Section 6.1 Confidence Intervals for the Mean(Large Samples)

Objective: SWBAT find a point estimate and a margin of error

And construct and interpret confidence intervals for the population mean

And determine the minimum sample size required when estimating p

• In this section you will learn how to use sample statistics to estimate the value of an unknown population parameter ц when the sample size is 30 or when the population is normally distributed. And the standard deviation σ is known.

Definition A point estimate Is a single value estimate for a population parameter The most unbiased point estimate of the population mean ц is the sample mean ҳ

Example 1

Finding a Point Estimate:Market researchers use the number of sentences per

advertisement as a measure of reliability for magazine advertisements. The following represents a random sample of the number of sentences found in 54 advertisements. Find a point estimate of the population mean ц.

9 20 18 16 9 16 16 9 11 13 22 16 5 18 6 6 5 1225 17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 726 6 11 12 11 15 6 12 14 11 4 9 18 12 12 17 11 20

Solution

• The sample mean of the data is

1

n

ii

XX

n 671=

54 12.4

So your point estimate for the mean length of all magazine advertisements is 12.4 sentences.

Try it yourself

Another random sample of the number of sentences found in 30 magazine advertisements is listed on page 280 use this sample to find the point estimate for ц.

a. Find the sample meanb. Estimate the mean sentence length of the

population.

Definition An interval estimate is an interval or range of values used to estimate a populationParameter.

To form an interval estimate use the point estimate as the center of the interval then add and subtract a margin of error. For instance if the point estimate is 12.4 and the margin of error is 2.1 then an interval estimate would be given by 12.4 ± 2.1 or 10.3 < ц < 14.5

Before finding an Interval estimate you first determine how confident you need to be that your interval estimate contains the population mean ц.

Definition The level of confidence c is the probability

that the interval estimate contains the population parameter.

The level of confidence c is the area under the standard normal curve between the critical values –zc and zc the

area remaining is 1- c and the area in each tail is ½ (1-c)

Margin of ErrorDefinition Given a level of confidence c , the margin of error sometimes also called the maximum error of estimate or error tolerance) E is the greatest possible distance between the point estimate and the value of the parameter it is estimating.

When n ≥ 30 the sample standard deviation s can be used in place of σ.

c X cE z zn

Example 2

Use the data in example 1 and a 95% confidence level to find the margin of error for the mean number of sentences in all magazine advertisements.

Solution : The z score that corresponds to a 95% confidence level is 1.96 This implies that 95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean. n=54 ≥ 30 We don’t know the population standard deviation. But because n ≥ 30 you can use s in place of σ.

Solution (cont.)

2

1

( )1333.24 5.0

1 53

n

ii

X Xs

n

Using the values 1.96cz 5.0 and n=54

5.0 1.96 54

1.3

c

s

E zn

Interpretation: You are 95% confident that the margin of error for the population mean is about 1.3 sentences.

Try it Yourself

Use the data in try it yourself 1 and a 95% confidence level to find the margin of error for the mean number of sentences in a magazine advertisement.

a.) Identify z, n, and s.b.) Find E using zc, σ≈ s , and n c.) State the margin of error.

Confidence Intervals for the Population Mean

Guidelines1. Find the sample statistic n and x

2. Specify σ if known . Otherwise if n≥30 Find the sample standard deviation s and use it as an estimate for σ.3. Find the critical value Zc that correspondsTo the given level of confidence.

4. Find the margin of error E

5. Find the left and right endpoints Left endpoint x - EAnd form the confidence intervals Right endpoint x + E Interval x – E < ц < x + E

2

1

( )

1

n

ii

X Xs

n

cE zn

1 =

n

ii

XX

n

Example 3:

Constructing a Confidence IntervalConstruct a 95% confidence interval for the mean

number of sentences in all magazine advertisements.

Solution in Examples 1 and 2 you found that the mean = 12.4 and E = 1.3 The confidence interval is as follows

X – E = 12.4 – 1.3 = 11.1 x+ E = 12.4 +1.3 = 13.7 11.1 < ц < 13.7

Example 5

Constructing a Confidence Interval, σ knownA college admissions director wishes to estimate

the mean age of all students enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From the past studies the standard deviation is known to be 1.5 years and the population tis normally distributed. Construct a 90% confidence interval of the population mean age.

