Ch. 6 Larson/Farber

Elementary Statistics

Larson Farber

Unit 6: Confidence Intervals

Ch. 6 Larson/Farber

Definition Review

Ch. 6 Larson/Farber

Big Picture – Confidence Intervals

A group of college students collected data on the speed of vehicles traveling through a construction zone on a state highway, where the posted speed was 25 mph. Assume that the standard deviation for the recorded speed of the vehicles is 3.5 mph. The recorded speed of 14 randomly selected vehicles is as follows:

20, 24, 27, 28, 29, 30, 32, 33, 34, 36, 38, 39, 40, 40

þ Assuming speeds are approximately normally distributed, how fast do you think the true mean speed of drivers in this construction zone is?

 32.14

6.18

14x

x

s

n

?

3.5

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6.18xs

32.14x

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Construction of a Confidence Interval

þ The construction of a confidence interval for the population mean depends upon three factors

– The point estimate of the population– The level of confidence – The standard deviation of the sample mean

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Confidence Intervals for the

Mean(large samples)

Section 6.1

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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

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Part I: 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

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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

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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 Means

When the sample size is at least 30, the sampling distribution for is normal.

-1.96 1.96

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zc

Solution: Finding the Margin of Error

zzcz = 0

0.95

0.0250.025

-zc = -1.96

95% of the area under the standard normal curve falls within 1.96 standard deviations of the mean.

zc = 1.96

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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 is greater than 30, the sample standard deviation, s, can be used for .

Maximum Error of Estimate

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Part 2: 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.

A random sample of 35 airfare prices (in dollars) for a one-way ticket from

Atlanta to Chicago.

99101107

102109

98

105103101

10598

107

10496

105

959894

100104111

11487

104

108101

87

103106117

94103101

10590

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Using zc = 1.96,

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.

s = 6.69n = 35

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Confidence Intervals for the Population Mean

A c-confidence interval for the population mean μ

•

• The probability that the confidence interval

contains μ is c.

where cx E x E E zn

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Part III: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.

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How could we get closer?

•101.77( )

99.55 103.99

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Construction of a Confidence Interval

þ The construction of a confidence interval for the population mean depends upon three factors

– The point estimate of the population– The level of confidence – The standard deviation of the sample mean

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How could we get closer?

•101.77( )

99.55 103.99

Two ways to get a smaller Confidence Interval:

• Lower confidence level (e.g. 75%)• Bigger Sample

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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

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Part IV: Sample Size

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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Confidence Intervals for the

Mean(small samples)

Section 6.2

What happens if we don’thave 30 observations?

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No

Normal or t-Distribution?Is n 30?

Is the population normally, or approximately normally, distributed? Cannot use the normal

distribution or the t-distribution. Yes

Is known?

No

Use the normal distribution with

If is unknown, use s instead.

cE zn

σYes

No

Use the normal distribution with

cE zn

σ

Yes

Use the t-distribution with

and n – 1 degrees of freedom.

cE tn

s

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þ Comparing three curves– The standard normal curve– The t curve with 14 degrees of freedom– The t curve with 4 degrees of freedom

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0t

n = 13d.f. = 12c = 90%

.90

The t-Distribution

-1.782 1.782

The critical value for t is 1.782. 90% of the sample means (n = 13) will lie between t = -1.782 and t = 1.782.

.05 .05

Sampling distribution

If the distribution of a random variable x is normal and n < 30, then the sampling distribution of is a t-distribution with n – 1 degrees of freedom.

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Confidence Interval–Small Sample

2. The maximum error of estimate is

1. The point estimate is = 4.3 pounds

Maximum error of estimate

In a random sample of 13 American adults, the mean waste recycled per person per day was 4.3 pounds and the standard deviation was 0.3 pound. Assume the variable is normally distributed and construct a 90% confidence interval for .

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Finding tc

If c = 0.90

n = 13 (df =12)

tc = ?

d.f. = n - 1

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http://surfstat.anu.edu.au/surfstat-home/tables/t.php

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Confidence Interval–Small Sample

2. The maximum error of estimate is

1. The point estimate is = 4.3 pounds

Maximum error of estimate

In a random sample of 13 American adults, the mean waste recycled per person per day was 4.3 pounds and the standard deviation was 0.3 pound. Assume the variable is normally distributed and construct a 90% confidence interval for .

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•4.3

Confidence Interval–Small Sample

4.15 < < 4.45

(

Left endpoint

4.152)

Right endpoint

4.448

With 90% confidence, you can say the mean waste recycled per person per day is between 4.15 and 4.45 pounds.

2. The maximum error of estimate is

1. The point estimate is = 4.3 pounds

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No

Normal or t-Distribution?Is n 30?

Is the population normally, or approximately normally, distributed? Cannot use the normal

distribution or the t-distribution. Yes

Is known?

No

Use the normal distribution with

If is unknown, use s instead.

cE zn

σYes

No

Use the normal distribution with

cE zn

σ

Yes

Use the t-distribution with

and n – 1 degrees of freedom.

cE tn

s

See Pg 329 of textbook

1. The Graduate Management Admission Test (GMAT) is a test required for admission into many masters of business administration (MBA) programs. Total scores on the GMAT are normally distributed and historically have a standard deviation of 113. Suppose a random sample of 8 students took the test, and their scores are recorded.

2. Sean is estimating the average number of Christmas Trees he will find in the windows of each store in the mall. He observes each of the 10 stores in the mall and records a sample mean of 15 trees with a standard deviation of 6.

