1. Statistical Inference for Two Samples
Chapter 10
Statistical Inference
for Two Samples

2. Learning Objectives Comparative experiments involving two samples
Test hypotheses on the difference in means of two normal distributions
Test hypotheses on the ratio of the variances or standard deviations of two normal distributions
Test hypotheses on the difference in two population proportions
Compute power, type II error probability, and make sample size decisions for two-sample tests
Explain and use the relationship between confidence intervals and hypothesis tests

3. Assumptions
Interested on statistical inferences on the difference in means of two normal distributions
Populations represented by X1 and X2
Expected Value

4. Assumptions Quantity Has a N(0, 1) distribution
Used to form tests of hypotheses and confidence intervals on μ1-μ2

5. Hypothesis Tests for a Difference in Means, Variances Known
Difference in means μ1-μ2 is equal to a specified value ∆0
H0: μ1-μ2 =∆0
H1: μ1-μ2 #∆0
Test statistic

6. Hypothesis Tests for a Difference in Means, Variances Known
Alternative Hypothesis
H1: μ1-μ2 #∆0
Rejection Criterion
z0> zα/2 or z0<-zα/2
H1: μ1-μ2 >∆0
z0> zα
H1: μ1-μ2<∆0
Z0< -zα

7. Choice of Sample Size Use of OC Curves
Use OC curves in Appendix Charts VIa, VIb, VIc, and VId
Abscissa scale of the OC curves

8. Choice of Sample Size Two-sided Sample Size One-sided Sample Size
Sample size n=n1=n2 required to detect a true difference in means ∆ of with power at least 1-β
Where ∆ is the true difference in means of interest
One-sided Sample Size

9. Type II Error
Follows the singe-sample case
Two-sided alternative

10. C.I. on a Difference in Means, Variances Known, and Choice of Sample Size
Confidence Interval
100(1-α)% C.I. on the difference in two means μ1-μ2

11. Choice of Sample Size Choice of Sample Size
Error in estimating μ1-μ2 by less than E at 100(1-α)% confidence

12. Example
Two machines are used for filling plastic bottles with a net volume of 16.0 ounces
The fill volume can be assumed normal, with standard deviation 1=0.020 and 2=0.025 ounces
A member of the quality engineering staff suspects that both machines fill to the same mean net volume, whether or not this volume is 16.0 ounces. A random sample of 10 bottles is taken from the output of each machine as follows

13. Questions Do you think the engineer is correct? Use =0.05
What is the P-value for this test?
What is the power of the test in part (1) for a true difference in means of 0.04?
Find a 95% confidence interval on the difference in means. Provide a practical interpretation of this interval.
Assuming equal sample sizes, what sample size should be used to assure that =0.05 if the true difference in means is 0.04? Assume that =0.05

14. Solution-Part 1
Parameter of interest is the difference in fill volume,
H0 : or
H1 : or
 = 0.05
The test statistic is
Reject H0 if z0 < z/2 = 1.96 or z0 > z/2 = 1.96
16.015, ,  = 0, , , n1 = 10, and n2 = 10
Since < 0.99 < 1.96, do not reject the null hypothesis

15. Solution-Part 2 and 3 2. P-value = 3. = 0  0 = 0
Hence, the power = 1  0 = 1

16. Solution-Part 4 4. Confidence interval
With 95% confidence, we believe the true difference in the mean fill volumes is between  and Since 0 is contained in this interval, we can conclude there is no significant difference between the means.

17. Solution-Part 5
5. Assume the sample sizes are to be equal, use  = 0.05,  = 0.05, and  = 0.08
Hence, n = 3, use n1 = n2 = 3

18. Hypotheses Tests for a Difference in Means, Variances Unknown
Tests of hypotheses on the difference in means μ1-μ2 of two normal distributions
If n1 and n2 exceed 40, use the CLT
Otherwise base our hypotheses tests and C.I. on the t distribution
Two cases for the variances

19. Case I: 12=22= 2: Pooled Test
Two normal populations with unknown means and unknown but equal variances
Expected value
Form an estimator of 2
Pooled estimator of 2, denoted by S2p
Test statistic

20. Hypotheses Tests Test hypothesis Test statistic
Where Sp is the pooled estimator of 

21. Critical Regions Alternative Hypothesis H1: μ1-μ2 #∆0 H1: μ1-μ2 >∆0
Rejection Criterion
t0>tα/2, n1+n2-2 or
t0<-tα/2, n1+n2-2
H1: μ1-μ2 >∆0
t0>tα, n1+n2-2
H1: μ1-μ2 <∆0
t0<-tα, n1+n2-2

22. Case 2: 12#22 Critical regions
Not able to assume that the unknown variances 12, 22 are equal
Test statistic
With v degrees of freedom
Critical regions
Identical to the case I
Degrees of freedom will be replaced by v

23. Confidence Interval on the Difference in Means
Case 12=22
100(1-)% CI on the difference in means μ1-μ2
Case 12#22
100(1- )% CI on the difference in means μ1-μ2

