Chapter 8 -- Confidence IntervalsChapter 8 -- Confidence Intervals

or p is unknown (another course: or p is unknown (another course: )) We want to estimate the parameterWe want to estimate the parameter Confidence Intervals are accurate estimates Confidence Intervals are accurate estimates

of the true value of the parameterof the true value of the parameter

Decisions to makeDecisions to make

How many to survey?How many to survey?

the entire population?the entire population?

just one object or person?just one object or person? How confident do we want to be?How confident do we want to be? What error can we live with?What error can we live with?

ConfidenceConfidence

CL = 1 - CL = 1 - = confidence level = confidence level CI = confidence intervalCI = confidence interval

ex. Let’s build a 95% Confidence Intervalex. Let’s build a 95% Confidence Interval 1 - 1 - = .95 (confidence level) = .95 (confidence level) = “unconfidence” = 0.05= “unconfidence” = 0.05 = .025= .025 z z = z = z.025.025 = 1.96 = 1.96

Confidence IntervalConfidence Interval

Confidence Interval

-6 -4 -2 0 2 4 6

/2

CL = 1 - /2

/2

Error BoundsError Bounds

EBP = error bound for a proportionEBP = error bound for a proportion EBM = error bound for a meanEBM = error bound for a mean

Error Bound for ProportionsError Bound for Proportions

P = X/nP = X/n P = proportion of successesP = proportion of successes X ~ N(np, npq)X ~ N(np, npq) P = X/n ~ N(np/n, npq/nP = X/n ~ N(np/n, npq/n22)) P ~ N(p, pq/n)P ~ N(p, pq/n) z = (p’ - p)/(pq/n)z = (p’ - p)/(pq/n)1/21/2

EBP = p’ - p = z*(p’q’/n)EBP = p’ - p = z*(p’q’/n)1/21/2

Error Bounds for MeansError Bounds for Means

EBM zn

EBM x

X ~ N(X ~ N(, , /n)/n)

zx

n

Error Bound For MeansError Bound For Means

Student-t DistributionStudent-t Distribution

txsn

EBM t sn

EBM x

Confidence Interval (C. I.)Confidence Interval (C. I.)

Means:Means:

(sample mean - EBM, (sample mean - EBM,

sample mean + EBM)sample mean + EBM)

Proportions:Proportions:

(p’ - EBP, p’ + EBP)(p’ - EBP, p’ + EBP)

C. I. For a Mean - Population C. I. For a Mean - Population Standard Deviation KnownStandard Deviation Known

For her project, Maria bought 50 small bags For her project, Maria bought 50 small bags of jelly beans. The average number of jelly of jelly beans. The average number of jelly beans in a bag was 25. It is known that the beans in a bag was 25. It is known that the standard deviation is 3 jelly beans. Maria standard deviation is 3 jelly beans. Maria wants to construct a 90% confidence wants to construct a 90% confidence interval for the true average number of jelly interval for the true average number of jelly beans in a small bag.beans in a small bag.

C. I. For a Mean - Population C. I. For a Mean - Population Standard Deviation KnownStandard Deviation Known

C.L. = 0.90 implies C.L. = 0.90 implies

z = 1.645z = 1.645

C.I. = (sample C.I. = (sample mean + EBM, mean + EBM, sample mean - sample mean - EBM )EBM )

= (25 - 0.70, 25 + 0.70)= (25 - 0.70, 25 + 0.70)

= (24.30 , 25.70)= (24.30 , 25.70) We are 90% confident that We are 90% confident that

the true average number of the true average number of jelly beans is in the range jelly beans is in the range 24.30 to 25.70. 24.30 to 25.70.

EBM zn

EBM 1645 350

.

EBM 0 70.

C. I. For a Mean - Population C. I. For a Mean - Population Standard Deviation Unknown Standard Deviation Unknown

Uses a new distribution called the Student-t Uses a new distribution called the Student-t (invented by William Gossett)(invented by William Gossett)

Notation: t Notation: t degrees of freedom degrees of freedom

degrees of freedom is abbreviated df.degrees of freedom is abbreviated df. Use Student-t when the population standard Use Student-t when the population standard

deviation is not known, the sample is deviation is not known, the sample is “small” (less than 30), and the population “small” (less than 30), and the population from which the sample comes is normal.from which the sample comes is normal.

C. I. For a Mean - Population C. I. For a Mean - Population Standard Deviation UnknownStandard Deviation Unknown

Seven fast food restaurants were surveyed concerning Seven fast food restaurants were surveyed concerning the number of calories in four ounces of french fries. the number of calories in four ounces of french fries. The data is 296, 329, 306, 324, 292, 310, 350 calories. The data is 296, 329, 306, 324, 292, 310, 350 calories. Construct a 95% confidence interval for the true Construct a 95% confidence interval for the true average number of calories in a four ounce serving of average number of calories in a four ounce serving of french fries. Assume the number of calories in french french fries. Assume the number of calories in french fries fries

follows a normal distribution.follows a normal distribution.

C. I. For a Mean - Population C. I. For a Mean - Population Standard Deviation Unknown Standard Deviation Unknown

Sample mean = 315.3Sample mean = 315.3 Sample stdev = 20.4Sample stdev = 20.4 Use Student-t with 6 Use Student-t with 6

df. (df = n - 1 = 7 - 1)df. (df = n - 1 = 7 - 1)

EBM t sn

EBM 2 45 20 47

. .

EBM 18 9.

t=2.45 comes from a t=2.45 comes from a Student-t table where Student-t table where

df = 6 and C.I. = 0.95df = 6 and C.I. = 0.95 C. I. =C. I. =

(sample mean - EBM,(sample mean - EBM,

sample mean + EBM)sample mean + EBM)

= (315.3-18.9, 315.3+18.9)= (315.3-18.9, 315.3+18.9)

=(296.4, 334.2)=(296.4, 334.2)

C. I. For a Binomial ProportionC. I. For a Binomial Proportion

At a local cabana club, 102 of the 450 At a local cabana club, 102 of the 450 families who are members have children families who are members have children who swam on the swim team in 1998. who swam on the swim team in 1998. Construct an 80% confidence interval for Construct an 80% confidence interval for the true proportion of families with children the true proportion of families with children who swim on the swim team in any year.who swim on the swim team in any year.

C. I. For a Binomial ProportionC. I. For a Binomial Proportion

C.L. = 0.80 implies C.L. = 0.80 implies

z = 1.28; n = 450 z = 1.28; n = 450

p’ = 102/450p’ = 102/450

q’ = 348/450 q’ = 348/450

EBP z p q n '* '/

EBP 128 102 450 348 450 450. ( / ) *( / ) /

EBP 0 0253.

C. I. For a Binomial ProportionC. I. For a Binomial Proportion

C. I. = (p’ - EBP, p’ + EBP)C. I. = (p’ - EBP, p’ + EBP)

((102/450)-0.0253,((102/450)-0.0253,

(102/450)+0.0253)(102/450)+0.0253)

= (0.2014, 0.2520)= (0.2014, 0.2520) We are 80% confident that the true We are 80% confident that the true

proportion of families that have children on proportion of families that have children on the swim team in any year is between 20% the swim team in any year is between 20% and 25%.and 25%.