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CHAPTER 4 (PART 2)

STATISTICAL INFERENCES

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4.2.3 Confidence Interval Estimates for Population Proportion

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EXAMPLE 4.4

A survey was conducted to estimate the proportion of the households with a personal computer. On of the 300 households surveyed, 75 had a personal computer.

Find a 95% confidence interval for the proportion in the population who has a personal computer.

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95% confidence interval

300

)75.0(25.096.125.0

ˆˆˆ

2

n

qpzp

]299.0,201.0[049.025.0

Thus, we can state with 95% confidence that proportion of all households with a personal computer is between 20.1% and 29.9%.

EXAMPLE 4.5

229 students answered the question in the class questionnaire about whether or not they support the building of the new Anali stadium. Of these 229, 167 said they did support the building of new stadium.

Estimate the proportion of students who support the building of the stadium using 99% confidence interval.

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EXERCISE 1

A random sample of 400 electronic components manufactured by a certain process are tested, and 30 are found to be defective.

Let p represent the proportion of components manufactured by this process that are defective. Find a 95% confidence interval for p.

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1 2 1 2

4.2.4 The (1 )100% Confidence Interval Estimates for the

Differences between Two Proportions, , ( 30, 30)P P n n

1 1 2 21 2

21 2

ˆ ˆ ˆ ˆ1 1ˆ ˆ

p p p pp p z

n n

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EXAMPLE 4.6

In a survey of 100 people from the city A, 55 people preferred reading Time Post. Meanwhile 46 from 150 people preferred reading Time Post in city B.

Find a 99% confidence interval estimate for the difference in the two proportions.

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SOLUTION

City A : n1 = 100 and x1 = 55

City B : n2 = 150 and x2 = 46

The sample proportions are;

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55.0100

55ˆ1 p

3067.0150

46ˆ 2 p

For 99% confidence interval,

Zα/2 = Z0.01/2 = Z0.005 = 2.576

Since 0 is not included in the interval, we can state with 99% confidence that there is a difference in the two proportions for people who preferred reading Time post.

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2

22

1

112/21

ˆˆˆˆ)ˆˆ(

n

qp

n

qpZpp

150

)6993.0(3067.0

100

)45.0(55.0576.2)3067.055.0(

01.02433.0

]2533.0,2333.0[

EXAMPLE 4.7

In a study to compare the effects of two pain relievers, it was found that of n1=200 randomly selected individuals instructed to use Excedrin, 93% indicated that it relieves their pain. Of n2=450 randomly selected individuals instructed to use the Tylenol, 96% indicated that it relieves their pain.

Find a 99% confidence interval for the difference in the proportions experiencing relief from pain for these two pain relievers.

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SOLUTION

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Samples of 400 printed circuit boards were selected form each of two production lines A and B. Line A produced 40 defectives, and line B produced 80 defectives.

Estimate the difference in the actual fractions of defectives for the two lines with a confidence coefficient of 0.9.

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EXERCISE 1

4.2.5 Error of Estimation and Determining the Sample size

Definition 4.3:

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2

/2

If is used as an estimate of , we can be 100(1- )% confident

that the error | | will not exceed a specified amount when the

sample size is

x

x E

zn

E

Definition 4.4:

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/2

/2

ˆIf is used as an estimate of , we can assert with

ˆ ˆ(1 )(1- )100% confidence that the error is less than .

ˆ ˆ(1 )If we set and solve for , the appropriate

sample size is

xp p

n

p pz

n

p pE z n

n

2

/2

ˆ ˆ (1 )z

n p pE

EXAMPLE 4.8

The dean of a university wants to use the mean of a random sample to estimate the average amount of time taken by the students to get from one class to the next class. She wants to assert with probability 0.95 that her error will be at most 0.25 minute. If she knows from the previous studies, minutes, how large a sample will she need?

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5.1

SOLUTION

Given 95% confidence interval, Zα/2 = 1.96

E = 0.25, σ = 1.5

Use formula

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2

25.0

)5.1(96.1

n

2976.138n

139n

EXAMPLE 4.7

A quality control engineer wants to estimate the fraction of defectives in a large

lot of film cartridges. From the

previous experience, he feels that

the actual fraction of defectives

should be somewhere around 0.05.

How large a sample should he take

if he wants to estimate the true fraction to within 0.01, using a 95% confidence interval?

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SOLUTION

Given 95% confidence interval, Zα/2 = 1.96

E = 0.01,

Use formula

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)95.0(05.001.0

96.12

n

76.1824n

1825n

05.0ˆ p

EXAMPLE 4.8

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A researcher wishes to estimate, with 95% confidence, the proportion of people who own a home computer. A previous study shows that 40% of those interviewed had a computer at home. The researcher took a sample size n = 2305. Calculate the error of estimation.

SOLUTION

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EXTRA EXERCISES

The mean and standard deviation of the speeds of the sample of 69 skaters at the end of the 6-meter distance were 5.753 and 0.892 meters per second, respectively.a)Find a 95% confidence interval for the mean velocity at the 6-meter mark. Interpret the interval.b)Suppose you wanted to repeat the experiment and you wanted to estimate this mean velocity correct to within 0.01 second, with probability 0.99. How many skaters would have to be included in your sample?

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EXTRA EXERCISES (from Exercise 1)

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