The sample mean, Different samples will produce different mean values , just like we got different mean values from tossing our dice.

The sample mean :•___________________________________________ ___________________________________________

•___________________________________________ ___________________________________________ ___________________________________________

X = n

XXXX n ...321


The sample mean :•is a random variable itself because it varies at random from sample to sample.

•___________________________________________ ___________________________________________ ___________________________________________

X = n

XXXX n ...321



•is normally distributed about the population mean ,

X = n

XXXX n ...321



•is normally distributed about the population mean , even if the population from which it is drawn is not normally distributed,

X = n

XXXX n ...321



•is normally distributed about the population mean , even if the population from which it is drawn is not normally distributed, provided the samples are large enough. Rule of thumb is n > 30.

X = n

XXXX n ...321

The Distribution of Sample MeansDifferent samples will produce different mean values , just like we got different mean values from tossing our dice.



In other words the sample means will ‘average out’ towards the population mean.




In other words the sample means will ‘average out’ towards the population mean. This result is called the ‘__________________’.




In other words the sample means will ‘average out’ towards the population mean. This result is called the ‘Central Limit Theorem’.




In other words the sample means will ‘average out’ towards the population mean. This result is called the

‘Central Limit Theorem’. i.e. X

In other words the sample means will ‘average out’ towards the

population mean. This result is called the ‘Central Limit Theorem’.

i.e. X

Distribution of Sample Means .

X

X

Mean of sample means

Std. deviation of distribution of sample means (standard error)

nX

X

*


X

X



i.e. . X



nX

X

*

Since sample means are normally distributed about the population mean,


X

X



i.e. . X



nX

X

Since sample means are normally distributed about the population mean, we can use the properties of a normal distribution curve


X

X



i.e. . X



nX

X

Since sample means are normally distributed about the population mean, we can use the properties of a normal distribution curve to predict the percentage of samples that will produce means within a particular distance from the population mean.


X

X



i.e. . X



nX

X


Example:


X

X



nX

X


Example:The results of a census of all 17 year-old males in NZ showed a mean height of = 177cm, with = 9cm.

If a random sample of 36 seventeen year-old NZ males is taken, calculate:

a)The expected value of the sample mean.

XXE )(

And,by the Central Limit Theorem, X the population mean.

____)( XE





XXE )(

And,by the Central Limit Theorem, X the population mean.

cmXE 177)(




XXE )(And,by the Central Limit Theorem, X the population mean.

cmXE 177)(

b) The standard deviation (standard error) of the sample mean.

nX

36

9

The results of a census of all 17 year-old males in NZ showed a mean height of = 177cm, with = 9cm.




cmXE 177)(


nX

36

9

= 1.5cm




cmXE 177)(


nX

36

9

= 1.5cm

c) What percentage of such samples would have a mean that is:


If a random sample of 36 seventeen year-old NZ males is taken:

________ _______ )177 if ,180174( zPXP

c) What percentage of such samples would have a mean that is:(i) Within 3cm of the population mean of 177cm?



XX

zPXP

180

174 )177 if ,180174(




XX

zPXP

177180

177174 )177 if ,180174(




n

z

n

PXP 177180

177174 )177 if ,180174(




36

9177180

36

9177174

)177 if ,180174( zPXP




5.1

177180

5.1

177174 )177 if ,180174( zPXP




5.1

177180

5.1

177174 )177 if ,180174( zPXP

2 2 zP

0.477242

= 0.9545 (4sf) So 95.45% of samples




________ -1 )182or 172( XPXXP

c) What percentage of such samples would have a mean that is:(ii) More than 5cm away from the population mean?



182172 -1 )182or 172( XPXXP

______________P-1 Z




182172 -1 )182or 172( XPXXP

XX

Z

177182177172P-1




182172 -1 )182or 172( XPXXP

n

Z

n

177182177172

P-1




182172 -1 )182or 172( XPXXP

36

9177182

36

9177172

P-1 Z




182172 -1 )182or 172( XPXXP

5.1

177182

5.1

177172P-1 Z

)3

13

3

13P(-1 Z

= 0.00086

.999140-1 So only about 0.09% of samples. Very rare.


Homework:

Do Achieving in Statistics: pages 31 & 32.

Extension

The point of today: Look at when we can draw conclusions about the population mean based on a sample mean.

STARTER: Look at applet that demonstrates the distribution of sample means:

SIM - onlinestatbook.com.SLASH.rvls.html.

• Work through the following examples as class (handout to fill in).

• Then do Sigma p184 – Ex. 11.5 (old version).

or Sigma p66 – Ex. 3.05 (new version)

Example:The census of all NZ seventeen year-old males from yesterday’s example was actually conducted back in 1987. It had mean of =177cm and of 9cm.A random sample of 36 seventeen year-old NZ males was selected just last year. This sample found a mean height of 180cm.

(a) What is the probability that a random sample of 36 students selected from a population with =177cm and 9cm would give a mean height greater than 180cm?

n

XzPXP

)177 if ,180(

36

9177180

zP


(b) Based on this answer, what percentage of samples would have means of 180cm or higher if the population mean was 177cm?

36

9177180

zP

2 zP

.022750

n

XzPXP

)177 if ,180(


(b) Based on this answer, what percentage of samples would have means of 180cm or higher if the population mean was 177cm?

