12.5 Differences between Means (s’s known) Two populations: (1, 1) & (2, 2) Two samples: one from each population Two sample means and sample sizes: n1 & n2 Compare two population means: H0: 1-2= (=0 in most cases) Alternatives: 1-2>; 1-2<; 1-2 We are testing whether the two populations are the same if delta=0.

Let’s go through a two sided alternative H0: 1-2=0 vs HA: 1-2≠0 Reject H0 if is too far from zero in either direction. How far from zero might be if 1-2=0? Sampling distribution of is asymptotically normal with mean 0 and standard deviation We need to know

Fact: If the sample means are from independent samples, then

Thus under certain assumptions: Correspondingly, a confidence interval for m1-m2 is

Assumptions 1 & 2 are known Normal populations or large sample sizes Under null hypothesis is (asymptotically) standard normal

Rejection Regions: 1-2> 1-2< 1-2 Rejection Regions Alternative Hypotheses 1-2> 1-2< 1-2 Rejection Regions z>z z<-z z>z/2 or z<-z/2

Example 12.4 Two labs measure the specific gravity of metal. On average do the two labs give the same answer? 1 -- Population mean by lab1 2 -- Population mean by lab2 H0: 1=2 vs HA: 12 1=0.02, n1=20, 2=0.03, n2=25,

95% Confidence Interval from –0.014 to 0.016

Two-tailed Hypotheses Test Two sample test Rejection region: |Z|>z0.025=1.96 Conclusion: Don’t reject H0.

Rejection Regions Alternative Hypotheses HA: 1>2 HA: 1<2 z>z z<-z z>z/2 or z<-z/2

Exercise An investigation of two kinds of photocopying equipment showed that a random sample of 60 failures of one kind of equipment took on the average 84.2 minutes to repair, while a random sample of 60 failures of another kind of equipment took on the average 91.6 minutes to repair. If, on the basis of collateral information, it can be assumed that s1=s2=19.0 minutes for such data, test at the 0.02 level of significance whether the difference between these two sample means is significant.

12.6 Differences Between Means (unknown equal variances) Large samples n130; n230 Small samples 1. 1=2 2. 12

Large Samples n130; n230 Estimate 1 and 2 by s1 and s2 Set

Rejection Regions Alternative Hypotheses HA: 1>2 HA: 1<2 z>z z<-z z>z/2 or z<-z/2

Small Samples 1=2= unknown Two populations are normal Standard error Estimate the common variance

Pooled standard deviation Using both s12 and s22 to estimate 2, we combine these estimates, weighting each by its d.f.. The combined estimate of 2 is sp2, the pooled estimate: Estimate  by sp

Two-Sample T-test T-test (t distribution with df=n1+n2-2) 100(1-)% CI Hypothesized m1-m2

Example 12.5 Compare blood pressures Two populations: common variance =0.05 n1=10, s1=16.2, n2=12, s2=14.3,

CI & test sp=15.2 df=10+12-2=20 Critical value t0.025=2.086 t statistic: reject H0 if |t|>2.086 Conclusion? Don’t Reject. CI: -122.086(6.51)=-12 13.6 -1.6 to 25.6

What happens when variances are not equal? Testing: H0: m1-m2=δ. Normal population s1 and s2 are not necessarily equal s1 and s2 unknown

Two sample t-test with unequal variances d.f. =min(n1-1, n2-1)

Exercise In a department store’s study designed to test whether or not the mean balance outstanding on 30-day charge accounts is the same in its two suburban branch stores, random samples yielded the following results: Use the 0.05 level of significance to test the null hypothesis m1-m2=0.

12.7 Paired Data 1982 study of trace metals in South Indian River. 6 random locations 5 3 1 6 2 4 T=top water zinc concentration (mg/L) B=bottom water zinc (mg/L) 1 2 3 4 5 6 Top 0.415 0.238 0.390 0.410 0.605 0.609 Bottom 0.430 0.266 0.567 0.531 0.707 0.716

One of the first things to do when analyzing data is to PLOT the data This is not a useful way to plot the data. There is not a clear distinction between bottom water and top water zinc—even though Bottom>Top at all 6 locations. Top Bottom

A better way Top Bottom Connect points in the same pair.

A better way Bottom=Top The plot suggests that mBottom>mTop. Is it true?

That is equivalent to ask: is it true that mdifference>0? 1 2 3 4 5 6 Top 0.415 0.238 0.390 0.410 0.605 0.609 Bottom 0.430 0.266 0.567 0.531 0.707 0.716 D=B-T 0.015 0.028 0.177 0.121 0.102 0.107 Ho: mD=0 vs HA: mD>0

First check the assumption that the population is normal

Doing a one-sided test Ho: mD=0 vs HA: mD>0 t0.05 at 5 d.f. is 2.015. So anything greater than 2.015 will be an evidence against H0. We reject H0: mB-T=0 in favor of HA: mB-T>0.

Another example The average weekly losses of man-hours due to accidents in 10 industrial plants before and after installation of an elaborate safety program: Plants 1 2 3 4 5 6 7 8 9 10 Before 45 73 46 124 33 57 83 34 26 17 After 36 60 44 119 35 51 77 29 24 11 diff(B-A) 9 13 2 5 -2 6 6 5 2 6 Is the safety program effective? (level=0.05)

Two Populations: Before and After Normal? Independent? No, No

Normal Probability Plots Small sizes Skew to right somehow

Normal Probability Plot for Difference Looks better

Consider the Differences Paired Observations:before and after the installation of safety program are from the same plants (dependent) Data from different plants may be independent Diff: 9 13 2 5 -2 6 6 5 2 6

Set up a Test—Paired T-Test ‘effective’ means the program reduces the accidents, i.e., before > after (D>0) =difference of average accidents H0: D=0 vs HA: D>0 The procedure is the same as the one- sample t-test Df=n-1

Rejection Regions for Paired T-test Alternative Hypotheses D> D< D Rejection Regions t>t t<-t t>t/2 or t<-t/2

Paired t-test One-tailed test Critical value: df=9, t0.05=1.833 Sample mean & standard deviation: t-statistic: Conclusion: reject H0 since t=4.03>1.833