Example 5 Cont.

Solution: Using n=20,mean =22.9,σ=1.5 and Zc = 1.645 the margin of error at the 90% confidence

interval is

The 90% confidence interval can be written as x ± E = 22.9 ± .6 or as follows 22.9 - .6 = 22.3 or 22.9 + .6 = 23.5 22.3 < ц < 23.5Interpretation: With 90% confidence you can say that

the mean age of all students lies between 22.3 and 23.5 years.

1.51.645 0.620cE z

n

Sample SizeFor the sample statistics as the level of

confidence increases, the confidence interval widens as the interval widens the precision of the estimate decreases. One way to improve the precision of an estimate without decreasing the level of confidence is to increase the sample size. But how large a sample size is needed to guarantee a certain level of confidence for a given margin of error.

Find a Minimum sample size to estimate ц

Given a confidence level c and a margin of error E find the minimum sample size the population mean needed to estimate the population mean ц.

If σ is unknown you can estimate it using s provided you have a preliminary sample with at least 30 members.

2cznE

Example 6

Determining a Minimum sample sizeYou want to estimate the mean number of

sentences in a magazine advertisement. How many magazine advertisements must be in the sample to be 95% confident that the sample mean is within one sentence of the population mean?

Solution

Using c= 0.95 zc = 1.96 , σ≈ s ≈ 5.0 (from example 2) and E =1. You can solve for the

Minimum sample size is 2

cznE

21.96 5.0

1

96.04

When necessary round up to the nearest whole number. So you could include at 97 magazine advertisements in your sample.

Try it Yourself

• How many magazine advertisements must be included in the sample if you want to be 95% confident that the sample mean is within 2 sentences of the population mean

zc = 1.96, E=2, and s ≈ 5.0

21.96 5.02

24.01

Point Estimate

DEFINITION:A point estimate is a single value estimate for a population parameter. The best point estimate of the population meanis the sample mean

Example: Point Estimate

The sample mean is

The point estimate for the price of all one way tickets from Atlanta to Chicago is $101.77.

A random sample of 35 airfare prices (in dollars) for a one-way ticket from Atlanta to Chicago. Find a point estimate for the population mean, .

99101107

102109

98

105103101

10598

107

10496

105

959894

100104111

11487

104

108101

87

103106117

94103101

10590

An interval estimate is an interval or range of values used to estimate a population parameter.

•101.77

Point estimate

( )•101.77

The level of confidence, x, is the probability that the interval estimate contains the population parameter.

Interval Estimates

0 z

Sampling distribution of

For c = 0.95

0.950.0250.025

95% of all sample means will have standard scores between z = -1.96 and z = 1.96

Distribution of Sample MeansWhen the sample size is at least 30, the sampling distribution for is normal.

-1.96 1.96

The maximum error of estimate E is the greatest possible distance between the point estimate and the value of the parameter it is, estimating for a given level of confidence, c.

When n 30, the sample standard deviation, s, can be used for .

Using zc = 1.96, s = 6.69, and n = 35,

You are 95% confident that the maximum error of estimate is $2.22.

Maximum Error of Estimate

Find E, the maximum error of estimate for the one-way plane fare from Atlanta to Chicago for a 95% level of confidence given s = 6.69.

Definition: A c-confidence interval for the population mean is

Confidence Intervals for

You found = 101.77 and E = 2.22

•101.77( )

Left endpoint

99.55

Right endpoint

103.99

With 95% confidence, you can say the mean one-way fare from Atlanta to Chicago is between $99.55 and $103.99.

Find the 95% confidence interval for the one-way plane fare from Atlanta to Chicago.

Sample SizeGiven a c-confidence level and an maximum error of estimate, E, the minimum sample size n, needed to estimate , the population mean is

You should include at least 43 fares in your sample. Since you already have 35, you need 8 more.

You want to estimate the mean one-way fare from Atlanta to Chicago. How many fares must be included in your sample if you want to be 95% confident that the sample mean is within $2 of the population mean?

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