Answer: Only 2.275%

36

9177180

zP

2 zP

.022750

n

XzPXP

)177 if ,180(

(b) Based on this answer, what percentage of samples would have means of 180cm or higher if the population mean was 177cm?Answer: Only 2.275%

(c) Sketch a normal distribution curve for the distribution of sample means from a population with and standard deviation of

n

XzPXP

)177 if ,180(

36

9177180

zP

2 zP

.022750

177cm 9cm

(b) Based on this answer, what percentage of samples would have means of 180cm or higher if the population mean was 177cm (like in 1987)?Answer: Only 2.275%


(d) So it is very ____________ that a randomly selected ________ taken from a _____________ with mean 177cm and standard deviation of 9cm would have a mean as high as this one.

2 zP

.022750

177cm 9cm



(d) So it is very unlikely that a randomly selected ________ taken from a _____________ with mean 177cm and standard deviation of 9cm would have a mean as high as this one.

2 zP

.022750

177cm 9cm



(d) So it is very unlikely that a randomly selected sample taken from a _____________ with mean 177cm and standard deviation of 9cm would have a mean as high as this one.

2 zP

.022750

177cm 9cm



(d) So it is very unlikely that a randomly selected sample taken from a population with mean 177cm and standard deviation of 9cm would have a mean as high as this one.

2 zP

.022750

177cm 9cm



(d) So it is very unlikely that a randomly selected sample taken from a population with mean 177cm and standard deviation of 9cm would have a mean as high as this one. Yet it did.(e) What is the most likely explanation?

2 zP

.022750

177cm 9cm

Do Sigma:

In old (2nd) edition: Pg. 184 – Ex. 11.4.

OR in NEW edition: Pg. 66 – Ex. 3.04.

LESSON 4 – C.I.s for Means 1

• Today’s theme: Solving problems involving Confidence Intervals for Means.

• Students do NuLake Ch 2.5 – Calculate Confidence Intervals of means.

http://www.youtube.com/watch?v=Ohz-PZqaMtk

Question: If the population mean height of 17 year-old NZ males is 177cm with of 9cm, within what interval would we expect the means of 95% of samples of size 36 to lie?

95%

Notice that the middle 95% of the area under normal curve means half on each side of the mean.



i.e. 47.5% (or 0.475) on each side.

47.5

%

47.5

%




Looking up 0.475 on the tables gives z = 1.96.

47.5

%

47.5

%





1.96-1.96

47.5

%

47.5

%


So when we calculate the mean from a random sample we expect that, 95% of the time, it will be within + 1.96 standard errors of the popn mean, .




1.96-1.96

47.5

%

47.5

%



Now, work out the lower and upper limits of the interval within which you’d expect 95% of sample means to lie if each sample has 36 people in it.

1.96-1.96

47.5

%

47.5

%




1.96-1.9647

.5%

47.5

%

Conclusion: So 95% of samples of size 36 from this population will produce means between _______cm and ________cm




1.96-1.9647

.5%

47.5

%

Conclusion: So 95% of samples of size 36 from this population will produce means between 174.06cm and 179.93cm

1.96-1.9647

.5%

47.5

%

Conclusion: So 95% of samples of 36 from this population will produce means between 174.06cm and 179.93cm

Problem:

1.96-1.9647

.5%

47.5

%


Problem:In real-life, we almost never know the population mean (or standard deviation).

1.96-1.9647

.5%

47.5

%


Problem:In real-life, we almost never know the population mean (or standard deviation).We only have enough resources to conduct ONE random sample and use it to estimate (infer) the population mean.

1.96-1.9647

.5%

47.5

%



How can our knowledge of the distribution of sample means help us here??

1.96-1.96

47.5

%

47.5

%




1.96-1.96

47.5

%

47.5

%



Answer: We construct an interval within which we think the population mean lies.

1.96-1.96

47.5

%

47.5

%




Estimate of margin of error

X

1.96-1.96

47.5

%

47.5

%



Estimate of margin of error.

This is known as a Confidence Interval.

X

Diagram on board





X

A 95% Confidence Interval for the population mean is an interval that has a 95% probability of containing the population mean.





X

A 95% Confidence Interval for the population mean is an interval that has a 95% probability of containing the population mean.A 95% confidence interval for is

X





X


XX 96.1





X


A 99% confidence interval for is n

X

_____

nX

96.1





X



X

.5762

nX

96.1



X

.5762

nX

96.1

Example 1:A soft drink is sold in bottles. The amount of drink in each bottle is normally distributed with a standard deviation of 40mL. The mean volume of drink in a random sample of 100 such bottles is 300mL. Construct a 95% confidence interval for the true mean volume of drink per bottle.


Solution:There is a 95% probability that the interval 300mL +1.96 standard errors contains the true population mean.

