Section 9.1: z-test and Confidence Interval for aDifference Between Two Population Means1Assume we observedX1, X2, · · · , Xm∼iidN(µ1, σ21)andY1, Y2, · · · , Yn∼ N(µ2, σ22),where σ21and σ22are known. Then,¯X =1mmXi=1Xi∼ N(µ1,σ21m)and¯Y =1nnXi=1Yi∼ N(µ2,σ22n).Then,¯X −¯Y ∼ N(µ1− µ2,σ21m+σ22n).2Case 1: Difference between two normal popula-tions.Write ¯x and ¯y are observed values of¯X and¯Y re-spectively. Then, the (1 − α)100% confidence in-terval for µ1− µ2is(¯x − ¯y) ± zα2sσ21m+σ22n.If m = n and we want the (1 − α)100% confidenceinterval for µ1− µ2less than w, then we needn ≥4z2α2(σ21+ σ22)w2.3LetZ =¯X −¯Y − ∆0rσ21m+σ22n.Then, under µ1− µ2= ∆0, we haveZ ∼ N(0, 1).Write z and the observed value of Z and let α bethe significance level. Then,(a) we reject H0and conclude Hawhen z > zαifwe testH0: µ1−µ2= ∆0(or ≤ ∆0) ↔ Ha: µ1−µ2> ∆0;(b) we reject H0and conclude Hawhen z < −zαifwe testH0: µ1−µ2= ∆0(or ≥ ∆0) ↔ Ha: µ1−µ2< ∆0;(c) we reject H0and conclude Hawhen |z| > zα/2if we testH0: µ1− µ2= ∆0↔ Ha: µ1− µ26= ∆0.4Case 2: Large Sample Case.When σ21and σ22are unknown, but both m and nare large (e.g. m, n > 40), then we approximatelyhaveZ =¯X −¯Y − ∆0rS21m+S22n∼approxN(0, 1)where S21is the sample variance of X1, X2, · · · , Xmand S22is the sample variance of Y1, Y2, · · · , Yn. Writes21and s22as the observed values of S21and S22re-spectively. Then, the (1 − α)100% confidence in-terval for µ1− µ2is(¯x − ¯y) ± zα2ss21m+s22n,and we reject H0and conclude H1ifz > zαin the testing problem (a)z < −zαin the testing problem (b)|z| > zα2in the testing problem (c)5First example of Section 9.1: example 9.1 on text-book. We observed m = 20 samples from the firergroup (cold rolled group) and n = 25 from the sec-ond group (two-sided galvanized group). Assumeσ1= 4.0 and σ2= 5.0. Observed values ¯x = 29.8and ¯y = 34.7.• Then, the 99% confidence interval for the groupdifference (µ1− µ2) is¯x − ¯y ± z0.005sσ21m+σ22n=29.8 − 34.7 ± 2.58 ×s4220+5225=[−8.26, −1.34].• To testH0: µ1= µ2↔ Ha: µ16= µ2,since|z| = |29.8 − 34.7q1620+2525| = 3.66 > z0.005= 2.58we reject H0and conclude µ16= µ2at α = 0.01significance level .6Second example of Section 9.1: Example 9.4 onthe textbook. Suppose we observed dataSample Sample SampleType Size Mean SDNo 663 2258 1519Yes 413 2637 1138• Since both m = 663 and n = 2636 are large, wecan use the large sample method which givesthe 95% confidence interval for µ1− µ2as2258 − 2637 ± 1.96 ×s15192663+11382413=[538.4, −219.6].• To testH0: µ1= µ2↔ Ha: µ16= µ2,since|z| = |2258 − 2637r15192663+11382413| = −4.66 > z0.025= 1.96we reject H0and conclude µ16= µ2at α = 0.05significance level.7Third example of Section 9.1: example 9.5 on text-book. The data give ¯x = 4.25, m = 78, s1= 1.30,¯y = 7.14, n = 88 and s2= 1.68. Since both m andn are large, we can use the large sample method.• The 95% confidence interval for µ1− µ2is4.25 − 7.14 ± (1.96)s1.30278+1.68288=[−3.34, −2.44].• To testH0: µ1= µ2↔ Ha: µ16= µ2,since|z| = |4.25 − 7.14r1.30278+1.68288| = 12.47 > z0.025= 1.96we reject H0and conclude µ16= µ2at α = 0.05significance