1. Comparing Two Population Means
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2. Overview m1 s12 m2 s22 Sample 1 x11, x12, ,,,, x1n1 Sample 2
Two sample problem
Comparing
Means
Paired samples
Independent sample
σ1 and σ2 known
σ1 and σ2 unknown
Proportions
Comparing Variances m1
s12 m2
s22
Sample 1
x11, x12, ,,,, x1n1
Sample 2
x21, x22, ,,,, x2n2

3. Analysis of Paired Samples

4. Paired Samples
The observations on the two populations are paired. (ex. repeated measures, before and after treatment)
Each pair of observations, (X1j, X2j), are taken under homogeneous conditions, but these conditions may change from one pair to another.
Use difference between paired values D= X1-X2
Advantage: Eliminating variation in a factor other than the difference between the two populations.

5. Example (Montgomery pp
Example (Montgomery pp. 219) We are interested in comparing two different types of tips for a hardness testing machine. This machine presses the tip into a metal specimen with a known force. By measuring the depth of the depression caused by the tip, the hardness of the specimen can be determined. Several specimens were selected at random and half tested with tip 1, half tested with tip 2 and the independent t-test was applied.
Problem of this procedure
The metal specimens might not be homogeneous in some way that might affect hardness (e.g. produced in different heats)
=> The observed difference between mean hardness readings for the two tip types includes hardness difference between specimens.
Solution
Make two hardness readings on each specimen, one with each tip
=> Paired Sample

6. Analysis
Let (X11, X21), (X12, X22),…,(X1n, X2n) be a set of n paired observations of (X1, X2), where
E[X1]= m1 , Var(X1)= s12 and E[X2]= m2 , Var(X2)= s22
Define Dj=X1j - X2j ( j =1,2,…,n)
=> Reducing the problem as a one sample problem
Assumption: Both X1 and X2 are normally distributed. Then,
Dj ~ N(mD , sD2)
Point estimator of mD = m1-m2:
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Point estimator of sD2:
Distribution :
t-dist. with d.f.= n-1

7. Confidence Intervals 100(1-a)% confidence interval for m1-m2
100(1-a)% upper confidence bound for m1-m2
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100(1-a)% lower confidence bound for m1-m2

8. Example (Montgomery pp222) The journal Human Factors (1962, pp
Example (Montgomery pp222) The journal Human Factors (1962, pp ) reports a study in which n=14 subjects were asked to parallel park two cars having very different wheel bases and turning radii. The time in seconds for each subject was recorded and is given below. Find the 90% CI for 1-2 assuming the normality .

9. 90% CI for 1-2 is found as follows
Note that thus CI includes zero. Thus, at the 90% level of confidence the data do not support the claim that the two cars have different mean parking times.

10. Hypothesis Testing for paired data
Test statistics for testing H0: m1-m2 =△0
Alternative hypothesis H1: m1-m2≠△0
Rejection Region: T0 > ta/2, n-1 or T0 < - t a/2, n-1
P-value: 2 P( T0 > | t0 | )
Alternative hypothesis H1: m1-m2>△0
Rejection Region: : T0 > ta, n-1
P-value: P( T0 > t0 )
Alternative hypothesis H1: m1-m2<△0
Rejection Region: : T0 < - ta, n-1
P-value: P( T0 < t0 )
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11. Example (Hayter p386) A new drug for inducing a temporary reduction in a patient’s heart rate is to be compared with a standard drug. A paired experiment is run whereby each of 40 patients is administered one drug on one day and the other drug on the following day. The spacing of the two experiments over two days ensures that there’s no “carryover” effect since the drugs are only temporary effective. Nevertheless, the order in which the two drugs are administered is decided in a random manner so that one patient may have the standard drug followed by the new drug and another patient may have the new drug followed by the standard drug. To compare the effects of two drugs, the percentage heart rate reductions for the standard drug xi and the new drug yi was recorded for the 40 subjects.

12. From the result of the experiment (Figure 9.13), we have
To compare the effects, we perform a hypothesis Testing at =0.01
We can reject H0 at = That is , there is evidence that the new drug has a different effect from the standard drug.

