EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 1

Estimating the Difference Between Two Means

Given two independent random samples, a point estimate the difference between μ1 and μ2 is given by the statistic

We can build a confidence interval for μ1 - μ2 (given σ1

2 and σ22 known) as follows:

21 xx

2

22

1

21

2/21212

22

1

21

2/21 )()(nn

zxxnn

zxx

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 2

Example 9.10 Page 286

Find a 96% Confidence Interval xbarA = 36 mpg σA = 6 nA = 50 xbarB = 42 mpg σB = 8 nB = 75 α=0.04 α/2 =0.02 Z0.02 = 2.055

Calculations:

6 – 2.055 sqrt(64/75 + 36/50) < (μB - μA) < 6 + 2.055 sqrt(64/75 + 36/50)

Results:

3.4224 < (μB - μA) < 8.5776

96% CI is (3.4224, 8.5776)

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 3

Differences Between Two Means: Variances Unknown

Case 1: σ12 and σ2

2 unknown but “equal”

Pages 287 and 288

Where,

Note v = n1 + n2 -2

212,2/2121

212,2/21

11)()(

11)()(

2121 nnStxx

nnStxx pnnpnn

2

)1()1(

21

222

2112

nn

SnSnSp

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 4

Differences Between Two Means: Variances Unknown (Page 290)

Case 2: σ12 and σ2

2 unknown and not equal

Where,

WOW!

2

22

1

21

,2/21212

22

1

21

,2/21 )()(n

s

n

stxx

n

s

n

stxx

1/

1/

)//(

2

2

222

1

2

121

22

221

21

nnS

nnS

nSnS

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 5

Estimating μ1 – μ2

Example (σ12 and σ2

2 known) :A farm equipment manufacturer wants to compare the average daily downtime of two sheet-metal stamping machines located in two different factories. Investigation of company records for 100 randomly selected days on each of the two machines gave the following results:

x1 = 12 minutes x2 = 10 minutes

s12 = 12 s2

2 = 8

n1 = n2 = 100

Construct a 95% C.I. for μ1 – μ2

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 6

Solution

95% CI Z.025 = 1.96

(12-10) + 1.96*sqrt(12/100 + 8/100) = 2 + 0.8765

1.1235 < μ1 – μ2 < 2.8765

Interpretation: If CI contains 0, then μ1 – μ2 may be either positive or negative (can’t say that one is larger than the other); however, since the CI for μ1 – μ2 is positive, we conclude μ1 must be larger than μ2 .

2

22

1

21

2/21212

22

1

21

2/21 )()(nn

zxxnn

zxx

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 7

μ1 – μ2 : σi2 Unknown

Example (σ12 and σ2

2 unknown but “equal”):Suppose the farm equipment manufacturer was unable to gather data for 100 days. Using the data they were able to gather, they would still like to compare the downtime for the two machines. The data they gathered is as follows:

x1 = 12 minutes x2 = 10 minutes

s12 = 12 s2

2 = 8

n1 = 18 n2 = 14

Construct a 95% C.I. for μ1 – μ2

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 8

SolutionGoverning Equations:

Calculations:

t0.025,30 = 2.042 sp2 = ((17*12)+(13*8))/30 = 10.267 sp = 3.204

2 + 2.042*3.204*sqrt(1/18 + 1/14) = 2 + 2.3314

-0.3314 < μ1 – μ2 < 4.3314

Interpretation:

Since this CI contains 0, we can’t conclude μ1 > μ2 .

2

)1()1(

21

222

2112

nn

SnSnS p

212,2/2121

212,2/21

11)(

11)(

2121 nnStxx

nnStxx pnnpnn

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 9

Paired Observations

Suppose we are evaluating observations that are not independent …

For example, suppose a teacher wants to compare results of a pretest and posttest administered to the same group of students.

Paired-observation or Paired-sample test …Example: murder rates in two consecutive years for several US cities. Construct a 90% confidence interval around the difference in consecutive years.

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 10

Calculation of CI for Paired DataExample 9.13 We have 20 pairs of values. We calculate the difference for

each pair. We calculate the sample standard deviation for the difference values. The appropriate equations are:

μd = μ1 – μ2

Based on the data in Table 9.1 Dbar = -0.87 Sd = 2.9773 n=20

We determine that a (1-0.05)100% CI for μd is: -2.2634 < μd < 0.5234

1

)( 2

n

dds id

)(1,2/n

std d

n

Interpretation: If CI contains 0, then μ1 – μ2 may be either positive or negative (can’t say that one is larger than the other). Since this CI contains 0, we conclude there is no significant difference between the mean TCDD levels in the fat tissue.

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 11

C.I. for Proportions

The proportion, P, in a binomial experiment may be estimated by

where X is the number of successes in n trials.

For a sample, the point estimate of the parameter is

The mean for the sample proportion is

and the sample variance

n

XP

n

xp

pp

n

pq

p

2

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 12

C.I. for Proportions

An approximate (1-α)100% confidence interval for p is:

Large-sample C.I. for p1 – p2 is:

Interpretation: If the CI contains 0 …

n

qpzp

2/

2

22

1

112/21 )(

n

qp

n

qpzpp

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 13

Interpretation of the Confidence Interval Significance

1. If the C.I. for p1 – p2 = (-0.0017, 0.0217), is there reason to believe there is a significant decrease in the proportion defectives using the new process?

2. What if the interval were (+0.002, +0.022)?

3. What if the interval were (-0.900, -0.700)?

EGR 252 Ch. 9 Lecture2 9th ed. JMB 2013 Slide 14

Determining Sample Sizes for Developing Confidence Intervals

Requires specification of an error amount е Requires specification of a confidence level

Examples in text Example 9.3 Page 273

• Single sample estimate of mean Example 9.15 Page 299

• Single sample estimate of proportion