A 95% C.I. for the population mean is:



A 95% C.I. for the population mean is: X

X + z ×Standard Error of the sample mean

+ Margin of Error

300 +

300 +100

4096.1

mL300 + 7.84mL

Margin of Error

E





+ Margin of Error

300 +

300 +100

4096.1

mL300 + 7.84mL

ANSWER: The 95% CI for the population mean is: _____mL < < _____mL





+ Margin of Error

300 +

300 +100

4096.1

mL300 + 7.84mL

ANSWER: The 95% CI for the population mean is: 292.2mL < < _____mL





+ Margin of Error

300 +

300 +100

4096.1

mL300 + 7.84mL

ANSWER: The 95% CI for the population mean is: 292.2mL < < 307.8mL



A 95% C.I. for the population mean is: X + Margin of Error

300 +100

4096.1

mL300 + 7.84mL




A 95% C.I. for the population mean is: X + Margin of Error

300 +100

4096.1

mL300 + 7.84mL


Now calculate the 99 % C.I. Will it be wider or narrower??


Solution:


mL300 + 7.84mL



A 99% C.I. is

300 +100

4096.1

300 +n

z

2/99.0

300 +100

40576.2

mL300 + 10.304mL

Margin of Error

E


Solution:


mL300 + 7.84mL



A 99% C.I. is

300 +100

4096.1

300 +n

z

2/99.0

300 +100

40576.2

mL300 + 10.304mL

ANSWER: The 99% CI for the population mean is: ____mL < < _____mL


Solution:


mL300 + 7.84mL



A 99% C.I. is

300 +100

4096.1

300 +n

z

2/99.0

300 +100

40576.2

mL300 + 10.304mL

ANSWER: The 99% CI for the population mean is: 289.7mL < < _____mL


Solution:


mL300 + 7.84mL



A 99% C.I. is

300 +100

4096.1

300 +n

z

2/99.0

300 +100

40576.2

mL300 + 10.304mL


Copy examples, then do NuLake Ex. 2.5: p8184

When we aren’t told the population standard deviation .

If we aren’t given the popn standard deviation , then use the sample standard deviation s as an estimate.

This is OK provided the sample size is large enough (n > 30).

LESSON 5 – C.I.s for Means 2

The purpose of today: Memorise definition of a confidence interval. Get confident at constructing confidence intervals for

population means.

To do today:1. Watch youtube clip: http://www.youtube.com/watch?v=Ohz-PZqaMtk

2. Interpret C.I. from yesterday’s e.g. in context.3. Finish NuLake 2.5.4. Do new Sigma p75 - Ex. 4.01: To end of Q14 compulsory.5. Q1517 are extra for experts.

2008 NCEA exam question:

Do new Sigma p75 - Ex. 4.01: To end of Q14 compulsory.Q1517 are extra for experts.

LESSON 6 – SAMPLE SIZE (MEANS)

• Today’s theme: Calculate the required sample size to meet a set of specified conditions for a Confidence Interval for the population MEAN.

• Do Sigma (old): Ex. 14.2 – pg. 230.

(New version: Ex. 4.02 – pg. 79)

Calculating the minimum sample size - means.

The confidence interval formula for estimating the population mean, , is:

The ____________, E, is _____________________________

________________________________________________.

nzX



The margin of error, E, is the _________________________

________________________________________________.

nzX



The margin of error, E, is the distance between the sample mean and the upper and lower limits of this interval.

Margin of Error, E =

For example, a confidence interval of 18cm < < 22cm, can also be expressed as 20cm + ___cm.

nzX

nz





For example, a confidence interval of 18cm < < 22cm, can also be expressed as 20cm + 2cm.

nz

nzX





For example, a confidence interval of 18cm < < 22cm, can also be expressed as 20cm + 2cm. The ___________ is __cm. Our estimate is “___________________”.

nz

nzX





For example, a confidence interval of 18cm < < 22cm, can also be expressed as 20cm + 2cm. The margin of error is 2cm. Our estimate is “___________________”.

nz

nzX





For example, a confidence interval of 18cm < < 22cm, can also be expressed as 20cm + 2cm. The margin of error is 2cm. Our estimate is “accurate to within 2cm”.Given a particular level of confidence ,

nz

nzX





For example, a confidence interval of 18cm < < 22cm, can also be expressed as 20cm + 2cm. The margin of error is 2cm. Our estimate is “accurate to within 2cm”.Given a particular level of confidence , we can calculate how big a sample is necessary to estimate to give a required accuracy or margin of error, E.

nz

nzX

E.g: A survey is to be conducted to determine the mean income of a group of workers. A pilot survey gives $100.




nz

E.g: A survey is to be conducted to determine the mean income of a group of workers. A pilot survey gives $100. How large must the sample be if the mean income is to be estimated to within $20 using a 95% confidence interval?




nz



Solution: A confidence interval for the mean income is:

For the income to be found to within $20, we need:

X 1·96

n

X 1·96

n

1·96 100

n < 20

1·96 100

n < 20


Solution: A confidence interval for the mean income is:

For the income to be found to within $20, we need:

1·96 100

n < 20

1·96 100

n < 20

20 196

n

X 1·96

n

X 1·96

n

22

20 196

n

n2

2

20

196

n > 96.04

Answer:A minimum sample size of 97 is needed.

Squaring both sides

When you’ve copied down this e.g:Do Sigma (new): Ex. 4.02 – pg. 79 (or Old version: Ex. 14.2 – pg. 230).FINISH FOR H.W.

Formula for calculating minimum sample size.

Where = Margin of Error.

i.e. half of C.I. width.