24. Example
The diameter of steel rods manufactured on two different extrusion machines is being investigated
Two random samples of of sizes n1=15 and n2=17 are selected, and the sample means and sample variances are , s12=0.35, , and s22=0.40, respectively
Assume that equal variances and that the data are drawn from a normal distribution
Is there evidence to support the claim that the two machines produce rods with different mean diameters? Use α=0.05 in arriving at this conclusion
Find the P-value for the t-statistic you calculated in part (1)
Construct a 95% confidence interval for the difference in mean rod diameter. Interpret this interval

25. Solution 1. Parameter of interest, 2. H0 : or 3. H1 : or 4.  = 0.05
5. Test statistic is
6. Reject the null hypothesis if t0 < where = 2.042 or t0 > where = 2.042
, , 0 = 0, , , n1 = 15, and n2 = 17,

26. Solution
8. Since 2.042 < < 2.042, do not reject the null hypothesis

27. Solution-Cont.
P-value = 2P ( 0.40), P-value > 0.80
95% confidence interval: t0.025,30 = 2.042
Since zero is contained in this interval, we are 95% confident that machine 1 and machine 2 do not produce rods whose diameters are significantly different

28. Paired t Test Special case of the two-sample t-tests
When the observations are collected in pairs
Each pair of observations is taken under homogeneous conditions
Conditions may change from one pair to another
Testing
H0: μD=∆0
H1: μD#∆0

29. Paired t Test Test statistic Rejection Region
D (bar) is the sample average of the n differences
Rejection Region
t0>tα/2, n-1 or t0<-tα/2, n-1
100(1-α)% C.I. on the difference in means in means

30. Example
Ten individuals have participated in a diet-modification program to stimulate weight loss
Their weight both before and after participation in the program is shown in the following list
Is there evidence to support the claim that this particular diet-modification program is effective in producing a mean weight reduction? Use α=0.05.
Subject
Before
After 1
195
187 2
213 3
247
221 4
201
190 5
175 6
210
197 7
215
199 8
246 9
294
278 10
310
285

31. Solution 1. Parameter of interest is the difference in mean weight, d
where di =Weight Before  Weight After.
2. H0 :
3. H1 :
4.  = 0.05
5. Test statistic is
6. Reject the null hypothesis if t0 > where = 1.833
, , n=10
8) Since > reject the null

32. Inferences on the Variances of Two Normal Populations
Both populations are normal and independent
Test the hypotheses
H0: 12=22
H1: 12≠22
Requires a new probability distribution, the F distribution

33. The F Distribution
Define rv F as the ratio of two independent chi-square r.v., each divided by its number of dof
F=(W/u) /(Y(v))
Follows the F distribution with u dof in the numerator and v dof in the denominator.
Usually abbreviated as Fu,v

34. The F Distribution Shape of pdf with two dof
Table V provides the percentage points of the F distribution
Note that f1-α,u,v =1/fα,v, u

35. Hypothesis Tests on the Ratio of Two Variances
Suppose H0: 12=22
S12 and S22 are sample variances
Test statistics
F0= S12 / S22
Suppose H1: 12#22
Rejection Criterion
f0>fα/2,n1-1,n2-1 or f0<f1-α/2,n1-1, n2-1

36. Example Two chemical companies can supply a raw material.
The concentration of a particular element in this material is important.
The mean concentration for both suppliers is the same, but we suspect that the variability in concentration may differ between the two companies
The standard deviation of concentration in a random sample of n1=10 batches produced by company 1 is s1=4.7 grams per liter, while for company 2, a random sample of n2=16 batches yields s2=5.8 grams per liter.
Is there sufficient evidence to conclude that the two population variances differ? Use α=0.05.

37. Solution 1. Parameters of interest are the variances of concentration,
4.  = 0.05
5. Test statistic is
6. Reject the null hypothesis if f0 < where = or f0 > where =3.12
7. n1=10, n2=16, s1= 4.7, and s2=5.8
8. Since < < 3.12 do not reject the null hypothesis

38. Hypothesis Tests on Two Population Proportions
Suppose two binomial parameters of interest, p1and p2
Large-Sample Test
Test statistic
Critical regions

39. β-Error
If the H1 is two sided, the β-error
Where

40. Confidence Interval on the Difference in Means
Two sided 100(1-α)% C.I. on the difference in the true proportions p1-p2

41. Example
Two different types of injection-molding machines are used to form plastic parts. A part is considered defective if it has excessive shrinkage or is discolored
Two random samples, each of size 300, are selected, and 15 defective parts are found in the sample from machine 1 while 8 defective parts are found in the sample from machine 2
Is it reasonable to conclude that both machines produce the same fraction of defective parts, using α=0.05?

42. Solution
Parameters of interest are the proportion of defective parts, p1 and p2
H0 :
H1 :
 = 0.05
Test statistic is
Reject the null hypothesis if z0 < where = 1.96 or z0 > where = 1.96
n1=300, n2=300, x1=15, x2=8, ,

43. Solution-Cont
Since 1.96 < 1.49 < 1.96 do not reject the null hypothesis  
P-value = 2(1P(z < 1.49)) =