Overview 8.1 Z Interval for the Mean 8.2 t Interval for the Mean 8.3 Z Interval for a Population Proportion 8.4 OMIT 8.5 Sample Size Considerations.

OverviewOverview

8.1 Z Interval for the Mean

8.2 t Interval for the Mean

8.3 Z Interval for a Population Proportion

8.4 OMIT

8.5 Sample Size Considerations

Introduction to confidence intervalsIntroduction to confidence intervals

Confidence interval estimate - consists of an

interval of numbers generated by a point

estimate together with an associated

confidence level specifying the probability

that the interval contains the parameter

Z Interval for the Population Z Interval for the Population Mean Mean μμ

May be constructed if EITHER of the following two conditions are met:

Case 1: The population is normally distributed, and the value of σ is known.

Case 2: The sample size is large (n ≥ 30), and the value of σ is known.

Z Interval for the Population Z Interval for the Population Mean Mean μμ continued continued

When a random sample of size n is taken from a population, a 100(1- )% confidence interval for μ is given by

lower bound = x –

upper bound = x +

where 1- is the confidence level.

The Z interval can also be written as x ±

and is denoted (lower bound, upper bound)

/ n

/ n

/ n

Z Interval for the PopulationZ Interval for the PopulationMean Mean μμ

Used only under certain conditions

Case 1: Population is normally distributed The value of σ is known

Case 2: n ≥ 30 The value of σ is known

Margin of ErrorMargin of Error

Denoted as E

Measure of the precision of the confidence interval estimate

For the Z interval

2

E Zn

Interpreting the Margin of ErrorInterpreting the Margin of Error

For a (1 - )100% confidence interval for μ

“We can estimate μ to within E units with (1 - )100% confidence.”

Example 8.8 - Interpreting the Example 8.8 - Interpreting the margin of errormargin of error

Find and interpret the margin of error E for the confidence interval for the mean sodiumcontent of the 23 breakfast cereals containing sodium in Example 8.5 page 395.

ExampleExample

The Washington State Department of Ecology reported that the mean lead contamination in trout in the Spokane River is 1 part per million (ppm), with a population standard deviation of 0.5 ppm. Suppose a sample of n = 100 trout has a mean lead contamination of = 1 ppm. Assume that σ = 0.5 ppm.

Determine whether Case 1 or Case 2 applies.Construct a 95% confidence interval for μ, the

population mean lead contamination in all trout in the Spokane River.

Interpret the confidence interval.

Example continuedExample continued

Solution

Case 1 or 2?

Is the distribution of the population known?Is the sample size large enough?

Example continuedExample continuedUse the formula for the confidence

interval.We are given n = 100, x= 1, and σ =

0.5. For a confidence level of 95%, Table 8.1 provides the value of Zα/2 = Z0.025 = 1.96. Plugging in the formula.

We are 95% confident that μ, the population mean lead contamination for all trout on the Spokane River, lies between 0.902 ppm and 1.098 ppm.

8.2 8.2 tt Interval for the Mean Interval for the Mean

t t DistributionDistribution

In real-world problems, σ is often unknown

Use s to estimate the value of σ

For a normal population

follows a t distribution

/

xts n

t t Distribution continuedDistribution continued

n - 1 degrees of freedom

Where x is the sample mean

μ is the unknown population mean

s is the sample standard deviation

n is the sample size

Characteristics of the Characteristics of the tt DistributionDistribution

Centered at zero

The mean of t is zero

Symmetric about its mean zero

As df decreases, the t curve gets flatter, and the area under the t curve decreases in the center and increases in the tails.

As df increases toward infinity, the t curve approaches the Z curve, and the area under the t curve increases in the center and decreases in the tails.

FIGURE 8.14FIGURE 8.14

t has area /2 to the right of it

Procedure for Finding Procedure for Finding tt

Step 1 Go across the row marked “Confidence

level” in the t table (Table D in the Appendix, page T-11) until you find the column with the desired confidence level at the top.

The t value is in this column somewhere.

Step 2 Go down the column until you see the

correct number of degrees of freedom on the left.

The number in that row and column is the desired value of t.

Example 8.11 - Finding Example 8.11 - Finding tt

Find the value of t that will produce a 95%

confidence interval for μ if the samplesize is n = 20.

Example 8.11 continuedExample 8.11 continued

Solution

Step 1

Go across the row labeled “Confidence level” in the t table (Figure 8.15) until we see the 95% confidence level.

t is somewhere in this column.


Solution

Step 2

df = n - 1 = 20 - 1 = 19.

Go down the column until you see 19 on the left.

The number in that row is t, 2.093.


FIGURE 8.15 t table (excerpt).

tt Interval for Interval for μμ

Random sample of size nUnknown mean μConfidence interval for μ

lower bound , upper bound

x is the sample mean t is associated with the confidence level n - 1 degrees of freedoms is the sample standard deviation.

/2 /x t s n

/2 /x t s n

tt Interval for Interval for μμ continued continued

The t interval may also be written as and is denoted (lower bound, upper bound)

The t interval applies whenever either of the following conditions is met:

Case 1: The population is normal. Case 2: The sample size is large (n ≥ 30).

/2 /x t s n

Margin of Error for the Margin of Error for the tt IntervalInterval

The margin of error E for a (1- )100% t interval for μ can be interpreted as follows:

“We can estimate to within E units with (1- )100% confidence.”

2

sE t

n

Example 8.14 - Margin of error Example 8.14 - Margin of error for the fourth-grader foot for the fourth-grader foot lengthslengthsSuppose a children’s shoe manufacturer is

interested in estimating the population mean length of fourth graders’ feet. A random sample of 20 fourth graders’ feet yielded the following foot lengths, in centimeters.7

22.4 25.5 23.7 21.7 23.4 22.8 24.1 22 22.5 24.1 21 22.7 23.2 25 21.6 24.7 23.1 24 20.9 23.5

Construct a 95% confidence interval for μ, the population mean length of all fourth graders’ feet.


