Spring 2020ME3620

Chapter V

ME3620 Theory of Engineering Experimentation

Spring 2020

Chapter V. Decision Making for Two Samples

1

Spring 2020ME3620

Chapter V

5–1 Introduction

In the previous chapter we considered hypothesis tests for a parameter (μ or σ2) of a single population.

In the first part of this chapter we will extend those results to the study of the case when we deal with two independent populations.

In general we will consider two populations, population 1 with μ1 and σ12 and

population 2 with μ2 and σ22. Inferences will be based on two random samples of sizes

n1 and n2.

2

Spring 2020ME3620

Chapter V

5–2 Inference on the Means of Two Populations, Variances Known

Consider two populations as in the figure below, for which the variances are known and it is necessary to obtain statistical inferences on the difference in the means.

• Consider two Independent Populations

• Population 1:- Mean: μ1- Variance: σ1

2

• Population 2:- Mean: μ2- Variance: σ2

2

• Inferences (conclusions) will be based on two random samples of sizes n1 and n2, respectively 3

Spring 2020ME3620

Chapter V

For one variable, Z is defined as:

The definition of Z can be extended for two variables as:

Under the following considerations:

ZTest statistic

4

Spring 2020ME3620

Chapter V

𝑋 and 𝑋 are calculated from the samples

𝜇 and 𝜇 are the expected means of the populations

𝜎 𝜎and are variances of the populations and areknown.

n1 and n2 are the sizes of the samples of populations 1 and 2, respectively.

5

Spring 2020ME3620

Chapter V

5–2.1 Hypothesis Testing on the Difference in Means, Variances Known

Confidence interval (CI) hypothesis will be studied on 5–2.3

2–Sided

Upper Tail

Lowe Tail

6

Spring 2020ME3620

Chapter V

Example. The screw production rate between two different types ofmachines are compared (normal distribution is assumed). If machinetype A produces more than 1000 screws/hr than the type B machine,then type A machine will be purchased. Based on the informationprovided, what is your advise?

Machine of Company A:n1 = 50 machines

= 14600 screws/hr= 2350 screws/hr

𝑋𝜎

Machine of Company B:n2 = 50 machines

= 13700 screws/hr= 2250 screws/hr

𝑋𝜎

7

Example. Problem 5–4.

Spring 2020ME3620

Chapter V

5–2.3 Confidence Interval on the Difference in Means, Variances Known

The difference in sample means, , is a point estimator of , and

has a standard normal distribution if the two populations are normal or is approximately standard normal if the conditions of the central limit theorem apply. That is:

𝑃 𝑧 𝑍 𝑧 1 α8

Spring 2020ME3620

Chapter V

𝑃 𝑧 𝑍 𝑧 1 α

𝑃 𝑧𝑋 𝑋 𝜇 𝜇

𝜎𝑛

𝜎𝑛

𝑧 1 α

Which can finally be rearranged as:

𝑃 𝑋 𝑋 𝑧𝜎𝑛

𝜎𝑛 𝜇 𝜇 𝑋 𝑋 𝑧

𝜎𝑛

𝜎𝑛 1 α

Thus, the 100 (1-α)% CI for 𝜇 𝜇 is defined by:

9

Spring 2020ME3620

Chapter V

Thus, the 100 (1-α)% CI for 𝜇 𝜇 is defined by:

This previous relation applies for the two sided case.

10

Spring 2020ME3620

Chapter V

For upper and lower bound cases the proper relations are

𝜇 𝜇 𝑋 𝑋 𝑧𝜎𝑛

𝜎𝑛

Fail to Reject Ho

𝑋 𝑋 𝑧𝜎𝑛

𝜎𝑛 𝜇 𝜇 Fail to Reject Ho

Upper Confidence bound:

If:

If:

Lower confidence bound:

11

Example. Problem 5–5. (a), (b), and (d)

Spring 2020ME3620

Chapter V 12

5–8 What If We Have Samples from More Than Two Populations?• Consider a bearing for which its operating life is the critical parameter.

• There are four different companies that can manufacture it and it is necessary to decide which company should be selected.

• In this case, since there are four levels (the manufacturing companies) and a single factor (bearing’s operating life), a procedure called Analysis of Variance (ANOVA) can be used to determine the best solution to the problem.

