Sample sample distribution

SAMPLE AND SAMPLE DISTRIBUTIONS CN303/3/ 1

SAMPLE AND SAMPLE DISTRIBUTIONS

OBJECTIVES

General Objective

To understand and the concept of sampling and sample distributions

Specific Objectives

At the end of the unit you should be able to:

Define the sampling distribution concept which is the base for inferentialstatistics.

Express the relationship between statistical samples and populationparameters.

Explain the concept of sampling distribution of sample means based onrandom sample taken with and without replacement from a population.

Calculate the mean, variance and standard deviation of the distribution ofthe sample means taken with or without replacement from a population.

State the criteria for big samples (n>30). Study the characteristics of the distributions of the means of samples

taken from a population. Use the central limit theorem to solve the probability problems involving

distribution of sample means for large number of samples.

UNIT 3


3.0 INTRODUCTION

As an engineer, you are required to find out the mean value of the service life fornewly developed light bulbs. One of the approaches is to randomly pick out, say50 light bulbs from the whole population of thousand bulbs produced and havethem tested. In doing so, you can approximate the mean value for the bulbs. Thismethod is known as sampling.

3.1 SAMPLE DISTRIBUTIONS

Every sample is a subset from a population. By studying the sample, it ispossible to find out the characteristics of the sample and eventually determinethe characteristics of the whole population. It would be ideal if the sample were aperfect miniature of the population in all characteristics. This ideal, however, isimpossible to achieve. The best that can be done is to select a sample that willbe representative with respect to some characteristics, preferably thosepertaining to the study.

For a sample to be a random sample, every member of the population musthave an equal chance to be selected. If selected without being biased, it willbecome the representative of the population.

INPUT


3.1.1 SAMPLE STATISTICS AND POPULATION PARAMETERS

Probability distribution concept can be applied for sample statistics. An exampleof sample statistics is the measurement for central tendency for a given sample

such as the mean

)(x or the variation such as standard deviation, S. Thepopulation mean, and the population standard deviation, are themeasurement for the central tendency of a sample. Below is a table for samplestatistics and population parameters:

Quantity Samplestatistics

Population parameters

Size N NMean

x

Variance 2s 2Standarddeviation

S

Proportion ^p p

3.1.2 DISTRIBUTION OF SAMPLE MEANS

If we select 100 samples of a specific size from a large population and computethe mean of the same variable for each 100 samples. The sample means,

10021 ..., xxx , constitute a sampling distribution of sample means.

If the samples are randomly selected with replacement, the sample means, formost part, will be somewhat different from the population mean . Thesedifferences are caused by sampling error.

Properties of the distribution of sample Means

1. The mean of the sample means will be the same as thepopulation mean.

2. The standard deviation of the sample means will be smaller thanthe standard deviation of the population, and will be equal to thepopulation standard deviation divided by the square root of thesample size


Example 3.1

1. Suppose a lecturer gave an eight point quiz to a small class of fourstudents. The results of the quiz were 2, 6, 4, and 8. Assume the fourstudents constitute the population.

Find i) The population , and draw the graph of the sample means.ii) xx , of the sample means

2. Assume that we have a population consisting of three numbers 1, 2, and3. The probability distributions for these numbers are

X 1 2 3P(x) 1/3 1/3 1/3

Find i) The population means, variance and standard deviationii) Now, if all samples of size 2 are taken with replacement, and the

mean of each sample is found, find:

a) The probability distribution for sample means, x , draw a tableb) The mean for the sample meansc) The variance and standard deviation for sample means

Solution to Example 3.1

1. The mean of the population is

54

8462

,

The standard deviation of the population is

236.24

)58()54()56()52( 2222


Below is the graph of the sample means. The graph appears to besomewhat normal, even though it is a histogram.

1 2 3 4

score

freq

uenc

y,

1

Now, if all samples of size 2 are taken with replacement, and the mean of eachsample is found, the distribution is shown next. (You can draw a tree diagram ifyou wish)

Sample Mean Sample Mean2, 2 2 6, 2 42, 4 3 6, 4 52, 6 4 6, 6 62, 8 5 6, 8 74, 2 3 8, 2 54, 4 4 8, 4 64, 6 5 8, 6 74, 8 6 8, 8 8

A frequency distribution of sample means is as follows.

X F

2 13 24 35 46 37 28 1


Below is the graph of the sample means. The graph appears to be somewhatnormal, even though it is a histogram.