Solution

n = 20, x = 23.095 s = 1.280. For a confidence level of 95%, t = 2.093.The margin of error of fourth-grade foot

length is

We are 95% confident that the population mean length of fourth graders’ feet lies between 22.496 and 23.694 cm.

/2

1.2802.093 0.599

20

sE t

n

8.3 Z Interval for a Population 8.3 Z Interval for a Population ProportionProportion

Point EstimatePoint Estimate

The sample proportion of successes

is a point estimate of the population proportion p.

p̂

number of successesˆ

sample size

xp

n

Central Limit Theorem for Central Limit Theorem for ProportionsProportionsThe sampling distribution of the sample

proportion p follows an approximately normal distribution with mean μp = p

standard deviation

When both the following conditions are satisfied: (1) np ≥ 5 and (2) n(1 - p) ≥ 5.

1

p

p p

n

Z Interval for Z Interval for ppMay be performed only if both the following

conditions apply: np ≥ 5 and n(1 - p) ≥ 5

Random sample of size n is taken from a binomial population with unknown population proportion p

The 100(1 - )% confidence interval for p is given by

/2

ˆ ˆ1ˆlower bound=p

p pZ

n

/2

ˆ ˆ1ˆupper bound=p

p pZ

n

Z Interval for Z Interval for pp continued continued

Alternatively

Where p is the sample proportion of successes, n is the sample size, and

depends on the confidence level

/2

ˆ ˆ1ˆ

p pp Z

n

Margin of Error for the Z Margin of Error for the Z Interval for Interval for pp

The margin of error E for a (1- )100% Z interval for p can be interpreted as follows:

“We can estimate p to within E with (1- )100% confidence.”

/2

ˆ ˆ1p pE Z

n

Example 8.19 - Polls and the Example 8.19 - Polls and the famous “plus or minus 3 famous “plus or minus 3 percentage points”percentage points”

There is hardly a day that goes by without some new poll coming out. Especially during election campaigns, polls influence the choice of candidates and the direction of their policies. In October 2004, the Gallup organization polled 1012 American adults, asking them, “Do you think there should or should not be a law that would ban the possession of handguns, except by the police and other authorized persons?” Of the 1012 randomly chosen respondents, 638 said that there should NOT be such a law.


a. Check that the conditions for the Z interval for p have been met. b. Find and interpret the margin of error E. c. Construct and interpret a 95% confidence interval for the population proportion of all American adults who think there should not be such a law.


Solution

Sample size is n = 1012

Observed proportion is

so

638ˆ 0.63

1012p

ˆ1 0.37p


Solution

a. We next check the conditions for the confidence interval:

and

ˆ 1012 0.63 637.56 5np

ˆ1 1012 0.37 374.44 5n p


Solution

b. The confidence level of 95% implies that our equals 1.96 (from Table 8.7).

The margin of error equals

/2

ˆ ˆ1 0.63 0.371.96 0.03

1012

p pE Z

n

Example 8.19 continuedExample 8.19 continuedSolutionc. The 95% confidence interval is point estimate ± margin of error

Thus, we are 95% confident that the population proportion of all American adults who think that there should not be such a law lies between 60% and 66%.

/2

ˆ ˆ1ˆ

ˆ

0.63 0.03

(lower bound 0.60, upper bound 0.66)

p pp Z

np E

8.5 Sample Size Considerations8.5 Sample Size Considerations

Sample Size for Estimating the Sample Size for Estimating the Population MeanPopulation Mean

The sample size for a Z interval that estimates the population mean μ to within a margin of error E with confidence 100(1- )% is given by

2

/2Zn

E

Sample Size for Estimating the Sample Size for Estimating the Population Mean continuedPopulation Mean continued

Where is the value associated with the desired confidence level (Table 8.1), E is the desired margin of error, and σ is the population standard deviation.

Round sample sizes calculations that are decimals to the next whole number.

Sample SizeSample SizeEstimating a Population Proportion with No Prior Information About p

Within a margin of error E with confidence 100(1- )% is given by

Where is the value associated with the desired confidence level, E is the desired margin of error, and 0.5 is a constant representing the most conservative estimate.

2

/20.5 Zn

E

Example 8.25 - Required Example 8.25 - Required sample size for pollssample size for polls

Suppose the Dimes-Newspeak organization would like to take a poll on the proportionof Americans who will vote Republican in the next Presidential election. How large asample size does the Dimes-Newspeak organization need to estimate the proportionto within plus or minus three percentage points (E = 0.03) with 95% confidence?

Example 8.25 continuedExample 8.25 continuedSolution

The 95% confidence implies that the value for is 1.96.

Since there is no information available about the value of the population proportion of all Americans who will vote Republican in the next election, we use 0.5 as our “worst case scenario” value of p:

2 2

/20.5 0.5 1.961067.11

0.03

Zn

E

ExampleExample 8.25 continued 8.25 continued

Solution

To estimate the population proportion of American voters who will vote Republican to within 3% with 95% confidence, they will need a sample of 1068 voters.

Don’t forget to round up!

Sample SizeSample Size

Estimating a Population Proportion When Prior Information About p Is Available

Within a margin of error E with confidence 100(1- )% is given by

Where is the value associated with the desired confidence level, E is the desired margin of error, and p is the sample proportion of successes available from some earlier sample.

2

/2ˆ ˆ1Z

n p pE

Overview 8.1 Z Interval for the Mean 8.2 t Interval for the Mean 8.3 Z Interval for a Population Proportion 8.4 OMIT 8.5 Sample Size Considerations.

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