2

E

zn

Sample-size question from 2007 NCEA External Exam

A random sample of size n is taken from a population having a known standard deviation σ. A 95% confidence interval for the population mean is calculated using the sample mean.

A second random sample of size 2n is taken from the same population and a 95% confidence interval for the population mean is calculated using its sample mean.

How many times greater is the width of the first confidence interval than the width of the second confidence interval?

Formula for calculating minimum sample size.

Where = Margin of Error.

i.e. half of Confidence Interval width.

2

E

zn

LESSON 7 – Intro to Confidence Intervals for Proportions

The points of today:• Introduction to Distribution of Sample PROPORTIONS.

• Construct confidence intervals for Population Proportions.

Notes on distn. of sample proportions (handout). Do handout on distribution of sample proportions

(Achieving in Statistics page 33). How to construct a C.I. for a proportion. HW: NuLake Ex. 2.6.

The Distribution of Sample Proportions

E.g. Political Opinion Polls - National vs Labour.

2 possible outcomes where is the proportion of successful outcomes in n trials.

If a sequence of n independent trials results in x successes, then x has a _________ distribution.




If a sequence of n independent trials results in x successes, then x has a Binomial distribution.

A point estimator of the popn proportion of successful trials, is the sample proportion .

With a sufficient sample size (rule of thumb n>30), the

distribution of sample proportions p is approximately

normal and…

n

xp




If a sequence of n independent trials results in x successes, then x has a Binomial distribution.

A point estimator of the popn proportion, is the sample proportion

With a sufficient sample size (rule of thumb n>30), the distribution of

sample proportions p is approximately normal and… ppE )(

np

)1(

1. Do handout on distribution of sample proportions.(Will do Q1 table on board as a class)

n

xp

By the Central Limit Theorem

Next slide:

The proofs of the formulae for mean and standard deviation of the distribution of sample proportions


sample proportions p is approximately normal and…

ppE )(np

)1(

Proof:

n

XEpE )(

)(1

XEn

nn

1

Since, for the Binomial Distribution, = n

Proof:

n

XVarpVar )(

X

nVar

1

XVarn

21

)1(1

2

n

n



sample proportions p is approximately normal and…

ppE )(np

)1(

Proof:

n

XEpE )(

)(1

XEn

nn

1


Proof:

n

XVarpVar )(

X

nVar

1

XVarn

21

)1(1

2 n

n


n

)1(

np

)1(

Example: Political opinion polls.500 New Zealanders aged 18 and over were selected at random for an

opinion poll.

Confidence Intervals for Proportions

Example: Political opinion polls.

500 New Zealanders aged 18 and over were selected at random for an opinion poll. They were asked to indicate whether Labour or National would be their preferred political party. 275 voted for National.

Find a 95% confidence interval for the true proportion of all NZers who favour National.

Solution: Our point estimate for is

There is a 95% probability that the interval 0.55 + 1.96 standard errors contains the true population proportion who would prefer National.

A 95% C.I. is

500

45.055.096.1

Confidence Intervals for Proportions

55.0

500

275p 5 5.0

55.0 Margin of

Error E

pn

ppz

)1(

_ _ _ _ _

p pz



Find a 95% confidence interval for the true proportion of all NZers who favoured National.



A 95% C.I. is

500

45.055.096.1

55.0

500

275p 5 5.0

55.0 Margin of

Error E

pn

ppz

)1(

_ _ _ _ _

p pz






A 95% C.I. is

500

45.055.096.1

55.0

500

275p 5 5.0

55.0Margin of

Error E

pn

ppz

)1(

0 4 3 6 1.0

p pz

ANSWER: The 95% CI for the proportion in favour of National is

______ < < _______






A 95% C.I. is

500

45.055.096.1

55.0

500

275p 5 5.0

55.0Margin of

Error E

pn

ppz

)1(

0 4 3 6 1.0

p pz


0.5064 < < _______



Find a 95% confidence interval for the true proportion of all NZers who favoured National .



A 95% C.I. is

500

45.055.096.1

55.0

500

275p 5 5.0

55.0Margin of

Error E

pn

ppz

)1(

0 4 3 6 1.0

p pz


0.5064 < < 0.5936

PONDER THIS:

Based on this opinion poll, does National have a STATISTICALLY SIGNIFICANT majority?

HW: Do NuLake Ex. 2.6 – CIs for proportions.

LESSON 8 – Practice constructing C.I.s for Proportions

The point of today:• Do lots of practice involving confidence intervals for

Population Proportions.

Go over any homework questions – NuLake p87,88: Ch 2.6 – C.I.s for proportions.

Then do Sigma pg. 232 – Ex. 14.3 (old version). or in new version: pg. 88 - Ex. 5.01. Finish for HW.

LESSON 9 – SAMPLE SIZE (PROPORTIONS)

• Today’s theme: Calculate the required sample size to meet a set of specified conditions for a Confidence Interval for the population PROPORTION.

• Key point – for minimum sample size, if not told p, assume p=0.5 as this gives the greatest margin of error (prepared for the worst).

Do Sigma: old edition – p235 – Ex. 14.4

or new edition – p91 - Ex. 5.02

Calculating the minimum sample size - proportions.

The confidence interval formula for estimating the population proportion, , is:

The margin of error, E, is the distance between the sample proportion and the upper and lower limits of this interval.


