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What if.... The two samples have different sample sizes (n)
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Results Psychology 110 150 140 135 Sociology 90 95 80 98
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Results Psychology 110 150 140 135 Sociology 90 95 80
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If samples have unequal n All the steps are the same! Only difference is in calculating the Standard Error of a Difference
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Standard Error of a Difference When the N of both samples is equal If N 1 = N 2 : Sx 1 - x 2 =
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Standard Error of a Difference When the N of both samples is not equal If N 1 = N 2 : N 1 + N 2 - 2
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Results Psychology 110 150 140 135 Sociology 90 95 80 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3
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N 1 + N 2 - 2 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3
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N 1 + N 2 - 2 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3 535 265
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N 1 + N 2 - 2 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3 535 265 72425 23525
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4 + 3 - 2 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3 535 265 72425 23525 4 3 43
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5 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3 535 265 72425 23525 4 3 43 71556.25 23408.33.25+.33
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5 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3 535 265 72425 23525 4 3 43 71556.25 23408.33.25+.33 197.08 (.58)
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5 X 1 = 535 X 1 2 = 72425 N 1 = 4 X 2 = 265 X 2 2 = 23525 N 2 = 3 535 265 72425 23525 4 3 43 71556.25 23408.33.25+.33 114.31 = 10.69
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Practice I think it is colder in Philadelphia than in Anaheim ( =.10). To test this, I got temperatures from these two places on the Internet.
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Results Philadelphia 52 53 54 61 55 Anaheim 77 75 67
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Hypotheses Alternative hypothesis –H 1 : Philadelphia < Anaheim Null hypothesis –H 0 : Philadelphia = or > Anaheim
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Step 2: Calculate the Critical t df = N 1 + N 2 - 2 df = 5 + 3 - 2 = 6 =.10 One-tailed t critical = - 1.44
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Step 3: Draw Critical Region t crit = -1.44
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Now Step 4: Calculate t observed t obs = (X 1 - X 2 ) / Sx 1 - x 2
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6 X 1 = 275 X 1 2 = 15175 N 1 = 5 X 1 = 55 X 2 = 219 X 2 2 = 16043 N 2 = 3 X 2 = 73 275 219 15175 16043 5 3 53 15125 15987.2 +.33 = 3.05
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Step 4: Calculate t observed -5.90 = (55 - 73) / 3.05 Sx 1 - x 2 = 3.05 X 1 = 55 X 2 = 73
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Step 5: See if t obs falls in the critical region t crit = -1.44 t obs = -5.90
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Step 6: Decision If t obs falls in the critical region: –Reject H 0, and accept H 1 If t obs does not fall in the critical region: –Fail to reject H 0
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Step 7: Put answer into words We Reject H 0, and accept H 1 Philadelphia is significantly ( =.10) colder than Anaheim.
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SPSS
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So far.... We have been doing independent samples designs The observations in one group were not linked to the observations in the other group
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Example Philadelphia 52 53 54 61 55 Anaheim 77 75 67
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Matched Samples Design This can happen with: –Natural pairs –Matched pairs –Repeated measures
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Natural Pairs The pairing of two subjects occurs naturally (e.g., twins)
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Matched Pairs When people are matched on some variable (e.g., age)
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Repeated Measures The same participant is in both conditions
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Matched Samples Design In this type of design you label one level of the variable X and the other Y There is a logical reason for paring the X value and the Y value
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Matched Samples Design The logic and testing of this type of design is VERY similar to what you have already done!
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Example You just invented a “magic math pill” that will increase test scores. On the day of the first test you give the pill to 4 subjects. When these same subjects take the second test they do not get a pill Did the pill increase their test scores?
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Hypothesis One-tailed Alternative hypothesis –H 1 : pill > nopill –In other words, when the subjects got the pill they had higher math scores than when they did not get the pill Null hypothesis –H 0 : pill < or = nopill –In other words, when the subjects got the pill their math scores were lower or equal to the scores they got when they did not take the pill
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Results Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2
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Step 2: Calculate the Critical t N = Number of pairs df = N - 1 4 - 1 = 3 =.05 t critical = 2.353
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Step 3: Draw Critical Region t crit = 2.353
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Step 4: Calculate t observed t obs = (X - Y) / S D
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Step 4: Calculate t observed t obs = (X - Y) / S D
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Step 4: Calculate t observed t obs = (X - Y) / S D X = 3.75 Y = 2.00
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Step 4: Calculate t observed t obs = (X - Y) / S D Standard error of a difference
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Step 4: Calculate t observed t obs = (X - Y) / S D S D = S D / N N = number of pairs
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S =
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Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2
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S = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1
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S = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1 D = 7 D 2 =13 N = 4
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S = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1 D = 7 D 2 =13 N = 4 7
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S = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1 D = 7 D 2 =13 N = 4 7 13
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S = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1 D = 7 D 2 =13 N = 4 7 13 4 4 - 1
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S = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1 D = 7 D 2 =13 N = 4 7 13 4 3 12.25
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.5 = Test 1 w/ Pill (X) Mel3 Alice5 Vera 4 Flo3 Test 2 w/o Pill (Y) 1 3 2 Difference (D) 2 1 D = 7 D 2 =13 N = 4 7 4 3.75
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Step 4: Calculate t observed t obs = (X - Y) / S D S D = S D / N N = number of pairs
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Step 4: Calculate t observed t obs = (X - Y) / S D.25=.5 / 4 N = number of pairs
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Step 4: Calculate t observed 7.0 = (3.75 - 2.00) /.25
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Step 5: See if t obs falls in the critical region t crit = 2.353
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Step 5: See if t obs falls in the critical region t crit = 2.353 t obs = 7.0
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Step 6: Decision If t obs falls in the critical region: –Reject H 0, and accept H 1 If t obs does not fall in the critical region: –Fail to reject H 0
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Step 7: Put answer into words Reject H 0, and accept H 1 When the subjects took the “magic pill” they received statistically ( =.05) higher math scores than when they did not get the pill
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SPSS
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Practice You just created a new program that is suppose to lower the number of aggressive behaviors a child performs. You watched 6 children on a playground and recorded their aggressive behaviors. You gave your program to them. You then watched the same children and recorded this aggressive behaviors again.