13. Example (Montgomery Page 224) The Federal Aviation Administration requires material used to make evacuation systems retain their strength over the life of the aircraft. In an accelerated life test, the principal material, polymer coated nylon weave, is aged by exposing it to 1580F for 168 hours. The tensile strength of the specimens of this material is measured before and after the aging process. The following data (in psi) are recorded.

14. Is there evidence to support the claim that the nylon weave tensile strength is the same before and after the aging process? Use a significance level of 0.01.
Calculate the P-value for this test.

15. Find a 99% CI on the mean difference in tensile strength and use it to answer the question from part A.
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16. Inference on the Means of Two Independent Populations, Variance Known
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17. Assumptions
X11, X12, …, X1n1 is a random sample of size n1 from population 1
X21, X22, …, X2n2 is a random sample of size n2 from population 2
The two populations are independent .
Variances of twp populations are known.
Both populations are normal, or if they are not normal, the conditions of the central limit theorem apply
Notations
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18. Point Estimator Point estimator of m1-m2: Standard Error of
Distribution
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19. Confidence Interval 100(1-a)% confidence interval for m1-m2
100(1-a)% upper confidence bound for m1-m2
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100(1-a)% lower confidence bound for m1-m2

20. Example (Montgomery pp204) Tensile strength tests were performed on two different grades of aluminum spars used in manufacturing the wing of a commercial transport aircraft. From past experience with the spar manufacturing process and the testing procedure, the standard deviations of tensile strengths are assumed to be known. The data obtained is given below. If m1 and m2 denote the true mean tensile strengths for the two grades of spars, find a 90% CI on the difference in mean strength m1-m2
90% CI: (12.22, 13.98)

21. Example (Montgomery pp 205) Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volume are known to be s1= 0.10 and s2 = 0.15 fluid ounces for the two machines, respectively. Two random samples of n1=12 bottles from machine 1 and n2 =10 bottles from machine 2 are selected, and the sample mean fill volumes are =30.61 and =30.34 fluid ounces. Assume normality.
Construct a 90% two-sided CI on the mean difference in fill volume. Interpret this interval.

22. Construct a 95% two-sided CI on the mean difference in fill volume
Construct a 95% two-sided CI on the mean difference in fill volume. Compare and comment on the width of this interval to the width of the interval in part A.
Construct a 95% upper confidence bound on the mean difference in fill volume. Interpret this interval.
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23. Hypothesis Testing Test statistics for testing H0: m1-m2=△0
Alternative hypothesis H1: m1-m2≠△0
Rejection Region: : Z0 > za/2 OR Z0 < - z a/2
P-value: 2 P( Z0 > | z0 | )
Alternative hypothesis H1: m1-m2>△0
Rejection Region: : Z0 > za
P-value: P( Z0 > z0 )
Alternative hypothesis H1: m1-m2<△0
Rejection Region: : Z0 < - za
P-value: P( Z0 < z0 )
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24. Example (Montgomery pp201) A product developer is interested in reducing the drying time of a primer paint. Two formulations of the paint are tested: formulation 1 is the standard chemistry and formulation 2 has a new drying ingredient that should reduce the drying time. From experience, it is known that the standard deviation of drying time is 8 minutes and this inherent variability should be unaffected by the addition of the new ingredient. Ten specimens are painted with formulation 1 and another 10 specimens are painted with formulation 2; the 20 specimens are painted in random order.
The two sample average drying times are =121 minutes and =112 minutes, respectively. What conclusions can the product developer draw about the effectiveness of the new ingredient, using a= 0.05 ?

25. Example (Montgomery pp205) Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that s1= s2 = 1.0 psi. From a random sample of size n1=10 and n2 =12, we obtain =162.7 and = The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic 1? Use a=0.05 in reaching a decision.

