math 101: elementary statistics estimating the di …...2018/11/15 · 1.it is claimed that a new...

RecallEstimating the Difference Between Two Population Means

Math 101: Elementary StatisticsEstimating the Difference of Two Population

Means

Department of Mathematics and Computer ScienceUniversity of the Philippines Baguio

November 6, 2018

Math 101: Elementary Statistics


Outline

1 Recall

2 Estimating the Difference Between Two Population Means



(1− α)100% Confidence Interval for µ

Case 1. When σ is known(X − zα/2

σ√n, X + zα/2

σ√n

)where zα/2 is the z-value leaving an area of α/2 to the right.




Case 2. When σ is unknown(X − tα/2

s√n, X + tα/2

s√n

)where tα/2 is the t-value with v = n − 1 degrees of freedom leaving anarea of α/2 to the right.




Case 3. When σ2 is unknown and n > 30(X − zα/2

s√n,X + zα/2

s√n

)where zα/2 is the z-value leaving an area of α

2 to the right.



Outline

1 Recall

2 Estimating the Difference Between Two Population Means



Estimating the Difference Between Two Population Means

If we have two population means µ1 and µ2 and standard deviation ofσ1 and σ2, respectively, a point estimator of the difference between µ1

and µ2 is the statistic X 1 − X 2.

Types of Samples

• independent samples

• paired samples



Based on Two Independent Samples

Case 1. σ21 and σ2

2 are known

(X 1 − X 2)− zα/2

√σ2

1

n1+σ2

2

n2, (X 1 − X 2) + zα/2

√σ2

1

n1+σ2

2

n2



Example.

A random sample size of size n1 = 25 taken from a normalpopulation with a standard deviation σ1 = 5 has a mean X 1 = 80. Asecond random sample size of n2 = 36, taken from a different normalpopulation with a standard deviation σ2 = 3 has a mean X 2 = 75. Find a94% confidence interval for µ1 − µ2.




Case 2. σ21 = σ2

2 but unknown

((X 1 − X 2)− tα/2(v)Sp

√1

n1+

1

n2, (X 1 − X 2) + tα/2(v)Sp

√1

n1+

1

n2

)

where Sp =

√(n1 − 1)S2

1 + (n2 − 1)S22

n1 + n2 − 2and v = n1 + n2 − 2



Example.

Students may choose between a 3-unit course in Physics withoutlaboratory and a 4-unit course with laboratory. The final writtenexamination is the same for each section. The mean score of a randomsample of 12 students in the section with lab is 84 with a standarddeviation of 4, and the mean score of another random sample of 18students in the section without lab is 77 with a standard deviation of 6.Find a 99% confidence interval for the difference between the meangrades for the two courses. Assume populations to be approximatelynormally distributed with equal variances.




Case 3. σ21 6= σ2

2 but unknown

(X 1 − X 2)− tα/2(v)

√S2

1

n1+

S22

n2, (X 1 − X 2) + tα/2(v)

√S2

1

n1+

S22

n2

where v =

(S2

1

n1+

S22

n2

)2

(S2

1

n1

)2

n1 − 1+

(S2

2

n2

)2

n2 − 1



Example.

The following data represent the running time of a random sample offilms produced by two motion companies:

Time (minutes)

Company 1 103 94 110 87 98Company 2 97 82 123 92 175 88 118

Compute a 90% confidence interval for the difference between themean running time of films produced by the two companies. Assume thatthe running times for each companies are approximately normallydistributed with unequal variances.



(1− α)100% Confidence Interval for µ1 − µ2

Remark 1.

These formulas hold strictly for independent samples selected fromnormal populations. However, they provide good approximate(1− α)100% confidence intervals when the distributions are not normalprovided that both n1 and n2 are greater than 30.




Remark 2.

Even if the population variances are considerably different, formula (b)will still provide a good estimate provided that n1 = n2 and bothpopulations are normal. Therefore, in a planned experiment, one shouldmake every effort to equalize the size of the samples.




Remark 3.

If σ21 and σ2

2 are unknown but n1 and n2 are greater than 30, use(X 1 − X 2)− zα/2

√S2

1

n1+

S22

n2, (X 1 − X 2) + zα/2

√S2

1

n1+

S22

n2



Example.

A statistics test was given to a random sample of 50 girls and anotherrandom sample of 75 boys. The mean score of the sample population ofthe girls is 80 with a standard deviation of 4 and the mean score of thesample population of the boys is 86 with a standard deviation of 6. Finda 95% confidence interval for the difference µB − µG .



Estimating the Difference Between Two Population Means

Paired sampling is achieved by “matching” or studying 2 relatedsamples. Matching may be achieved by:

• using the same subject in the 2 samples

• pairing of subjects with respect to any extraneous variable whichmight affect or influence the outcome.



Based on Two Related/Paired Samples

Based on Two Related/Paired Samples(d − tα/2(v)

Sd√n, d + tα/2(v)

Sd√n

)where di = xi − yi

d =

n∑i=1

di

n

v = n − 1

Sd =

√√√√√nn∑

i=1

d2i −

(n∑

i=1

di

)2

n(n − 1)

n = number of pairs



Examples

1. It is claimed that a new diet will reduce a person’s weight by 4.5kilograms on the average in a period of 2 weeks. The weights of arandom sample of 7 people who followed this diet were recordedbefore and after a two-week period:

Person 1 2 3 4 5 6 7

Weight Before 58.5 60.3 61.7 69.0 64.0 62.6 56.7

Weight After 60.0 54.9 58.1 62.1 58.5 59.9 54.4

Compute a 95% confidence interval for the mean difference in theweight. Assume the distribution of weights to be approximatelynormal.



Exercises

2. Twenty college freshmen were divided into 10 pairs, each member ofthe pair having approximately the same IQ. One of each pair wasselected at random and assigned to a mathematics section usingprogrammed materials only. The other member of each pair wasassigned to a section in which the professor lectured. At the end ofthe semester each group was given the same examination and thefollowing results were recorded.

Pair 1 2 3 4 5 6 7 8 9 10

ProgrammedMaterials 76 60 85 58 91 75 82 64 79 88

Lectures 81 52 87 70 86 77 90 63 85 83

Find a 98% confidence interval for the mean difference in scores ofthe two learning procedures. Assume normality.


math 101: elementary statistics estimating the di …...2018/11/15 · 1.it is claimed that a new...

Documents