012345

2 3 4 5 6 7 8sample mean

frequency

The mean of the sample means, denoted by 51680

168...32

x which is the

same as the population mean. Hence xThe standard deviation of the sample means denoted by

581.116

)58..()53()52( 222

x which the same as the population

standard deviation is divided by 2 : 581.12

236.2

x

Note: if all possible sample of size n are taken with replacement from the samepopulation, the mean of the sample means, denoted by

x , equals to the

population mean ; and the standard deviation of the sample means, denotedby

x , equals n

.

2. Population mean,23/6)321(3/1)3/1(3)3/1(2)3/1(1)()( xxpXE

Population variance, )]([)( 222 XEXEs = )(3)(2)(1 3

12312

312

= )941(31

= 314

322

3142 2

Therefore 32


ii)

Sample Mean,

x1, 1 1.01, 2 1.51, 3 2.02, 1 1.52, 2 2.02, 3 2.53, 1 2.03, 2 2.53, 3 3.0

The probability distribution for sample means, x

x 1.0 1.5 2.0 2.5 3.0

P( x ) 1/9 2/9 3/9 2/9 1/9

You can draw a histogram for sample means,

x against P(

x ) as in ‘activity 3A’and then find the mean for the sample means, )()( xpxxEx =

)(3)(5.2)(2)(5.1)(1 91

92

93

92

31

= 918

=2= (population mean)

Variance for the sample means, x is)()]([)(( 2222 xpxxExEx

= )(0.2)(5.1)(1 932

922

912

= 313

312

3132 2 x


Standard deviation for sample means, 31: xx

Look: 31x =

nn

n

2&,2 3

222

32


ACTIVITY 3A

TEST YOUR UNDERSTANDING BEFORE PROCEEDING TO THE NEXTINPUT…!

1. Let the population consist of the digits 1, 2 and 3. Find the populationmean and the population standard deviation.

2. 10000 female students are found to have a mean weight of 63 kg with astandard deviation of 7 kg. 100 samples of size 36 are taken, withoutreplacement, from the above. Estimate the mean and standard deviationof the sample-means.


FEEDBACK TO ACTIVITY 3A

1. ,2 x = 32

2. = 63 and x = 1.17


3.2 THE CENTRAL LIMIT THEOREM

As the sample size n increases, the shape of the distribution of the samplemeans taken with replacement from a population with mean and standarddeviation will approach the normal distribution. As previously shown, thisdistribution will have a mean and a standard deviation

n

The central limit theorem can be used to answer questions about sample meansin the same manner that the normal distribution can be used to answer questionsabout individual values. The only difference is that a new formula must be usedfor the z values.

n

Xz

Notice that X is the sample mean, and the denominator is the

standard error of the mean. It is important to remember two things when usingthe central limit theorem:

When the original variable is normally distributed, the distribution of the samplemeans will be normally distributed, for any sample size n.When the distribution of the original variable departs from normality, a samplesize of 30 or more is needed to use the normal distribution to approximate thedistribution of the sample means. The larger the sample, the better theapproximation will be.

……………………………………………………………………………………………..

INPUT


NOTE

Since the sample size is 30 or larger, the normality assumption is not necessary,

as in the example above. When do we usen

Xz/

or

Xz ?

The formulan

Xz/

should be used to gain information about a sample mean

whereas the formula

Xz is used to gain information about an individual

data value obtained from the population. See the example below.

……………………………………………………………………………………………

Example 3.2

1. Students in semester 1 and 2 in Polytechnics spend an average of 25hours sleeping in a week. Assume the variable is normally distributed andthe standard deviation is 3 hours. If 20 students from semester 1 and 2are randomly selected, find the probability that the mean of the number ofhours they sleep will be greater than 26.3 hours.

2. The average age of motorcycles registered in polytechnics is 8 years, or96 months. Assume the standard deviation is 16 months. If a randomsample of 36 motorcycles is selected, find the probability that the mean oftheir ages is between 90 and 100 months.

3. The average number of pounds of meat a person consumes a year is218.4 pounds. Assume that the standard deviation is 25 pounds and thedistribution is approximately normal.

i) Find the probability that a person selected at random consumesless than 224 pounds per year.

ii) If a sample of 40 individuals is selected, find the probability thatthe mean of the sample will be less than 224 pounds per year.



1. Since the variable is approximately normally distributed, the distribution ofsample means will be approximately normal, with a mean of 25. Thestandard deviation of the sample means is

671.0203

nx

The distribution of the means is shown above, with the appropriate area shaded.

