confidence intervals. types of estimates even when samples are taken using proper sampling...

CONFIDENCE INTERVALS

Types of EstimatesEven when samples are taken

using proper sampling techniques, there is still room for sampling error.

Statistics are actually estimates of true population values (parameters)

Types of Estimates2 basic types

Point Estimate• A specific numerical value that estimates a parameter

• Ex: is a point estimate for . “The average salary for a school nurse is $32000.”

x

Types of Estimates

Interval Estimate• An entire range of values used to estimate a parameter

• Ex: “The average salary for a school nurse is between $29,000 - $35,000 a year.”

• or “ The average salary for a school nurse is $32,000 3000.”

Interval Estimates 3000 in last example is refered

to as the margin of error.

An advantage to interval estimates compared to point estimates is that there is a better chance that the actual parameter falls within that range.

What is a Confidence Interval?An interval estimate of a parameter

obtained from a sample with a certain probability the estimate will contain the parameter.

For example: If the 95% confidence interval for the salary of nurses is between $29000 - $35000 then that means there is a 95% chance that the true mean lies within that range.

CALCULATING CONFIDENCE INTERVALS FOR THE MEAN (WHEN IS KNOWN & n 30)

How is a confidence interval determined?Need 2 values:

1. The Maximum Error of Estimate (E)o The largest possible difference between

a point estimate and the actual parameter itself.

o Formula:

2. The sample mean:

/ 2E zn

x

Common Confidence Intervals ()Most problems will ask for the

90%, 95%, or 99% confidence interval.

Formula for a Confidence Interval:

/ 2 / 2z zxn n

x

Determining the z-scores:A 90% C.I. is comparable to the

middle 90% on the normal distribution

.45

.45

z = -1.64 z =1.64

Z-scores for common confidence intervals90% C.I. - z = 1.64

95% C.I. - z = 1.96

99% C.I. - z = 2.58

Example 1:The president of a University

wishes to estimate the average age of students presently enrolled. From past studies, the standard deviation is known to be 2 years. A sample of 50 students is selected and the mean is 23.2 years. Find the 95% C.I. and the 99% C.I. of the population mean.

Part 1: = 95%

23.2x 1.96z 2 50n

2 223.2 1.96 23.2 1.96

50 50

/ 2 / 2z zxn n

x

22.65 23.75

Part 2: = 99%

23.2x 2.58z 2 50n

2 223.2 2.58 23.2 2.58

50 50

/ 2 / 2z zxn n

x

22.47 23.9

DETERMINING SAMPLE SIZE FOR ACCURACY

Is it large enough???

Determining sample sizeSample size is the key element in

determining accuracy when comparing sample means to population means.

Example: If you want your sample mean to be within $1000 of your population mean then you must ensure to take a large enough sample.

How large is large enough?Formula to determine sample

size:2

/ 2zn

E

Example 1:A college president wants a

professor to estimate the average age of the students. The professor decides the estimate should be accurate within 1 year and 99% confident. From a previous study, the standard deviation is known to be 3 years. How large a sample is required?

Example 1: solution

2.58z 3 1E

2

/ 2zn

E

22.58 3

59.9 601

n

CONFIDENCE INTERVALS FOR UNKOWN & n < 30

Requires the use of the T-Distribution

Comparison to the Normal Distribution:Similariti

es

1. Bell-Shaped

2. Symmetric About the mean

3.Mean, Median, Mode = 0 and located in the center of the distribution

Differences

1. Variance is > 1

2. Actually a family of curves rather than just

one curve

3. As n gets larger the t-distribution

approaches the normal

distribution

Degrees of Freedom•Changes the shape of the t-distribution•The # of values that are free to vary after a sample statistic has been computed. •Tells the researcher (or the student) which specific curve from the family to use. •The # of Degrees of Freedom is always equal to n - 1

Formula for Confidence Interval

Same basic format of other formula but using different chart!

/ 2 / 2

s st t

nx

nx

Example 1:A recent study of 25 students

showed that they spent an average of $18.53 for gasoline per week. The standard deviation of the sample was $3. Find the 99% confidence interval of the true mean.

Example 1: Solution

n = 25 D. of f. = 24t = 2.797

3s 18.53x

/ 2 / 2

s st t

nx

nx

3 32.797 2.797

2518.53 18.53

25

16.85 20.21

When to use t or z?????Do you know

the population standard

deviation?

yesUse the normal distribution (z-

scores)

no

Is n 30? yesUse the normal distribution (z-

scores)

no

Use t-distribution (t-

scores)

CONFIDENCE INTERVALS FOR PROPORTIONS

Example of a proportionIn a study, 200 people were

asked if they were satisfied with their job or profession. 162 said they were.

Point estimate for a sample proportion:

16281%

200p

1 81% 19%q

Formula for Confidence Interval

Where ‘p’ represents the true proportion of people from the population

/ 2 / 2

p q p qp z p p z

n n

Example 1:A survey of 80 recent fatal traffic

accidents showed that 46 were alcohol- related. Find the 95% confidence interval of the true proportion of people who die in alcohol-related accidents.

Example 1: Solution

n = 80z = 1.96

.425q 46.575

80p

.575 .425 .575 .425.575 1.96 .575 1.96

80 80p

.47 .68p

/ 2 / 2

p q p qp z p p z

n n

DETERMINING SAMPLE SIZE FOR PROPORTIONS

THE FORMULA:

If no approximation for p-hat is known, you should use .5

2

/ 2zn p q

E

Example 1:An educator desires to estimate,

within .03 the true proportion of high school students who study at least 1 hour each school night. He wants to be 98% confident. How large a sample is necessary? (From a previous study it is known that 60% of 250 students did)

Example 1: Solution

.60p .40q .03E 2.33z

2

/ 2zn p q

E

2

2.33.60 .40 1448

.03n

Example 2:We wish to estimate the

proportion of students who own a cell phone. We want to be 95% confident and accurate within 5% of the true proportion. Find the minimum sample size necessary.

Example 2: Solution

.50p .50q .05E 1.96z

2

/ 2zn p q

E

2

1.96.50 .50 385

.05n