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.

Results Philadelphia 52 53 54 61 55 Anaheim 77 75 67

Hypotheses Alternative hypothesis Null hypothesis H1: Philadelphia < Anaheim Null hypothesis H0:  Philadelphia = or >  Anaheim

Step 2: Calculate the Critical t df = N1 + N2 - 2 df = 5 + 3 - 2 = 6  = .10 One-tailed t critical = - 1.44

Step 3: Draw Critical Region tcrit = -1.44

Now Step 4: Calculate t observed tobs = (X1 - X2) / Sx1 - x2

X1= 275 X12= 15175 N1 = 5 X1 = 55 X2= 219 X22= 16043 N2 = 3 X2 = 73 219 275 16043 15175 3 5 5 3 5 + 3 - 2

= 3.05 X1= 275 X12= 15175 N1 = 5 X1 = 55 X2= 219 X22= 16043 15987 15175 15125 3 5 .2 + .33 5 3 6 = 3.05

Step 4: Calculate t observed -5.90 = (55 - 73) / 3.05 Sx1 - x2 = 3.05 X1 = 55 X2 = 73

Step 5: See if tobs falls in the critical region tcrit = -1.44 tobs = -5.90

Step 6: Decision If tobs falls in the critical region: Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words We Reject H0, and accept H1 Philadelphia is significantly ( = .10) colder than Anaheim.

So far. . . . We have been doing independent samples designs The observations in one group were not linked to the observations in the other group

Example Philadelphia 52 53 54 61 55 Anaheim 77 75 67

Matched Samples Design This can happen with: Natural pairs Matched pairs Repeated measures

Natural Pairs The pairing of two subjects occurs naturally (e.g., twins)

Matched Pairs When people are matched on some variable (e.g., age)

Repeated Measures The same participant is in both conditions

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

Matched Samples Design The logic and testing of this type of design is VERY similar to what you have already done!

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?

Hypothesis One-tailed Alternative hypothesis H1: 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 H0: 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

Results Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2

Step 2: Calculate the Critical t N = Number of pairs df = N - 1 4 - 1 = 3  = .05 t critical = 2.353

Step 3: Draw Critical Region tcrit = 2.353

Step 4: Calculate t observed tobs = (X - Y) / SD

Step 4: Calculate t observed tobs = (X - Y) / SD

Step 4: Calculate t observed tobs = (X - Y) / SD X = 3.75 Y = 2.00

Step 4: Calculate t observed tobs = (X - Y) / SD Standard error of a difference

Step 4: Calculate t observed tobs = (X - Y) / SD SD = SD / N N = number of pairs

S =

S = Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y)

S = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 S =

S = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 D = 7 D2 =13 N = 4 S =

S = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 D = 7 D2 =13 N = 4 7 S =

S = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 D = 7 D2 =13 N = 4 7 S = 13

S = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 D = 7 D2 =13 N = 4 7 S = 13 4 4 - 1

S = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 D = 7 D2 =13 N = 4 7 S = 13 12.25 4 3

.5 = Difference (D) 2 1 Test 1 w/ Pill (X) Mel 3 Alice 5 Vera 4 Flo 3 Test 2 w/o Pill (Y) 1 3 2 D = 7 D2 =13 N = 4 7 .5 = .75 4 3

Step 4: Calculate t observed tobs = (X - Y) / SD SD = SD / N N = number of pairs

Step 4: Calculate t observed tobs = (X - Y) / SD .25=.5 / 4 N = number of pairs

Step 4: Calculate t observed 7.0 = (3.75 - 2.00) / .25

Step 5: See if tobs falls in the critical region tcrit = 2.353

Step 5: See if tobs falls in the critical region tcrit = 2.353 tobs = 7.0

Step 6: Decision If tobs falls in the critical region: Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words Reject H0, and accept H1 When the subjects took the “magic pill” they received statistically ( = .05) higher math scores than when they did not get the pill

SPSS

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

Thus, there are 3 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

Practice Does drinking milkshakes affect (alpha = .05) your weight? To see if milkshakes affect a persons weight you collected data from 5 sets of twins. You randomly had one twin drink water and the other twin drank milkshakes. After 3 months you weighed them.

Results Water Twin A 186 Twin B 200 Twin C 190 Twin D 162 Twin E 175 Milkshakes 195 202 196 165 183

Hypothesis Two-tailed Alternative hypothesis H1: water = milkshake Null hypothesis H0: water = milkshake

Step 2: Calculate the Critical t N = Number of pairs df = N - 1 5 - 1 = 4  = .05 t critical = 2.776

Step 3: Draw Critical Region tcrit = -2.776 tcrit = 2.776

Step 4: Calculate t observed tobs = (X - Y) / SD

(D) -9 -2 -6 -3 -8 D = -28 D2 =194 N = 6 -28 3.04 = 194 5 5 - 1

Step 4: Calculate t observed tobs = (X - Y) / SD 1.36=3.04 / 5 N = number of pairs

Step 4: Calculate t observed -4.11 = (182.6 – 188.2) / 1.36 X = 182.6 Y = 188.2 SD = 1.36

Step 5: See if tobs falls in the critical region tcrit = -2.776 tcrit = 2.776 tobs = -4.11

Step 6: Decision If tobs falls in the critical region: Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words Reject H0, and accept H1 Milkshakes significantly ( = .05) affect a persons weight.

