Lecture 10: Comparing two populations: proportions Problem: Compare two sets of sample data: e.g. is the proportion of As in this semester 152 the same as last Fall? Methods: Extend the methods introduced for situations involving one sample to the new situation with two samples. We will learn how to use two sample proportions for: constructing a confidence interval estimate of the difference between the corresponding population proportions, and testing a claim made about the two population proportions. Data requirements We have sample proportions from two independent simple random samples. For each of the two samples, the number of successes is at least 5 and the number of failures is at least 5.
COMPARING PROPORTIONS IN LARGE SAMPLES Examples: Compare probability of H on two coins. Compare proportions of republicans in two cities. 2 populations: p1=proportion of S (successes) in population 1, p2=proportion of S in population 2. GOAL: Determine if p1=p2 based on two samples. Perform two Binomial experiments (one in each population) 1 ST sample: x successes in m ind. trials, get sample prop. of S: ; 2 nd sample: y successes in n ind. trials, get sample prop. of S:. p ˆ1 pˆ 2 x = m y = n Test Ho: p1= p2 vs Ha: p1 p2 or Ha: p1> p2 or Ha: p1< p2
TESTING HYPOTHESES PROCEDURE Test on significance level α. STEP1. Ho: p1= p2 vs Ha: p1 p2 or Ha: p1> p2 or Ha: p1< p2 STEP 2. Test statistic: where, is the pooled or combined proportion Under the Ho, the test statistic has standard normal distribution for large samples. 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: p1 p2 Reject Ho if z > z α/2 ; Ha: p1 > p2 Reject Ho if z > z α ; Ha: p1 < p2 Reject Ho if z < - z α. pˆ ˆ 1 p2, 1 1 pˆ (1 pˆ )( + ) m n STEP 5. Answer the question in the problem. z = x + y ˆp pˆ =. m + n
EXAMPLE A sample of 180 college graduates was surveyed. 100 of them men and 80 women, and each was asked if they make more or less than $40,000 per year. The following data was obtained. $40,000 < $40,000 Total Men: 60 40 100 Women: 30 50 80 Total 90 90 180 Are men more likely to make more than $40,000 than women? Use α=0.05. Soln. Let p1 = true proportion of men making over $40k; p2 = true proportion of women making over $40k;
EXAMPLE, contd. STEP1. Ho: p1= p2 vs Ha: p1> p2 60 30 60 + 30 pˆ 1 = = 0.6, pˆ ˆ 2 = = 0.375, p =. 100 80 100 + 80 STEP 2. Test statistic: z pˆ pˆ 0.6 0.375 1 2 = = = 1 1 1 1 pˆ (1 pˆ )( + ) 0.5(0.5)( + ) m n 100 80 3. STEP 3. Critical value= z α =z 0.05 =1.645. STEP 4. DECISION. z = 3 > 1.645, reject Ho. STEP 5. Men are more likely than women to make over $40k.
EXAMPLE contd. Find the p-value for the test P-value = P(Z>z) = P(Z>3) = 0.0013 Since the p-value is smaller than the significance level, we reject Ho.
Example: For the sample data listed in the Table below, use a 0.05 significance level to test the claim that the proportion of black drivers stopped by the police is greater than the proportion of white drivers who are stopped. Soln. Let p1 = true proportion of white drivers stopped; p2 = true proportion of black drivers stopped;
EXAMPLE, contd. STEP1. Ho: p1= p2 vs Ha: p1< p2 147 24 147 + 24 pˆ 1 = = 0.105, pˆ ˆ 2 = = 0.120, p = = 0.1069. 1400 200 1400 + 200 STEP 2. Test statistic: z pˆ pˆ 0.105 0.120 1 2 = = = 1 1 1 1 pˆ (1 pˆ )( + ) 0.1069(0.8931)( + ) m n 1400 200 STEP 3. Critical value= z α =z 0.05 = - 1.645. STEP 4. DECISION. z = -0.64 > -1.645, do not reject Ho. 0.64. STEP 5. Black men are not more likely to be stopped than white men by the police.
EXAMPLE contd. Find the p-value for the test P-value = P(Z < z) = P(Z < -0.64) = 0.2611 The p-value =0.2611 > 0.05 (significance level), so we do not reject Ho.
Independent and dependent samples Two samples are independent if the sample values selected from one population are not related to or somehow paired or matched with the sample values selected from the other population. Examples: weights of students in different univ., test results of students in different towns, yields on different fields, etc. Two samples are dependent (or consist of matched pairs) if the members of one sample can be used to determine the members of the other sample. Examples: Test results for students before and after a study session, weight of a group of people before and after a weight loss program, predicted and true max temps for several days in a given month in Reno, etc.
COMPARING MEANS: INDEPENDENT SAMPLES 1 ST sample: x1, x2,, x m from population with mean µx; 2 nd sample: y1, y2,, y n 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 we can 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. z = x σ m 2 x y σ + n 2 y.