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Practice Did your program significantly lower ( =.05) the number of aggressive behaviors a child performed?
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Results Time 1 (X) Child118 Child211 Child319 Child46 Child510 Child614 Time 2 (Y) 16 10 17 4 11 12
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Hypothesis One-tailed Alternative hypothesis –H 1 : time1 > time2 Null hypothesis –H 0 : time1 < or = time2
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Step 2: Calculate the Critical t N = Number of pairs df = N - 1 6 - 1 = 5 =.05 t critical = 2.015
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Step 4: Calculate t observed t obs = (X - Y) / S D
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1.21 = (D) 2 1 2 2 D = 8 D 2 =18 N = 6 8 18 6 6 - 1 Time 1 (X) Child118 Child211 Child319 Child46 Child510 Child614 Test 2 (Y) 16 10 17 4 11 12
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Step 4: Calculate t observed t obs = (X - Y) / S D.49=1.21 / 6 N = number of pairs
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Step 4: Calculate t observed 2.73 = (13 - 11.66) /.49 X = 13 Y = 11.66 S D =.49
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Step 5: See if t obs falls in the critical region t crit = 2.015 t obs = 2.73
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Step 6: Decision If t obs falls in the critical region: –Reject H 0, and accept H 1 If t obs does not fall in the critical region: –Fail to reject H 0
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Step 7: Put answer into words Reject H 0, and accept H 1 The program significantly ( =.05) lowered the number of aggressive behaviors a child performed.
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SPSS
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New Step Should add a new page Determine if –One-sample t-test –Two-sample t-test If it is a matched samples design If it is a independent samples with equal N If it is a independent samples with unequal N
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Thus, there are 4 different kinds of designs Each design uses slightly different formulas You should probably make up ONE cook book page (with all 7 steps) for each type of design –Will help keep you from getting confused on a test
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Practice A research study was conducted to examine whether or not there were differences between older and younger adults on perceived life satisfaction. A pilot study was conducted to examine this hypothesis. Ten older adults (over the age of 70) and ten younger adults (between 20 and 30) were give a life satisfaction test (known to have high reliability and validity). Scores on the measure range from 0 to 60 with high scores indicative of high life satisfaction; low scores indicative of low life satisfaction. Older Adults= 44.5; S = 8.68; n = 10 Younger Adults = 28.1; S = 8.54; n = 10
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Practice Dr. Willard is studying the effects of a new drug (Drug-Y) on learning improvement following traumatic brain injury. To study this, Dr. Willard takes six rats and lesions a part of the brain responsible for learning. He puts each rat in a maze and counts the number of times it takes each rat to navigate through the maze without making a mistake. Dr. Willard then puts each rat on a regimen of Drug-Y for one week. After one week, he places each rat in a similar maze and counts the number of times it takes each rat to navigate through the maze without a mistake. Examine if Drug-Y had a positive impact on rats performance.
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tTime 1Time 2 Ben2824 Splinter2926 George3022 Jerry3330 Fievel3429 Patches3228
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Practice A sample of ten 9th grades at James Woods High School can do an average of 11.5 pull-ups (chin-ups) in 30 seconds, with a sample standard deviation of s = 3 The US Department of Health and Human Services suggests that 9th grades be able to do a minimum of 9 pull-ups in 30 seconds, if not, they're watching too much Family Guy. Is this sample of 9th grades able to do significantly (alpha =.01) more pull-ups than the number recommended by the US Department of Health and Human Services?
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t obs = 4.257 t crit = 2.101 There age is related to life satisfaction.
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t = (31 - 26.5)/0.764 = 4.500/0.764 = 5.890 Because the obtained t-Value is larger that the critical t- Value, the mean difference between the number of maze navigation's at Time 1 and Time 2 is statistically significant. Thus, we can conclude that Drug-Y lead to a statistically significant decrease in the number of times it took rats to navigate through a maze without making a mistake.
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The obtained t-Value is (11.5 - 9)/1 = 2.5/1 = 2.500 6) Because the obtained t-Value (2.500) is less than the critical t-Value (2.689), the difference between the mean number of pull-ups that 9th grades from James Woods High School can do is not significantly greater than the number of pull-ups recommended by the US Department of Health and Human Services.
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Cookbook
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