• Thus ANOVA is used to compare means when there are more than two levels of a single factor.

• The levels of the factor are also called treatments.

Spring 2020ME3620

Chapter V 13

5–8 What If We Have Samples from More Than Two Populations?• As an example, consider the case in which the tensile strength of paper is thought to be

function of the hardwood concentration of the pulp used to produce the paper.• An experiment using 5%, 10%, 15% and 20% hardwood concentration is to be performed • 6 specimens for each concentration level (treatment) are produced.• In summary, there will be 6 observations (replicates) of each level (treatment)• The tensile strength is measured randomly and the results are summarized in the following

table:

Spring 2020ME3620

Chapter V 14

• Thus, in general for a single factor experiment with a number of levels (treatments) a, and for which b observations (replicates) for each treatment are performed, the data gathered can be arranged in matrix form as shown in the following table:

Treatment AveragesTotalsObservations

Spring 2020ME3620

Chapter V 15

• These observations can be presented as:

Yij Random variable representing the (ij) observation.μ Overall mean.τiTreatment Effect (variation of the overall mean due to particular treatment)ε Error in measurement.

• This equation can be re-written as:

Where, μi = μ + τi .

Spring 2020ME3620

Chapter V 16

5

10

15

20

25

0 5 10 15 20 25

Y1,1

ε1,1

Y2,4

ε2,4

Treatment (level of the factor)[Harwood Concentration (%)]

Para

met

er re

spon

se to

trea

tmen

t[T

ensi

le st

reng

th o

f pap

er (p

si)]

Spring 2020ME3620

Chapter V 17

• To define the ANOVA for testing the equality of a population means it is necessary to introduce the following variables

yi represents the total of the n observations under the ith treatment.𝑦 represents the average of the observations under the ith treatment.y represents the grand total of all the observations.

𝑦

represents the grand mean of all the observations.

• That is:

N = an is the total number of observations.

𝑦 𝑦 𝑦 𝑦 /n 𝑖 1,2, … , 𝑎

𝑦

𝑦 𝑦 𝑦

/N

Spring 2020ME3620

Chapter V 18

• We are interested in determining whether a different treatment (level) has an effect on the factor that it is being studied.

• That is, if the mean of the factor for each treatment is the same as the overall mean, then the treatment does not affect the factor. Therefore, the null hypothesis states that all treatments are the same and equal to zero

• If H0 is true, each observation consists of the overall mean μ plus a given random error.

• The way in which it is determined whether the overall mean is the same as the mean for each treatment is by comparing the overall variance with the sum of each treatment’s variance plus the random error.

𝐻 :𝜏 𝜏 ⋯ 𝜏

𝐻 :𝜏 0 for at least one i.

Spring 2020ME3620

Chapter V 19

• The total variance can be then thought as being the addition of the variance due to treatment effect plus the variance due to error in measurement

Total Variance = Variance due toTreatment effect

Variance due toerror in measurement+

Spring 2020ME3620

Chapter V 20

• Therefore, the ANOVA sum of squares identity is

• The determination of each term in the previous expression can be simplified as follows:

Spring 2020ME3620

Chapter V 21

• Thus, in order to determine whether the effect of a treatment is statistically significant, we determine whether SSE (the error of sum squares) is statistically significant:

• This analysis is done using the procedure summarized next:

Spring 2020ME3620

Chapter V 22

• Where the test statistic, F0, can be determined using the following table

MSTreatmentsMean square of treatmentsMSEMean square of errors

Spring 2020ME3620

Chapter V 23

• In order to systematically determine the values of SST, SSTreatments, and then MSTreatments and MSE to complete the previous table, an additional table can be easily created as follows:

TreatmentObservation

TotalsAverages

1 2 … n1 y1,1 y1,2 … y1,n y1 𝑦 =y1/n2 y2,1 y2,2 … y2,n y2 𝑦 =y2/n… … … … … … …a ya,1 ya,2 … ya,n ya 𝑦 =ya/n

• Example. Problem 5–77, 5–78. Determine whether the treatment has an effect on the overall process, use P–value and rejection criterion approaches.