COMPARING MEANS: INDEPENDENT SAMPLES CASE 2: σx and σy not known, but assumed equal. STEP 2. Test statistic: 2 s p 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. t = s p x y 1 1 + m n 2 1 { 2 2 s = ( m 1) s + ( n 1) s }. p m + n 2 x y,
COMPARING MEANS: INDEPENDENT SAMPLES CASE 3: σx and σy not known, and may not be assumed equal. STEP 2. Test statistic: t = x Under Ho, the degrees of freedom for the t distribution may be approximated by df=min(m-1, n-1) (i.e. smaller of m-1 and n-1). 2 sx + m y s 2 y n. 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. x Then, n=15, = 180, s x =50, m=13, =150, s y =30. Assumed equality of variances/st.dev. σx=σy y
EXAMPLE1 contd. STEP1. Ho: µx = µy (medicine not effective) vs Ha: µx > µy (med. effective) STEP 2. Pooled variance: 2 2 2 2 ( m 1) s ( 1) 2 x + n sy (15 1)50 + (13 1)30 s p = = = 1761.54. m + n 2 15 + 13 2 Standard deviation Test statistic: s p = s = 1761.54 = 41.97 2 p t x y 180 150 = = = 1.8863. 1 1 1 1 sp + 41.97 + m n 15 13 STEP 3. Critical value=t 0.01 =2.479, df=26. STEP 4. t=1.8863 not > 2.479, do not reject Ho. STEP 5. Not enough evidence to conclude that the medicine is effective.
Example 2. Sample statistics are shown for the distances of the home runs hit in record-setting seasons by Mark McGwire and Barry Bonds. Use a 0.05 significance level to test the claim that the distances come from populations with different means. McGwire Bonds n 70 73 x 418.5 403.7 s 45.5 30.6 Soln. Let µx= mean distance for McGwire; µy= mean distance for Bonds. CASE3. σx and σy are not known, and can not be assumed equal.
EXAMPLE2 contd. STEP1. Ho: µx = µy (same mean distances) vs Ha: µx µy (different mean distances) Test statistic: t x y 418.5 403.7 = = = 2 2 2 2 s sy 45.5 30.6 x + + m n 70 73 2.273. STEP 3. Critical value= t 0.025 = 1.994, df=69 (min(69, 72)). STEP 4. t=2.273 > 1.994, reject Ho. STEP 5. There is enough evidence to conclude that the mean distances of the home runs for the two players are different.
Independent and dependent samples Recall: Two samples are independent if the sample values selected from one population are not related to or somehow paired or matched with the sample values selected from the other population. Two samples are dependent (or consist of matched pairs) if the members of one sample can be used to determine the members of the other sample.
PAIRED t-test: comparing dependent samples Observations come as matched pairs (X,Y). X and Y are NOT independent, X and Y are dependent. Examples. X is score on a test before studying hard; Y is score on the test after studying hard for the same student; X is score on a test or in sports before training program, Y score after training program; X is weight before weight loss program, Y is weight after the program; X and Y are heights of twins or siblings.
PAIRED t-test: HYPOTHESES Hypotheses of interest: does training make a difference? µx = score before training; µy = score after training. Ho: µx = µy vs Ha: µx < µy (no difference) (score after training is higher) Data are pairs of observations: (x1, y1), (x2, y2),, (xn, yn). Typically, we work with differences: d=x-y, phrase hypotheses in terms of differences: µd = true mean difference. In terms of differences: Hypotheses e.g. Ho: µd = 0 vs Ha: µd < 0 Data: d1, d2,, dn. obs before after difference 1 x1 y1 d1=x1-y1 2 x2 y2 d2=x2=y2.... n xn yn dn=xn-yn
PAIRED t-test: TEST PROCEDURE To test Ho, we do one sample t-test. Need sample mean and standard deviation of d s: Compute the test statistic: n 1 d = di s = n n 2 ( di d ) and 2 i= 1 d. Under Ho the test statistic has t(n-1) distribution. i= 1 t n 1 Make decision in exactly the same way as for the one sample t- test. = s d d / n.
PAIRED t-test: an example The amount of lactic acid in the blood was examined for 10 men, before and after a strenuous exercise, with the results in the following table. (a) Test if exercise changes the level of lactic acid in blood. Use significance level α=0.01. (b) Find a 95% CI for the mean change in the blood lactose level. Before 15 16 13 13 17 20 13 16 14 18 After 33 20 30 35 40 37 18 26 21 19
PAIRED t-test: lactic acid example contd. Solution. Take d= After level before level of lactic acid. Data for d: 18, 4, 17, 22, 23, 17, 5, 10, 7, 1. Sample stats: STEP1. Ho: µd = 0 vs Ha: µd 0 STEP 2. Test statistic: d = s d = 2 12.4 and 63.156. d 12.4 t = = = 4.93. s / n 7.95 / 10 STEP 3. Critical value? df=n-1=9, t α/2 =t 0.005 =3.25. d STEP 4. DECISION: t = 4.93 > 3.25 = t 0.005, so reject Ho. STEP 5. There is enough evidence to conclude that exercise changes lactic acid level.
Example 2: Are Forecast Temperatures Accurate? The following Table consists of five actual low temperatures and the corresponding low temperatures that were predicted five days earlier. Use a 0.05 significance level to test the claim that there is a difference between the actual low temperatures and the low temperatures that were forecast five days earlier.
Example 2: contd. Computed from the data: = 13.2, s d = 10.7, n = 5 µ d = mean daily difference between the predicted and the observed min temperatures. H 0 : µ d = 0 H 1 : µ d 0 Step 2. Test statistic: d d 13.2 t = = = 2.759. s / n 10.7 / 5 d STEP 3. Critical value? df=n-1=4, t α/2 =t 0.025 =2.776. STEP 4. DECISION: t = -2.759 > -2.776, so do not reject Ho. STEP 5. There is no significant difference between the mean predicted and observed min daily temperatures.