The z-value isn

Xz

= 94.1671.03.1

203

253.26

The area between 0 and 1.94 is 0.4738. Since the desired area is in the tail,subtract 0.4738 from 0.5000. Hence 0.5000 – 0.4738 = 0.0262, or 2.62%.One can conclude that the probability of obtaining a sample mean larger than26.3 hours is 2.62% (i.e., %62.2)3.26)( XP )

25 26.3


2. The desired area is shown in the figure below:

The two z-values are

25.236/169690

1

z and 50.136/1696100

2

z

The two areas corresponding to the z values 0f -2.25 and 1.50, respectively, are0.4878 and 0.4332. Since the z-values are on opposite sides of the mean, findthe probability of adding the areas: 0.478 + 0.4332 = 0.921, or 92.1%.

Hence, the probability of obtaining a sample mean between 90 and 100 monthsis 92.1% i.e., P(90< X <100) = 92.1%.

90 96 100


3. (i) Since the question asks about an individual person, the formula

Xz is used. The distribution is shown in the figure below.

The z value is 22.025

4.218224

Xz

The area between 0 and 0.22 is 0.0871; this area must be added to 0.5000 to getthe total area to the left of z = 0.22.

0.0871 + 0.5000 = 0.5871

Hence, the probability of selecting an individual who consumes less than 224pounds of meat per year is 0.5871, or 58.71% ( i.e., P(X<224) = 0.5871.

218.4 224Distribution of individual datavalues for the population


(ii) Since the question concerns the mean of a sample with a size of

40, the formulan

Xz/

is used. The area is shown in the figure

below:

The z value is

42.1

4025

4.218224/

n

Xz

The area between z = 0 and z = 1.42 is 0.422; this value must be added to0.5000 to get the total area.

0.422 + 0.5000 = 0.9222

Hence, the probability that the mean of a sample of 40 individuals is less than224 pounds per year is 0.9222, or 92.22%. That is P( 9222.0)224 X

218.4 224


Comparing the two probabilities, one can see that the probability of selecting anindividual who consumes less than 224 pounds of meat per year is 58.71%, butthe probability of selecting a sample of 40 people with a mean consumption ofmeat that is less than 224 pounds per year is 92.22%. This rather largedifference is due to the fact that the distribution of sample means is much lessvariable than the distribution of individual data values.


ACTIVITY 3B


1. The average salary for workers at an electronic factory is RM13.50 perhour. Assume that the standard deviation is RM2.90 per hour and thedistribution is approximately normal. If X is the mean salary per hour for arandom sample of the workers at the factory, find the mean and standarddeviation for a sample distribution X if the sample size is (a) 30 workers,and (b) 75 workers

2. The average weight of sugar sachets is 32 grams. Assume the standarddeviation is 0.3 gram. If a random sample of 20 sachets is selected, findthe probability that the mean of their weight is between 31.8 and 31.9grams.

3. Analysis of 150 compressive strength results gave a mean strength of 32N/mm2 and standard deviation 6.5 N/mm2. Given that 10 samples of 12results are considered, find the number of samples with mean strengthgreater than 33 N/mm2.

4. Asbestos-cement sheets are manufactured with a mean length 2400 mmand standard deviation 3 mm. Given that 20 batches consisting of 3 dozensheets are considered, determine

(a) the probability that a batch (chosen at random) has a meanlength between 2399.5 mm and 2400.6 mm

(b) the number of batches with mean length less than 2399.3 mm.


FEEDBACK TO ACTIVITY 3B

1. a) 53.0,50.13, xx RMX b) 33.0,50.13, xx RMX

2. 0.667 or 66.7%

3. x =32 N/mm2, x =1.81 N/mm2, P( 2912.0)33 x 3 samples

4. (a) P(2399.5< x <2400.6) = 0.7262 (b) P( x <2399.3) = 0.0808


3.3 DISTRIBUTION OF THE SAMPLE MEANS

a. Distribution of the sample means with replacement

Statement 1

The shape of the distribution of the sample means X taken with replacementfrom a known population with mean and standard deviation , regardless ofthe sample size (n), will approach the normal distribution. As previously shown,this distribution will have a mean and a standard deviation

n

Statement 2

If the sample is taken from any population with known and , and the samplesize is very large (n30), the distribution of sample mean is almost normal withmin and standard deviation that is ),( 2

nNx

b. Distribution of the sample means without replacement

The formula for the standard error of the mean,n , is accurate when the sample

are drawn with replacement or without replacement from a very large or infinitepopulation. Since sampling with replacement is for the most part unrealistic, acorrection factor is necessary for computing the standard error of the mean forsamples drawn without replacement from a finite population. Compute thecorrection factor by using the following formula:

INPUT


1NnN where N is the population size and n is the sample size.

This correction factor is necessary if relatively large samples aretaken from a small population, because the sample mean will then be moreaccurately estimate the population means and there will be less error in theestimation. Therefore, the standard error of the mean must be multiplied by thecorrection factor to adjust it for large samples taken from a small population. Thatis

1

NnN

nx

Finally the formula for the z value becomes

1.

