COMPARING MEANS: INDEPENDENT SAMPLES 1 ST sample: x1, x2, …, xm from population with mean μx; 2 nd sample: y1, y2, …, yn from population with mean μy; GOAL: Determine if μx = μy based on the two samples. Test Ho: μx = μy vs Ha: μx ≠ μy or Ha: μx > μy or Ha: μx < μy Procedure depends on what can we assume about variability of the populations: σx and σy. CASE1. σx and σy are known. CASE2. σx and σy are not known, but may be assumed equal σx=σy CASE3. σx and σy are not known, and can not be assumed equal. Test statistics are developed for each of the 3 cases.

COMPARING MEANS: INDEPENDENT SAMPLES CASE 1: σx and σy known Test on significance level α. STEP1. Ho: μx = μy vs Ha: μx ≠ μy or Ha: μx > μy STEP 2. Test statistic: Under the Ho, the test statistic has standard normal distribution. STEP 3. Critical value? For one-sided test z α, for two-sided z α/2. STEP 4. DECISION-critical/rejection region(s) depends on Ha. Ha: μ ≠ μo Reject Ho if |z|> z α/2 ; Ha: μ > μo Reject Ho if z > z α ; Ha: μ < μo Reject Ho if z < - z α. STEP 5. Answer the question in the problem.

COMPARING MEANS: INDEPENDENT SAMPLES CASE 2: σx and σy not known, but assumed equal. STEP 2. Test statistic: where is a pooled estimate of the common variance Under the Ho, the test statistic has t distribution with df = m+n-2. STEP 3. Critical value? One-sided test t α, two-sided t α/2. STEP 4. DECISION-critical/rejection region(s) depends on Ha. Ha: μ ≠ μo Reject Ho if |t|> t α/2 ; Ha: μ > μo Reject Ho if t > t α ; Ha: μ < μo Reject Ho if t < - t α.

COMPARING MEANS: INDEPENDENT SAMPLES CASE 3: σx and σy not known, and may not be assumed equal. STEP 2. Test statistic: Under Ho, the degrees of freedom for the t distribution may be approximated by df=min(m-1, n-1). STEP 3. Critical value? One-sided test t α, two-sided t α/2. STEP 4. DECISION-critical/rejection region(s) depends on Ha. Ha: μ ≠ μo Reject Ho if |t|> t α/2 ; Ha: μ > μo Reject Ho if t > t α ; Ha: μ < μo Reject Ho if t < - t α.

EXAMPLE1 A medication for blood pressure was administered to a group of 13 randomly selected patients with elevated blood pressure while a group of 15 was given a placebo. At the end of 3 months, the following data was obtained on their Systolic Blood Pressure. Control group, x: n=15, sample mean = 180, s=50 Treated group, y: m=13, sample mean =150, s=30. Test if the treatment has been effective. Assume the variances are the same in both groups and use α=0.01. Soln. Let μx= mean blood pressure for the control group; μy= mean blood pressure for the treatment group. Then, n=15, = 180, s x =50, m=13, =150, s y =30. Assumed equality of variances/st.dev. σx=σy

EXAMPLE1 contd. STEP1. Ho: μx = μy (medicine not effective) vs Ha: μx > μy (med. effective) STEP 2. Pooled variance: Standard deviation Test statistic: STEP 3. Critical value=t 0.01 =2.479, df=26. STEP 4. t= not > 2.479, do not reject Ho. STEP 5. Not enough evidence to conclude that the medicine is effective.

(1-α)100% CONFIDENCE INTERVAL FOR (μx- μy) CASE 1: σx and σy known CASE 2: σx and σy not known, but assumed equal. Use t distribution with df=m+n-2. CASE 3: σx and σy not known, and may not be assumed equal. Use t distribution with df=min(m-1, n-1).

EXAMPLE1 contd. Construct a 95% CI for the difference in the means of blood pressures for the two groups (μx - μy). Soln. We already know n=15, = 180, s x =50, m=13, =150, s y =30, s p = CASE 2. 95% CI, so α=0.05, so α/2=0.025, t(26) = % CI is: NOTE: The interval contains zero. Intuitively, that conforms our decision that there is no difference in means between the medicine and the placebo.

MINITAB EXERCISE Is there a significant difference between test scores on the 1st and 2nd test? Use data in example176.xls (see class web site).

MINITAB EXAMPLE contd. Two Sample T-Test and Confidence Interval Two sample T for exm1 vs exm2 N Mean StDev SE Mean exm exm % CI for mu exm1 - mu exm2: ( -0.8, 17.1) T-Test mu exm1 = mu exm2 (vs not =): T = 1.85 P = DF = 41