For example, a confidence interval of 0.37 < < 0.43, can also be expressed as 0.4 + 0.03. The margin of error is _____.

nzp

)1(

nz

)1(

Calculating the minimum sample size.




For example, a confidence interval of 0.37 < < 0.43, can also be expressed as 0.4 + 0.03. The margin of error is 0.03.

nzp

)1(

nz

)1(

Calculating the minimum sample size.




For example, a confidence interval of 0.37 < < 0.43, can also be expressed as 0.4 + 0.03. The margin of error is 0.03. Our estimate is “accurate to within 0.03”.

The sample size depends on three factors:1.The level of confidence required, .2.The true value of , which will often be unknown.3.The accuracy required. i.e. the margin of error, E, we are willing to accept.

nzp

)1(

nz

)1(


For example, a confidence interval of 0.37 < < 0.43, can also be expressed as 0.4 + 0.03. The margin of error is 0.03. Our estimate is “accurate to within 0.03”.


ExampleAn international airline is thinking of making smoking illegal

on its aircraft. Before making the decision it wishes to estimate the proportion of smokers in the population of passengers on its planes by taking a random sample. How big a sample must it take to be 95% sure that the value so obtained does not differ from the true proportion by more than 0.05?

nz

)1(


ExampleAn international airline is thinking of making smoking illegal

on its aircraft. Before making the decision it wishes to estimate the proportion of smokers in the population of passengers on its planes by taking a random sample. How big a sample must it take to be 95% sure that the value so obtained does not differ from the true proportion by more than 0.05?

Solution: A 95% confidence interval for the proportion of smokers on all planes, is:

For the proportion to be found to within 0.05, we need:

np

)1(96.1

Margin of Error

< 0.05

ExampleAn international airline is thinking of making smoking illegal on its aircraft. Before making the decision it wishes to estimate the proportion of smokers in the population of passengers on its planes by taking a random sample. How big a sample must it take to be 95% sure that the value so obtained does not differ from the true proportion by more than 0.05?



PROBLEM!

np

)1(96.1

05.0)1(

96.1

n

Margin of Error < 0.05




PROBLEM! We don’t have a value for π. That’s the very thing we’re trying to estimate!!

np

)1(96.1

05.0)1(

96.1

n





PROBLEM! We don’t have a value for π. That’s the very thing we’re trying to estimate!! To get around this problem we have 3 options:

np

)1(96.1

05.0)1(

96.1

n


An international airline is thinking of making smoking illegal on its aircraft. Before making the decision it wishes to estimate the proportion of smokers in the population of passengers on its planes by taking a random sample. How big a sample must it take to be 95% sure that the value so obtained does not differ from the true proportion by more than 0.05?




1. Use a value of p that has held in the past (previous samples).

2. Take a small pilot survey, and use the sample proportion p from that as an estimate of .

np

)1(96.1

05.0)1(

96.1

n





1.Use a value of p that has held in the past (previous samples).

2.Take a small pilot survey, and use the sample proportion p from that as an estimate of .

3.Use =0.5. This allows for the greatest possible error because the maximum possible value of (1-) occurs when both and (1-) are = ½

i.e. when (1-) = 0.5 × 0.5

= 0.25

np

)1(96.1

05.0)1(

96.1

n

Margin of Error

< 0.05






i.e. when (1-) = 0.5 × 0.5 = 0.25

Back to this example:

We’re given no information on the value of , so let = 0.5.

05.0)1(

96.1

n

05.0)1(

96.1

n

Margin of Error





i.e. when (1-) = 0.5 × 0.5= 0.25



05.0)1(

96.1

n

05.0 )5.01(5.0

96.1

n

05.0 25.0

96.1 n

3. Use =0.5. This allows for the greatest possible error because the maximum possible value of (1-) occurs when both and (1-) are = ½

i.e. when (1-) = 0.5 × 0.5 = 0.25



05.0)1(

96.1

n

05.0 )5.01(5.0

96.1

n

05.0 25.0

96.1 n

22

05.0 25.096.1

n

Squaring both sides

2

2

05.0

25.096.1

n

n > 384.16…

Answer:A sample size of 385 passengers is needed.

Do Sigma p235 – Ex. 14.4 (old version)

p91 – Ex. 5.02 (new version)

Homework: NuLake pg. 96: Q6477.

LESSON 10 – Differences between means 1

The point of today:

Construct confidence intervals for the difference between 2 population means.

• Do NuLake 2.7: pg. 8993.

2007 NCEA exam – C.I.s

Confidence Intervals for the Difference Between 2 Means


Involves comparison between the means of two populations (e.g. males & females).


Involves comparison between the means of two populations (e.g. males & females). We select a random sample from each group and calculate the 2 means, subtracting to get the difference.

We then use this difference to estimate the difference between the means of the 2 populations from which the samples were drawn.

The expected difference between the 2 sample means, is the true difference between the 2 population means: (Central Limit Theorem)

i.e. Mean difference between sample means= diff. between popn means.