NnN

n

Xz

When the population is large and the sample is small, the correction factor isgenerally not used, since it will be very close to 1.000. Therefore

nx .


Example 3.3

1. The average price of houses in Jitra is RM157000 and is rather skewed.Assume the standard deviation is RM29500. If x is the mean price for asample of 400 houses selected at random, find the probability:a) That the sample mean is between RM154000 and 160000.b) That the mean price for this sample is below RM154000.

2. The average time taken by line workers in an electronic firm to assemblethe electronic components is 80 hours with the standard deviation of 8hours. Find the probabilities (P) of the mean assembly time if a randomsample consisting of 16 workers is selected.

a. )8278( xPb. )8476( xPc. )8674( xP

3. The average service hour of 400 batteries is 800 with the standarddeviation of 45. If a random sample of 45 batteries is selected, what is theprobability that the sample mean is between 790 and 810 hours.

4. The data shows the number of children belonging to a group of 50Polytechnic lecturers.

No. ofchildren

0 1 2 3 4

No. oflecturers

1 18 24 4 3

a. Find the mean and the standard deviation of the data above.b. If a sample of 10 lecturers is taken, find the mean number of children

of this sample that is more than 2.



1. Although the price of houses in Jitra is skewed and not normally distributed,the sample mean price is rather normal due to the big sample size (n=400).

Therefore the central limit theorem is applicable.

Given =157000 and =RM29500.157000RMx

1475400

29500 RMnx Therefore

)1475,157000( 2Nx

a. )160000154000( xP

)1475

1570001600001475157000

1475)157000154000(

xP

=P(-2.03 )03.2 z=0.976

b. 0212.0)03.2(1475

1570001545001475157000)154500(

ZPxPxP

a. )8278( xPb. )8476( xP


2. Although the sample size is small (n=16), the time distribution toassemble the components is normally distributed. Therefore the distributionof the sample mean is normally distributed with mean = 80 hours and thestandard deviation

168x = 2 hours.

a.

2

8082280

28078)8278( xPxP

= )11( ZP=0.6826

b.

2

8084280

28076)8476( xPxP

= )22( ZP= 0.9544

c.

2

8086280

28074)8674( xPxP

= )33( ZP= 0.9974

3. The probability that the mean sample is between 790 and 810 hours is0.9066.

4. The probability distribution is:

No. of children(x) 0 1 2 3 4Relative frequency, p(x) 0.02 0.36 0.48 0.08 0.06

a) )(xxp = 28.1)06.0(4)08.0(3)48.0(2)36.0(1)02.0(0

72.0)06.0(4)08.0(3)48.0(2)36.0(1)02.0(0)()( 22222222 xpx

8445.072.0


b) Due to large samples (N = 50) and 10 lecturers were selected withoutreplacement, the sampling distribution for sample means is almost normalwith 8.1x and

2424.0150

105010

8485.01

NnN

nx Therefore

)2424.0,8.1( 2Nx

2033.0)83.0(2424.0

8.122424.0

8.1)2(

ZPxPxP


ACTIVITY 3C


1. The heights of 2500 men are normally distributed with a mean of 170 cm and astandard deviation of 7 cm. If random samples are taken of 30 men, predictthe standard deviation and the mean of sampling distribution of means, ifsampling is done (a) with replacement, and (b) without replacement.

2. A group of 1000 ingots of metal have a mean mass of 7.4 kg and a standarddeviation of 0.4 kg. Find the probability that a sample of 50 ingots chosen atrandom from the group, without replacement, will have a combined mass of (a)between 360 and 377.5 kg, and (b) more than 375 kg.

3. Determine the mean and standard deviation of the set of numbers 1, 2, 4, 5,and 6, correct to three decimal places. By selecting all possible differentsamples of size 2 which can be drawn with replacement (25 pairs) determine

(a) the mean of the sampling distribution of means, and (b) the standard errorof the means, correct to three decimal places.

4. Determine the standard error of the means for problem 3, if sampling is withoutreplacement, correct to three significant figures.

5. The length of 1500 bolts is normally distributed with a mean of 22.4 cm and astandard deviation of 0.048 cm. If 30 samples are drawn at random from thispopulation, each of size 36 bolts, determine the mean of the sampling

distribution and the standard error of the means when sampling is done withreplacement.

6. Determine the standard error of the means in problem 5, if sampling is donewithout replacement, correct to 4 decimal places.


7. If a random sample of 64 lamps is drawn from a batch, determine theprobability that the mean time to failure will be less than 785 hours, correct to 3decimal places.