26. Inference on the Means of Two Populations, Variance Unknown
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27. Assumptions
X11, X12, …, X1n1 is a random sample of size n1 from population 1
X21, X22, …, X2n2 is a random sample of size n2 from population 2
The two populations are independent .
Variances of twp populations are unknown.
Both populations are normal, or if they are not normal, the conditions of the central limit theorem apply

28. Case 1: The variances are assumed equal
(s12 = s22 = s2)
Combine the two sample variance S12 and S22 to form an estimator of s2
: pooled variance estimator
The test statistic is
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29. Confidence Interval 100(1-a)% confidence interval for m1-m2
100(1-a)% upper confidence bound for m1-m2
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100(1-a)% lower confidence bound for m1-m2

30. Example (Montgomery pp215) An article in the journal Hazardous Waste and Hazardous Materials (Vol.6, 1989) reported the results of an analysis of the weight of calcium in standard cement and cement doped with lead. Reduced levels of calcium would indicate that the hydration mechanism in the cement is blocked and would allow water to attack various location in the cement structure. Ten samples of standard cement had an average weight percent calcium of =90.0 with a sample standard deviation of s1=5.0 and 15 samples of the lead-doped cement had an average weight percent calcium of =87.0 with a sample standard deviation of s2=4.0 . We will assume that weight percent calcium is normally distributed and find a 95% CI on the difference in means, m1-m2, for the two types of cement. Assume equal variance for the two populations.

32. Hypothesis Testing Test statistics for testing H0: m1-m2=△0
Alternative hypothesis H1: m1-m2≠△0
Rejection Region: T0 > ta/2, n1+n2-2 OR T0 < - t a/2, n1+n2-2
P-value: 2 P( T0 > | t0 | )
Alternative hypothesis H1: m1-m2>△0
Rejection Region: : T0 > ta,n1+n2-2
P-value: P( T0 > t0 )
Alternative hypothesis H1: m1-m2<△0
Rejection Region: : T0 < - ta, n1+n2-2
P-value: P( T0 < t0 )
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33. Example (Montgomery, pp208) Two catalysts are being analyzed to determine how they affect the mean yield of a chemical process. Specifically, catalyst 1 is currently in use but catalyst 2 is acceptable. Because catalyst 2 is cheaper, it should be adopted, providing it does not change the process yield. A test is run in the pilot plant and results in the data shown in the following table. Is there any difference between the mean yields? Use a=0.05 and assume equal variances.

34. Example (Montgomery pp217) A consumer organization collected data on two types of automobile batteries, A and B. The summary statistics for 12 observations of each type are =36.51, =34.21, sA=1.43 and sB = Assume that the data are normally distributed with sA= sB
Is there evidence to support the claim that type A battery mean life exceeds that of type B? Use a significance level of 0.01 in answering this question.

35. Calculate the P-value for this test.
Construct a one-sided 99% confidence bound for the difference in mean battery life. Explain how this interval confirms your finding in part A.
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36. Case 2: s12 ≠ s22 The test statistic
is distributed approximately as t with degrees of freedom given by
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37. Confidence Interval 100(1-a)% confidence interval for m1-m2
100(1-a)% upper confidence bound for m1-m2
This is a basic course blah, blah, blah…
100(1-a)% lower confidence bound for m1-m2

38. Hypothesis Testing Test statistics for testing H0: m1-m2=△0
Alternative hypothesis H1: m1-m2≠△0
Rejection Region: T0 > ta/2, n OR T0 < - t a/2, n
P-value: 2 P( T0 > | t0 | )
Alternative hypothesis H1: m1-m2>△0
Rejection Region: : T0 > ta,v
P-value: P( T0 > t0 )
Alternative hypothesis H1: m1-m2<△0
Rejection Region: : T0 < - ta, v
P-value: P( T0 < t0 )
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39. Example (Montgomery pp218) Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears measured in foot-pounds is an important characteristic. A random sample of 10 gears from supplier 1 results in = and s1=22.5 and another random sample of 16 gears from the second supplier results in = and s2=21.
Is there evidence to support the claim that supplier 2 provides gears with higher mean impact strength? Use a=0.05 and assume that both populations are normally distributed but the variances are not equal.

40. Calculate the P-value for this test.
Do the data support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot pounds higher than that of supplier 1? Make the same assumptions as in part A.
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41. Do the data support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot pounds higher than that of supplier 1? Make the same assumptions as in part A.
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