21 )( 21 XXE

We select a random sample from each group and calculate the 2 means, subtracting to get the difference.



i.e. Mean difference between sample means= diff. between popn means

21

Sample Mean (point

estimate)

Sample Size

Popn Mean

Variance of Sample Means

)( 21 XXE

We select a random sample from each group and calculate the 2 means, subtracting to get the difference.



i.e. Mean difference between sample means= diff. between popn means

Sample Mean (point

estimate)

Sample Size

Popn Mean


21 XX

2X

1X n1

n2

―

21 )( 21 XXE



i.e. Mean difference between sample means = diff. between popn meansSample

Mean (point estimate)

Sample Size

Popn Mean


n1 n2

―21 XX

2X

1X

21 )( 21 XXE





Sample Size

Popn Mean


n1 1

n2

―

1

21

n

21 XX

2X

1X

21 )( 21 XXE





Sample Size

Popn Mean


n1 1

n2 2

―

1

21

n

21 XX

2X

1X

2

22

n

21 )( 21 XXE





Sample Size

Popn Mean


n1 1

n2 2

― 1 – 2

1

21

n

21 XX

2X

1X

2

22

n

2

22

1

21

nn

)( 21 XXE 21

Sample Mean (point

estimate)

Sample Size

Popn Mean


n1 1

n2 2

― 1 – 221 XX

2X

1X1

21

n

2

22

n

2

22

1

21

nn

So the Standard Error of the difference between 2 sample means is:

Sample Mean (point

estimate)

Sample Size

Popn Mean


n1 1

n2 2

― 1 – 221 XX

2X

1X1

21

n

2

22

n

2

22

1

21

nn


2

22

1

21

21 nnXX

Sample Mean (point

estimate)

Sample Size

Popn Mean


n1 1

n2 2

― 1 – 221 XX

2X

1X1

21

n

2

22

n

2

22

1

21

nn

NOTE:1. The 2 samples must be INDEPENDENT of one another.

2. When finding a confidence interval for the difference between 2 means, we use the popn parameters 1 and 2.

If not told these, we can use the sample SD’s s1 and s2,

provided the sample sizes are large enough (n>30).

So 2

22

1

21

21 nnXX

NOTE:1. The 2 samples must be INDEPENDENT of one another.

2. When finding a confidence interval for the difference between 2 means, we use the popn parameters 1 and 2.

If not told these, we can use the sample SD’s s1 and

s2, provided the sample sizes are large enough.

3. A 95% Confidence Interval tells us that 95% of such intervals will CONTAIN the difference between the POPULATION MEANS.

So 2

22

1

21

21 nnXX


2

22

1

21

21 nnXX

Confidence Intervals for Difference Between 2 Means


2

22

1

21

21 nnXX

Example:

If a random sample of 49 women has a mean life of 76 years with a standard deviation of 8 years and a random sample of 64 men has a mean life of 72 years with a standard deviation of 9 years.

(a) Find a 95% confidence interval for the difference between the mean lifetimes of all women and all men.

(b) What can we conclude about the mean lifespans of all men and all women on the basis of this confidence interval? Justify your answer.


Example:



Solution:

For the women:


Example:



Solution:

For the women:

n1 = 49



Example:



Solution:

For the women: 1Xn1 = 49 =76


Example:



Solution:

For the women: 1Xn1 = 49 =76 s1 = 8


Example:



Solution:

For the women:

For the men:

1Xn1 = 49 =76 s1 = 8

n2 = 64


Example:



Solution:

For the women:

For the men:2X

1Xn1 = 49 =76 s1 = 8

n2 = 64 =72


Example:



Solution:

For the women:

For the men:2X

1Xn1 = 49 =76 s1 = 8

n2 = 64 =72 s2 = 9


Example:



Solution:

For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9

1X

2X


Example:



Solution:

For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9

1X

2X

= 4 yrs

21 XX = 76 – 72

Solution:

For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9

A 95% Confidence Interval for 1-2, the difference between the population mean lifetimes of women and men is:

1X

2X

= 4 yrs

21 XX = 76 – 72

21 XX + z ×Standard Error of

21 XX

4

Use the sample standard deviations – OK if sample is large enough

For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9


1X

2X

= 4 yrs

21 XX = 76 – 72


21 XX

4

4 Margin of

Error E


For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9


1X

2X

= 4 yrs

21 XX = 76 – 72


21 XX

4

4


Margin of

Error E

For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9


1X

2X

= 4 yrs

21 XX = 76 – 72


21 XX

4

4ANSWER: The 95% CI for the difference between the population mean lifetimes of women and men is 0.857yrs < (1-2)< 7.143yrs



ANSWER: Since the interval does not contain a difference of ZERO, there are sufficient grounds to say that there is a difference between the mean lifespans of the populations of all men and all women.

For the women: n1 = 49 =76 s1 = 8

For the men: n2 = 64 =72 s2 = 9


1X

2X

= 4 yrs

21 XX = 76 – 72


21 XX

4

4ANSWER: The 95% CI for the difference between the population mean lifetimes of women and men is 0.857yrs < (1-2)< 7.143yrs



ANSWER: Since the interval does not contain a difference of ZERO, there are sufficient grounds to say that there is a difference between the mean lifespans of the populations of all men and all women. TRY WITH A 99% C.I.

Difference between 2 means exercises

• Do NuLake Ch 2.7: pg. 8993

LESSON 11 – Differences between means 2

The point of today:

Construct confidence intervals for the difference between 2 population means.

