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Lecture 10: Con dence Intervals Part 1

(Sections 7.1 and 7.2)

Exercise 1. Let’s suppose we wanted to nd the average amount of time De Anza College students spend watching baseball per week. Since it is impractical to ask every single student at De

Anza for this piece of information, we decide to take a simple random sample of 64 students. The

average time for these 64 students is 4 hours. It is also known that the standard deviation for the

entire college is 1 hour.

(a) How likely is it that the average for the entire college is exactly 4 hours?

It is

very unlikely that the average is exactly four hours.

(b) How likely is it that the average is close” to 4 hours?

This depends on your de nition of the word close”.

so big, it is likely that 4 hours is close”.

Since the sample size

is

In order to address part (b) more precisely, we will use the concept of con dence intervals.

De nition: A con dence interval (or interval estimate) is a range (or an interval) of val-

ues used to estimate the true value of a population parameter. A con dence interval is sometimes

abbreviated as CI.

Example: An example of a con dence interval for the mean baseball watching time would be

(3, 5).

Exercise 2. What is another example of a con dence interval?

A con dence interval is associated with a con dence level, such as 0.95.

De nition: The con dence level is the probability 1 – (often expressed as the equivalent

percentage value) that the con dence interval actually does contain the population parameter,

assuming that the estimation process is repeated a large number of times.

Example: For a 95% con dence interval,

= 1 – 0.95 = 0.05.

To construct con dence intervals, we will use critical values.

De nition: De nition: For a con dence level c, the critical value zc is the number such that

the area under the standard normal curve between -zc and zc equals c.

Exercise 3. Find the critical value zc that corresponds to a 95% con dence level.

A 95% con dence interval corresponds to

Here is a picture:

1

=

1- 0.95 = 0.05.

Class Exercise 1. Find the critical value zc that corresponds to a 90% con dence level.

Class Exercise 2. Find the critical value zc that corresponds to a 98% con dence level.

The following technical de nition is fairly deep so read it over a few times when you go home:

De nition: A level C con dence interval for a population mean is an interval computed from

sample data by a method that has probability C of producing an interval containing the true value

of a population mean.

Using the class exercise and the de nition above, a 95% con dence interval for is:

p

p

(x – 1:959963986 = n, x + 1:959963986 = n).

The above interval is typically written the following way:

p

x 1:959963986 = n

(*).

Remark: By the class exercise above, 95% of all 95% con dence intervals contain the population

mean.

Exercise 4. Let’s suppose we wanted to nd the average amount of time De Anza College students spend watching baseball per week. Since it is impractical to ask every single student at De

Anza for this piece of information, we decide to take a simple random sample of 64 students. The

average time for these 64 students is 4 hours. It is also known that the standard deviation for the

entire college is 1 hour. Construct a 95% con dence interval for the population mean.

2

By (*), the 95% con dence interval is x

Substituting x =

p

1:96 =

n.

4, = 1, and n = 64 yields

p

4 1.9599639861/ 64.

The last expression simpli es to (3.755, 4.245). The 95% con dence interval for the average

amount of time De Anza College students spend watching baseball per week is (3.755, 4.245).

The con dence interval for estimating a population mean is:

x zc pn .

Here are the three requirements of a sample in order to construct a con dence interval:

1) The sample is a simple random sample.

2) The value of the population standard deviation is known.

3) Either or both of these conditions is satis ed: The population is normally distributed or n > 30

Exercise 5. Suppose you wanted to estimate the mean SAT Math score for the more than 385,000

high school seniors in California. Only about 49% of California students take the SAT. These selfselected seniors are planning to attend college and so are not representative of all California seniors.

You know better than to make inferences about the population based on the sample data. At considerable e ort and expense, you give the test to a simple random sample (SRS) of 500 California

high school seniors. The mean for your sample is x = 461. Assume that = 100. Construct and

interpret a 99% con dence interval for the mean score in the population of all 385,000 seniors.

Are the three requirements satis ed?

The rst step is to nd the margin of error. To do that, we need to nd zc .