Practice You wonder if psychology majors have higher IQs than sociology majors ( = .05) You give an IQ test to 4 psychology majors and 4 sociology majors

Results Psychology 110 150 140 135 Sociology 90 95 80 98

Step 1: Hypotheses Alternative hypothesis Null hypothesis H1: psychology > sociology Null hypothesis H0: psychology = or < sociology

Step 2: Calculate the Critical t df = N1 + N2 - 2 df = 4 + 4 - 2 = 6  = .05 One-tailed t critical = 1.943

Step 3: Draw Critical Region tcrit = 1.943

Now Step 4: Calculate t observed tobs = (X1 - X2) / Sx1 - x2

X1= 535 X12= 72425 N1 = 4 X1 = 133.75 X2= 363 X22= 33129 N2 = 4 X2 = 90.75 9.38 = 363 535 72425 33129 4 4 4 (4 - 1)

Step 4: Calculate t observed 4.58 = (133.75 - 90.75) / 9.38 Sx1 - x2 = 9.38 X1 = 133.75 X2 = 90.75

Step 5: See if tobs falls in the critical region tcrit = 1.943 tobs = 4.58

Step 6: Decision If tobs falls in the critical region: Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words We Reject H0, and accept H1 Psychology majors have significantly ( = .05) higher IQs than sociology majors.

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.

Results Philadelphia 52 53 54 61 55 Anaheim 77 75 67

Hypotheses Alternative hypothesis Null hypothesis H1: Philadelphia < Anaheim Null hypothesis H0:  Philadelphia = or >  Anaheim

Step 2: Calculate the Critical t df = N1 + N2 - 2 df = 5 + 3 - 2 = 6  = .10 One-tailed t critical = - 1.44

Step 3: Draw Critical Region tcrit = -1.44

Now Step 4: Calculate t observed tobs = (X1 - X2) / Sx1 - x2

= 3.05 X1= 275 X12= 15175 N1 = 5 X1 = 55 X2= 219 X22= 16043 15987 15175 15125 3 5 .2 + .33 5 3 6 = 3.05

Step 4: Calculate t observed -5.90 = (55 - 73) / 3.05 Sx1 - x2 = 3.05 X1 = 55 X2 = 73

Step 5: See if tobs falls in the critical region tcrit = -1.44 tobs = -5.90

Step 6: Decision If tobs falls in the critical region: Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words We Reject H0, and accept H1 Philadelphia is significantly ( = .10) colder than Anaheim.

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.

Practice Did your program significantly lower ( = .05) the number of aggressive behaviors a child performed?

Results Time 1 (X) Child1 18 Child2 11 Child3 19 Child4 6 Child5 10 Time 2 (Y) 16 10 17 4 11 12

Hypothesis One-tailed Alternative hypothesis H1: time1 > time2 Null hypothesis H0: time1 < or = time2

Step 2: Calculate the Critical t N = Number of pairs df = N - 1 6 - 1 = 5  = .05 t critical = 2.015

Step 4: Calculate t observed tobs = (X - Y) / SD

1.21 = (D) 2 1 -1 Time 1 (X) Child1 18 Child2 11 Child3 19 Child4 6 Test 2 (Y) 16 10 17 4 11 12 D = 8 D2 =18 N = 6 8 1.21 = 18 6 6 - 1

Step 4: Calculate t observed tobs = (X - Y) / SD .49=1.21 / 6 N = number of pairs

Step 4: Calculate t observed 2.73 = (13 - 11.66) / .49 X = 13 Y = 11.66 SD = .49

Step 5: See if tobs falls in the critical region tcrit = 2.015 tobs = 2.73

Step 6: Decision If tobs falls in the critical region: Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words Reject H0, and accept H1 The program significantly ( = .05) lowered the number of aggressive behaviors a child performed.

SPSS

Practice An early hypothesis of schizophrenia was that it has a simple genetic cause. In accordance with the theory 25% of the offspring of a selected group of parents would be expected to be diagnosed as schizophrenic. Suppose that of 140 offspring, 19.3% were schizophrenic. Test this theory.

Goodness of fit chi-square Make sure you compute the Chi square with the frequencies. Chi square observed = 2.439 Critical = 3.84 These data are consistent with the theory!

Practice In the 1930’s 650 boys participated in the Cambridge-Somerville Youth Study. Half of the participants were randomly assigned to a delinquency-prevention pogrom and the other half to a control group. At the end of the study, police records were examined for evidence of delinquency. In the prevention program 114 boys had a police record and in the control group 101 boys had a police record. Analyze the data and write a conclusion.

Chi Square observed = 1.17 Chi Square critical = 3.84 Phi = .04 Note the results go in the opposite direction that was expected!

Practice A research study was conducted to examine the 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. Determine if age is related to life satisfaction.

Older Adults Younger Adults 45 34 38 22 52 15 48 27 25 37 39 41 51 24 46 19 55 26 36

Age is related to life satisfaction. Older Younger Mean = 44.5 Mean = 28.1 S = 8.682677518 S = 8.543353492 S2 = 75.388888888 S2 = 72.988888888 tobs = 4.257; t crit = 2.101 Age is related to life satisfaction.

Practice Sleep researchers decide to test the impact of REM sleep deprivation on a computerized assembly line task. Subjects are required to participate in two nights of testing. On each night of testing the subject is allowed a total of four hours of sleep. However, on one of the nights, the subject is awakened immediately upon achieving REM sleep. Subjects then took a cognitive test which assessed errors in judgment. Did sleep deprivation lower the subjects cognitive ability?

REM Deprived Control Condition 26 20 15 4 8 9 44 36 13 3 38 25 24 10 17 6 29 14

tobs = 6.175  tcrit = 1.83 Sleep deprivation lowered their cognitive abilities.

SPSS Problem #2 7.37 7.11