• Do Sigma pg. 239 – Ex. 14.5 (old version).

or pg. 83 – Ex. 4.03 (new version)

STARTER:

GO THROUGH PROBLEM FROM HW AS A CLASS.

Do Sigma pg. 239 – Ex. 14.5 (old version). or pg. 83 – Ex. 4.03 (new version)

LESSON 12The distribution of the sample Total.

The point of today:

Construct confidence intervals for the combined total of a sample of items.

• Example

• 2009 NCEA paper (AS90642): Q1b & c.

• Probabilities for sample totals: Ex. 3.03 (pg. 64) – complete for HW.

This is where you are asked to give a confidence interval for the combined total of a sample of n items (or of the entire population of N items).(E.g. total weight of a sample of eight Y13 males).

Confidence Intervals for the Sample Total, Tn

This is where you are asked to give a confidence interval for the combined total of a sample of n items (or of the entire population of N items).(E.g. total weight of a sample of eight Y13 males). You will be told the mean value per item and the standard deviation.

These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .


X


These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .Type 1: Based on , a known population mean.

Type 2: Based on , the mean from a random sample.


X

X


These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .Type 1: Based on , a known population mean. (look at today).

Type 2: Based on , the mean from a random sample. (look at next lesson)


X

X


These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .Type 1: Based on , a known population mean.This is where you are given the mean value per item in the population and asked to construct a confidence interval for the total value of a sample of n items.E.g. Seventeen year-old NZ males have a known mean weight of 80kg, with a standard deviation of 5kg.Construct a 99% CI for the combined total weight of a random sample of 8 students.


X

These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .Type 1: Based on , a known population mean.This is where you are given the mean value per item in the population and asked to construct a confidence interval for the total value of a sample of n items.E.g. Seventeen year-old NZ males have a known mean weight of 80kg, with a standard deviation of 5kg.Construct a 99% CI for the combined total weight of a random sample of 8 students.

Solution: The distribution of the total weight of 8 students is the sum of 8 identically distributed random variables.

Here we know the population mean weight per seventeen year-old male, and the standard deviation, .So we can simply add the means and add the variances.

X

E.g. Seventeen year-old NZ males have a known mean weight of 80kg, with a standard deviation of 5kg.Construct a 99% CI for the combined total weight of a random sample of 8 students.



Distribution of a Total of n independent items:

If X1, X2,………..Xn are n independent sample values, then the sample total is

Tn = X1 + X2,……….+ Xn





Tn = X1 + X2,……….+ Xn

Expected value of total: E[Tn] = E [X1 + X2,……….+ Xn ]

= E[X1]+……… + E[Xn]




Tn = X1 + X2,……….+ Xn


= E[X1]+……… + E[Xn]

Variance of estimates of the total:

Var[Tn] =Var [X1 + X2,……….+ Xn ]

= Var[X1]+……… + Var[Xn]

n

So we can simply add the means and add the variances.



Tn = X1 + X2,……….+ Xn


= E[X1]+……… + E[Xn]

Variance of estimates of the total: Var[Tn] =Var [X1 + X2,……….+ Xn ]

= Var[X1]+……… + Var[Xn]

= nσ2 (if all have equal SD)

So the std. deviation of estimates of the total is:

nT

n

Tn = X1 + X2,……….+ Xn


= E[X1]+……… + E[Xn]


= Var[X1]+……… + Var[Xn]



Back to the example: Total weight of sample of 8 males:

E(T8) =

n

nT

Tn = X1 + X2,……….+ Xn


= E[X1]+……… + E[Xn]


= Var[X1]+……… + Var[Xn]




E(T8) = 8(80)

n

nT

Tn = X1 + X2,……….+ Xn


= E[X1]+……… + E[Xn]


= Var[X1]+……… + Var[Xn]




E(T8) = 8(80) = 640kg.

Var(T8) = 8(52) = 200.

n

nT

Tn = X1 + X2,……….+ Xn


= E[X1]+……… + E[Xn]


= Var[X1]+……… + Var[Xn]




E(T8) = 8(80) = 640kg.

Var(T8) = 8(52) = 200. So σ8 =

200

n

= 14.14213562kg

nT


= Var[X1]+……… + Var[Xn]



Back to the example: Total weight of sample of 8 males:E(T8) = 8(80) = 640kg.

Var(T8) = 8(52) = 200. So σ8 =

99% CI for T is E(T8)

= 640

= 640kg (4sf)

Tz

...14.14 576.2

kg43.36ANSWR: The 99% CI for T8, the total weight of the sample of 8 males is:

603.6kg <T< 676.4kg (all to 4sf)

200 = 14.14213562kg

nT

1. Do 2009 NCEA AS90642 – Q1 (b) and (c)

2. Do Sigma:

- Old (2nd edition): pg. 183 – Ex. 11.3.

- or New: pg. 64 – Ex. 3.03.

LESSON 13Confidence Intervals for Population

Totals

• STARTER: Revise the definition of a Confidence Interval.

• Notes on CI for population totals.

Do NCEA AS90642 – 2009 paper: Q2c. Do NuLake p98-100 (mixed problems). Do NuLake practice assessment (p101).


These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .Type 1: Based on , a known population mean.

Type 2: Based on , the mean from a random sample.


X

X


These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .Type 1: Based on , a known population mean. (looked at last lesson).