So, zc =

Applying the formula,

E = z =2 pn =

The con dence interval is (x – E, x + E).

So, the con dence interval is

We are 99% con dent that the population mean is in the interval:

Class Exercise 3. You work for a consumer advocate agency and want to nd the mean repair

cost of a washing machine. As part of your study, you randomly select 40 repair costs and nd the

3

mean to be $120.00. The population standard deviation is $17.50. Construct and interpret a 95%

con dence interval for the population mean repair cost. Answer: (114.58, 125.42)

Class Exercise 4. Hoping to lure more shoppers downtown, a city builds a new public parking

garage in the central business district. The city plans to pay for the structure through parking

fees. During a two-month period (44 weekdays), daily fees collected average $126. Assume =

$15. Write and interpret a 90% con dence interval for the mean daily income this parking garage

will generate. Answer: (122.28, 129.72)

Exercise 6. A publisher wants to estimate the mean length of time (in minutes) all adults spend

reading newspapers. To determine the estimate, the publisher takes a random sample of 15 people

and obtains the following results: 11, 9, 8, 10, 10, 9, 7, 11, 11, 7, 6, 9, 10, 8, and 10. From

past studies, the publisher assumes is 1.5 minutes and that the population of times is normally

distributed. Construct and interpret a 90% con dence interval for the mean length of time.

The rst step is to nd the margin of error. To do that, we need to nd zc .

So, zc =

= 9.067.

x = 11+9+8+10+10+9+7+11+11+7+6+9+10+8+10

15

4

Applying the formula,

E = zc pn =

Again, the con dence interval is (x – E, x + E).

So, the con dence interval is

We are 90% con dent that the population mean is in the interval:

(8.430, 9.704).

Class Exercise 5. A manufacturer of high-resolution video terminals must control the tension

on the mesh of ne wires that lies behind the surface of the viewing screen. Too much tension will

tear the mesh and too little will allow wrinkles. The tension is measured by an electrical device

with output readings in millivolts (mV). Some variation is inherent in the production process.

Careful study has shown that when the process is operating properly, the standard deviation of

the tension readings is = 43 mV. Here are the tensions readings from an SRS of 20 screens

from a single day’s production: 269.5, 297, 269.6, 283.3, 304.8, 280.4, 233.5, 257.4, 317.5, 327.4,

264.7, 307.7, 310.0, 343.3, 328.1, 342.6, 338.8, 340.1, 374.6, and 336.1. Construct and interpret a

90% con dence interval for the mean tension of all the screens produced on this day. Answer:

(290.51, 322.13)

Class Exercise 6. A professor is interested in knowing the average IQ of members of his hometown

of Boston. The mean for Boston is not known, but it is known that the standard deviation is 15.

In order to make an estimate, the professor nds the IQ of 25 of his friends. The average IQ of his

friends is 120. Find the 95% con dence interval for the average IQ of Bostonians.

De nition: The margin of error (or error bound) is one half of the width of the con dence

interval.

Exercise 7. What is the margin of error in the previous exercise?

The margin of error is

p

1.9599639861/

64

or 0.245.

Formula: In general, the a% con dence interval for the population mean is given by

x invNorm( 1+a=2 100 ) pn .

(**)

Exercise 8. What is the formula for the 92% con dence interval for the population mean?

From above, the formula for the a% con dence interval for the population mean is

5

Since a =

92, the con

x invNorm( 1+a=2 100 ) pn .

dence interval is:

x

invNorm( 1+922=100 ) pn .

The above expression simpli es to:

x

1.750686071 pn .

The 92% con dence interval is given by the following formula: x

1.750686071 pn .

Class Exercise 7. Find the formula for the 94% con dence interval for the population mean.

Answer: x 1.88079361 pn

Formula: The error bound for the a% con dence interval for the population mean is

invNorm( 1+a=2 100 ) pn .

Exercise 9. Using the formula above, what is the error bound for the 97% con dence interval for

the population mean if = 4 and n = 15?

Substituting 97 for a, 4 for , and 15 for n gives:

invNorm( 1+972=100 ) p415

6

2.241.

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