3. Patrick wonders about the average number of servings of eggnog at the Holiday Party. He knows that typically this variable has a standard deviation of 2.2 servings. He records a sample mean of 4 servings for a sample of 50 people.

1. The Graduate Management Admission Test (GMAT) is a test required for admission into many masters of business administration (MBA) programs. Total scores on the GMAT are normally distributed and historically have a standard deviation of 113. Suppose a random sample of 8 students took the test, and their scores are recorded. (We know population is normally distributed, so we can use Z even though n <30)

2. Sean is estimating the average number of Christmas Trees he will find in the windows of each store in the mall. He observes each of the 10 stores in the mall and records a sample mean of 15 trees with a standard deviation of 6. (We do not know population is normally distributed, so must use t with 9 degrees of freedom because we have a small sample and we do not know sigma)

3. Patrick wonders about the average number of servings of eggnog at the Holiday Party. He knows that typically this variable has a standard deviation of 2.2 servings. He records a sample mean of 4 servings for a sample of 50 people. (We do not know population is normally distributed, but we know sigma is 2.2 so we can use Z.)

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Confidence Intervals for Population Proportions

Section 6.3

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What if we are interested in a population proportion or percentage?

For example: What percentage of the population likes spinach?

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Required Condition: If np >= 5 and nq >=5 the sampling distribution for p-hat is normal.

Confidence Intervals forPopulation Proportions

is the point estimate for the proportion of failures where

The point estimate for p, the population proportion of successes, is given by the proportion of successes in a sample

(Read as p-hat)

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Confidence Intervals for Population Proportions

The maximum error of estimate, E, for a x-confidence interval is:

A c-confidence interval for the population proportion, p, is

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Confidence Interval for pIn a study of 1907 fatal traffic accidents, 449 were alcohol related. Construct a 99% confidence interval for the proportion of fatal traffic accidents that are alcohol related.

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1. The point estimate for p is

2. 1907(.235) > 5 and 1907(.765) > 5, so the sampling distribution is normal.

Confidence Interval for pIn a study of 1907 fatal traffic accidents, 449 were alcohol related. Construct a 99% confidence interval for the proportion of fatal traffic accidents that are alcohol related.

3.

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0.21 < p < 0.26

(

Left endpoint

.21•.235

)

Right endpoint

.26

With 99% confidence, you can say the proportion of fatal accidents that are alcohol related is between 21% and 26%.

Confidence Interval for pIn a study of 1907 fatal traffic accidents, 449 were alcohol related. Construct a 99% confidence interval for the proportion of fatal traffic accidents that are alcohol related.

Ch. 6 Larson/Farber

If you have a preliminary estimate for p and q, the minimum sample size given a x-confidence interval and a maximum error of estimate needed to estimate p is:

Minimum Sample Size

If you do not have a preliminary estimate, use 0.5 for both .

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You will need at least 4415 for your sample.

With no preliminary estimate use 0.5 for

Example–Minimum Sample Size

You wish to estimate the proportion of fatal accidents that are alcohol related at a 99% level of confidence. Find the minimum sample size needed to be be accurate to within 2% of the population proportion.

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With a preliminary sample you need at least n = 2981 for your sample.

Example–Minimum Sample SizeYou wish to estimate the proportion of fatal accidents that are alcohol related at a 99% level of confidence. Find the minimum sample size needed to be be accurate to within 2% of the population proportion. Use a preliminary estimate of p = 0.235.

Ch. 6 Larson/Farber

Ch. 6 Larson/Farber

Example #1 (pg 310)

Market researchers use the number of sentences per advertisement as a measure of readability for magazine advertisements. Suppose for the 50 advertisements we determine that the average number of sentences (xbar) is 12.4 and the standard deviation is 5.0:

Compute the 95% confidence interval for the mean mu.

– Question #1: Do we know the population standard deviation?

– Question #2: What is the interval?

Ch. 6 Larson/Farber

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Solution

xbar = 12.4s = 5.0n = 50 c = 0.95 Zc = 1.96 (for c = 0.95)

E = 1.96 * 5 / √50 = 1.4

12.4 – 1.4 < μ < 12.4 + 1.411.0 < μ < 13.8 Answer: We are 95% confident that the mean number of

sentences in the POPULATION is between 11.0 and 13.8.

Ch. 6 Larson/Farber

Example #2 (Pg. 327)

You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162 degrees with a standard deviation of 10 degrees. You know that the distribution of temperature is normally distributed.

A. Find the 95% confidence interval for the mean temperature.

B. Find the 99% confidence interval for the mean temperature.

Ch. 6 Larson/Farber

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95% Confidence Intervalxbar = 162.0s = 10.0n = 16 c = 0.95 tc = 2.132 (df = 15, c = .95)

E = 2.132 * 10 / √16 = 5.3

162 – 5.3 < μ < 162 + 5.3156.7 < μ < 167.3 Answer: We are 95% confident that the average temperature

of all the coffee in the POPULATION is between 157 and 167 degrees. 

Ch. 6 Larson/Farber

What about 99%?

xbar = 162.0s = 10.0n = 16 c = 0.99 tc = 2.947 (df = 15, c = .99)

E = 2.947 * 10 / √16 = 7.4

162 – 7.4 < μ < 162 + 7.4154.6 < μ < 169.4 Answer: We are 95% confident that the average temperature

of all the coffee in the POPULATION is between 154.6 and 169.4 degrees.