Type 2: Based on , the mean from a random sample. (look at today)


X

X


These problems come in 2 types, depending on whether you’re given the population mean, or the mean from a sample .


XType 2: Based on , the mean from a random sample:X

Type 2: Based on , the mean from a random sample:

This is where you are asked to construct a confidence interval for the total value of N items but the population mean per item is unknown.

Instead we are told , the mean from a sample. Then an estimate of the total value of N items is:

To construct a CI for a total based on a sample:

1. Construct a confidence interval for the population mean per item,

2. Multiply the lower and upper bounds of the interval by N , the number of items.

Example:68 Year 13 male students are to be selected at random from throughout NZ to win a prize of an overseas holiday after NCEA exams.

X

XXN


This is where you are asked to construct a confidence interval for the total value of N items but the population mean per item is unknown.

Instead we are told , the mean from a sample. Then an estimate of the total value of N items is:




Example:68 Year 13 male students are to be selected at random from throughout NZ to win a prize of an overseas holiday after NCEA exams. The organisers need to estimate the likely total weight of the students, due to weight restrictions on the aircraft.

X

XXN

Then an estimate of the total value of N items is:




Example:68 Year 13 male students are to be selected at random from throughout NZ to win a prize of an overseas holiday after NCEA exams. The organisers need to estimate the likely total weight of the students, due to weight restrictions on the aircraft.The mean and SD of the popn of all Year 13 males is unknown so they conduct a pilot study by selecting a random sample of 30.

XN

Then an estimate of the total value of N items is:




Example:68 Year 13 male students are to be selected at random from throughout NZ to win a prize of an overseas holiday after NCEA exams. The organisers need to estimate the likely total weight of the students, due to weight restrictions on the aircraft.The mean and SD of the popn of all Year 13 males is unknown so they conduct a pilot study by selecting a random sample of 30. This sample has a mean weight of 76Kg with standard deviation of 7Kg.Construct a 96% CI for the expected total of weight of 68 randomly selected Year 13 students.

XN




Example:

68 Year 13 male students are to be selected at random from throughout NZ to win a prize of an overseas holiday after NCEA exams. The organisers need to estimate the likely total weight of the students, due to weight restrictions on the aircraft.

The mean and SD of the popn of all Year 13 males is unknown so they conduct a pilot study by selecting a random sample of 30. This sample has a mean weight of 76Kg with standard deviation of 7Kg.Construct a 96% CI for the expected total of weight of 68 randomly selected Year 13 students.

Solution:


Example:68 Year 13 male students are to be selected at random from throughout NZ to win a prize of an overseas holiday after NCEA exams. The organisers need to estimate the likely total weight of the students, due to weight restrictions on the aircraft.The mean and SD of the popn of all Year 13 males is unknown so they conduct a pilot study by selecting a random sample of 30. This sample has a mean weight of 76Kg with standard deviation of 7Kg.Construct a 96% CI for the expected total of weight of 68 Year 13 students.

Solution:1.Construct a 96% confidence interval for the popn mean : Interval is given by:

X

nz

76

30

7054.276

625.276



So 96% CI for popn mean weight, is:

nz

76

30

7054.276

625.276



So 96% CI for popn mean weight, is: 73.375kg < < 78.625kg

nz

76

30

7054.276

625.276

The mean and SD of the popn of all Year 13 males is unknown so they conduct a pilot study by selecting a random sample of 30. This sample has a mean weight of 76Kg with standard deviation of 7Kg.Construct a 96% CI for the expected total of weight of 68 Year 13 students.



2.Multiply the lower and upper bounds of the interval by N , the number of items.

96% CI for the expected total weight of the 68 Y13s is: (N × lower limit for ) < TN < (N × upper limit for )

nz

76

30

7054.276

625.276

Construct a 96% CI for the expected total of weight of 68 Year 13 students.



2.Multiply the lower and upper bounds of the interval by N , the number of items.

96% CI for the expected total weight of the 68 Y13s is: (N × lower limit for ) < TN < (N × upper limit for )

= (68 × 73.375) < TN < (68 × < 78.625)

= 4990kg < T68 < 5347kg answer

nz

76

30

7054.276

625.276

1. C.I. for Population Totals:Do 2009 NCEA paper (AS90642): Q2c.

2. Preparation for test: Do NuLake p98-100 (Mixed

problems).

Do NuLake practice assessment (p101)

Sample-size question from 2007 NCEA External Exam

A random sample of size n is taken from a population having a known standard deviation σ. A 95% confidence interval for the population mean is calculated using the sample mean.

A second random sample of size 2n is taken from the same population and a 95% confidence interval for the population mean is calculated using its sample mean.

How many times greater is the width of the first confidence interval than the width of the second confidence interval?

LESSON 14 – ASSESSMENTWhat to study: Do NuLake mixed problems (p98) – merit level qs.

NuLake practice assesment (p101)

More practice (Achieved & Merit):

Do Sigma Confidence Intervals Review exercise: Old: p241 – Ex. 14.6 New: p95 – Ex. 5.03

CIs for totals (Excellence) – past papers:2009 Q2c2008 Q62006 Q7

Statistics and Modelling Course 2011 Topic: Confidence Intervals Achievement Standard 90642...

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