Statistical Inference for Two Samples Chapter 10 Statistical Inference for Two Samples

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

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

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

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

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

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

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

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

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

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

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

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

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, 16.005,  = 0, 0.025, 0.02, n1 = 10, and n2 = 10 Since -1.96 < 0.99 < 1.96, do not reject the null hypothesis

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

Solution-Part 4 4. Confidence interval With 95% confidence, we believe the true difference in the mean fill volumes is between 0.0098 and 0.0298. Since 0 is contained in this interval, we can conclude there is no significant difference between the means.

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

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

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

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

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

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

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

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 8.73, s12=0.35, 8.68, 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

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 7. 8.73, 8.68, 0 = 0, 0.35, 0.40, n1 = 15, and n2 = 17,

Solution 8. Since 2.042 < 0.230 < 2.042, do not reject the null hypothesis

Solution-Cont. P-value = 2P 2( 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

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

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

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

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 7. 17, 6.41, n=10 8) Since 8.387 > 1.833 reject the null

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

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

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

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

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.

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 = 0.265 or f0 > where =3.12 7. n1=10, n2=16, s1= 4.7, and s2=5.8 8. Since 0.265 < 0.657 < 3.12 do not reject the null hypothesis

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

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

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

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?

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, 0.05, 0.0267

Solution-Cont Since 1.96 < 1.49 < 1.96 do not reject the null hypothesis   P-value = 2(1P(z < 1.49)) = 0.13622