8. Determine the probability that the mean time to failure of a random sample of16 lamps will be between 790 hours and 810 hours, correct to 3 decimalplaces.

9. For a random sample of 64 lamps, determine the probability that the meantime to failure will exceed 820 hours, correct to 2 significant figures.


FEEDBACK TO ACTIVITY 3C

1. (a) x 1.278 cm (b) x 1.271 cm, x 170 cm

2. The mean of the sampling distribution of means = x 7.4 kgThe standard error of the means, x 0.0552 kg(a) 0.9966 (b) 0.0351

3. ,855.1,6000.3 (a) 600.3x , (b) x 1.312

4. x = 1.136

5. x =22.4 cm, x =0.08 cm

6. x = 0.0079 cm

7. 0.023

8. 0.497

9. 0.0038


SELF ASSESSMENT 3

You are approaching success. Try all the questions in this self-assessmentsection and check your answers on the next page. If you encounter anyproblems, consult your instructor. Good luck.

1. If the samples of a specific size are selected from a population and themeans are computed, what is this distribution of means called?

2. What is the mean of the sample means?3. What does the central limit theorem say about the shape of the distribution

of sample means?4. What formula is used to gain information about a sample mean when the

variable is normally distributed or when the sample size is 30 or more?

For exercise below, assume that the sample is taken from a largepopulation and the correction factor can be ignored.

5. The mean serum cholesterol of a large population of overweightadults is 220 Mg/dl and the standard deviation is 16.3 mg/dl. If a sample ofadults is selected. Find the probability that the mean will be between 220and 222 mg/dl.

6. The mean weight of 18 year old females is 126 pound, and the standarddeviation is 15.7. If the sample of 25 females is selected, find theprobability that the mean of the sample will be greater than 128.3 pounds.Assume the variable is normally distributed.


7. The average price of the pound of sliced bacon is RM2.02. Assume thestandard deviation is RM0.08. If a random sample of 40 one-poundpackages is selected, find the probability that the mean of the sample willbe less than RM2.00.

8. The mean score on a dexterity test for 12 year old is 30. The standarddeviation is if a psychologist admitters the test to a class of 22 student,find the probability that the mean of the sample will be between 27 and 31.Assume the variable is normally distributed.

9. The average age of lawyers is 43.6 years, with a standard deviation of 5.1years. If the law firm employs 50 lawyers, find the probability that theaverage age of the group is greater than 44.2 years old.

10. Procter & Gamble reported that an American family of 4 washes anaverage of one ton (2000 pounds) of clothes each year. If the standarddeviation of the distribution is 187.5 pounds, find the probability that themean of the randomly selected sample of 50 families or four will bebetween 1980 and 1990 pounds.

11. The average time it taken a group of adults to complete a certainachievement test is 46.2 minutes. The standard deviation is 80 minutes.Assume the variable is normally distributed

a) Find the probability that a randomly selected adult willcomplete the test in less than 43 minutes.

b) Find the probability that if 50 randomly selected adults takethe test, the mean time it takes the group to complete thetest will be less than 43 minutes.

c) Does it seem reasonable that an adult would finish the test inless than 43 minutes? Explain

d) Does it seem reasonable that the mean of the 50 adultscould be less than 43 minutes?


12. The average cholesterol content of a certain brand of eggs is 215milligrams and the standard deviation is 15 milligrams. Assume thevariable is normally distributed.

a) If a single egg is selected, find the probability that thecholesterol content will be more than 220 milligrams.

b) If a sample of eggs is selected, find the probability that themean of the sample will be larger than 220 milligrams.

13. The average labor cost for car repairs for a large chain of car repair shop isRM 48.25. The standard deviation is RM 4.20. Assume the variable isnormally distributed.

(a) If a store is selected at random, find the probability that thelabour cost will range between RM 46 and RM 48

(b) If stores are selected at random, find the probability that themean of the sample will be between RM 46 and RM 48.© Which answer is larger? Explain why.


FEEDBACK TO SELF-ASSESSMENT 3

Have you tried the questions??? If “YES”, check your answers now.

1. The distribution is called the sampling distribution of sample means.2. The mean of the mean is equal to the population mean.3. The distribution will be approximately normal when the sample size is

large.

4. z =n

x/

5. 0.24866. 0.23277. 0.05718. 0.82399. 0.203310.0.125411.a) 0.3446

b) 0.0023c) Yes , since it is within one standard deviation of the mean.d) very unlikely

12. a) 0.3707 b) 0.0475

13. a) 0.1815 b) 0.3854 c) Means are less variable than individual data.

Sample sample distribution

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