Introduction to Statistics for the Social Sciences SBS200, COMM200, GEOG200, PA200, POL200, or SOC200 Lecture Section 001, Spring 2016 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays & Fridays Welcome

By the end of lecture today 3/23/16 One-sample t-tests Two sample t-test Standard deviation and variance Pooled Variance

Before next exam (April 8th) Please read chapters 1 - 11 in OpenStax textbook Please read Chapters 2, 3, and 4 in Plous Chapter 2: Cognitive Dissonance Chapter 3: Memory and Hindsight Bias Chapter 4: Context Dependence

Homework On class website: Please complete homework worksheet #20 Two-sample t hypothesis tests Due: Friday, March 25th

Labs continue this week, Lab sessions Everyone will want to be enrolled in one of the lab sessions Labs continue this week, Designing Project 3

We are looking to compare two means Study Type 2: t-test Comparing Two Means? Use a t-test We are looking to compare two means http://www.youtube.com/watch?v=n4WQhJHGQB4

We are looking to compare two means Study Type 2: t-test Comparing Two Means? Use a t-test We are looking to compare two means http://www.youtube.com/watch?v=n4WQhJHGQB4

Hypothesis testing: one sample t-test Let’s jump right in and do a t-test Hypothesis testing: one sample t-test Is the mean of my observed sample consistent with the known population mean or did it come from some other distribution? We are given the following problem: 800 students took a chemistry exam. Accidentally, 25 students got an additional ten minutes. Did this extra time make a significant difference in the scores? The average number correct by the large class was 74. The scores for the sample of 25 was Please note: In this example we are comparing our sample mean with the population mean (One-sample t-test) 76, 72, 78, 80, 73 70, 81, 75, 79, 76 77, 79, 81, 74, 62 95, 81, 69, 84, 76 75, 77, 74, 72, 75

µ = 74 µ Hypothesis testing Step 1: Identify the research problem / hypothesis Did the extra time given to this sample of students affect their chemistry test scores Describe the null and alternative hypotheses One tail or two tail test? Ho: µ = 74 µ = 74 H1:

We use a different table for t-tests Hypothesis testing Step 2: Decision rule = .05 n = 25 Degrees of freedom (df) = (n - 1) = (25 - 1) = 24 two tail test This was for z scores We use a different table for t-tests

two tail test α= .05 (df) = 24 Critical t(24) = 2.064

µ = 74 Hypothesis testing = = 868.16 = 6.01 24 x (x - x) (x - x)2 76 72 78 80 73 70 81 75 79 77 74 62 95 69 84 76 – 76.44 72 – 76.44 78 – 76.44 80 – 76.44 73 – 76.44 70 – 76.44 81 – 76.44 75 – 76.44 79 – 76.44 77 – 76.44 74 – 76.44 62 – 76.44 95 – 76.44 69 – 76.44 84 – 76.44 = -0.44 = -4.44 = +1.56 = + 3.56 = -3.44 = -6.44 = +4.56 = -1.44 = +2.56 = -0.44 = +0.56 = -2.44 = -14.44 = +18.56 = -7.44 = +7.56 0.1936 19.7136 2.4336 12.6736 11.8336 41.4736 20.7936 2.0736 6.5536 0.3136 5.9536 208.5136 344.4736 55.3536 57.1536 Step 3: Calculations µ = 74 Σx = N 1911 25 = = 76.44 N = 25 = 6.01 868.16 24 Σx = 1911 Σ(x- x) = 0 Σ(x- x)2 = 868.16

µ = 74 Hypothesis testing = 76.44 - 74 1.20 2.03 . Step 3: Calculations µ = 74 = 76.44 N = 25 s = 6.01 76.44 - 74 = 76.44 - 74 1.20 2.03 critical t 6.01 25 Observed t(24) = 2.03

Hypothesis testing Step 4: Make decision whether or not to reject null hypothesis Observed t(24) = 2.03 Critical t(24) = 2.064 2.03 is not farther out on the curve than 2.064, so, we do not reject the null hypothesis Step 6: Conclusion: The extra time did not have a significant effect on the scores

Hypothesis testing: Did the extra time given to these 25 students affect their average test score? Start summary with two means (based on DV) for two levels of the IV notice we are comparing a sample mean with a population mean: single sample t-test Finish with statistical summary t(24) = 2.03; ns Describe type of test (t-test versus z-test) with brief overview of results Or if it had been different results that *were* significant: t(24) = -5.71; p < 0.05 The mean score for those students who where given extra time was 76.44 percent correct, while the mean score for the rest of the class was only 74 percent correct. A t-test was completed and there appears to be no significant difference in the test scores for these two groups t(24) = 2.03; n.s. Type of test with degrees of freedom n.s. = “not significant” p<0.05 = “significant” n.s. = “not significant” p<0.05 = “significant” Value of observed statistic 17

26.08 < µ < 33.92 mean + z σ = 30 ± (1.96)(2) 95% 26.08 < µ < 33.92 mean + z σ = 30 ± (1.96)(2) 99% 24.84 < µ < 35.16 mean + z σ = 30 ± (2.58)(2)

Melvin Melvin Mark Difference not due sample size because both samples same size Difference not due population variability because same population Yes! Difference is due to sloppiness and random error in Melvin’s sample Melvin

6 – 5 = 4.0 .25 Two tailed test 1.96 (α = .05) 1 1 = = .25 16 4 √ 4.0 z- score : because we know the population standard deviation Ho: µ = 5 Bags of potatoes from that plant are not different from other plants Ha: µ ≠ 5 Bags of potatoes from that plant are different from other plants Two tailed test 1.96 (α = .05) 1 1 = = .25 6 – 5 16 4 √ = 4.0 .25 4.0 -1.96 1.96

Because the observed z (4.0 ) is bigger than critical z (1.96) These three will always match Yes Yes Probability of Type I error is always equal to alpha Yes .05 1.64 No Because observed z (4.0) is still bigger than critical z (1.64) 2.58 No Because observed z (4.0) is still bigger than critical z(2.58) there is a difference there is not there is no difference there is 1.96 2.58

89 - 85 Two tailed test (α = .05) n – 1 =16 – 1 = 15 -2.13 2.13 t- score : because we don’t know the population standard deviation Two tailed test (α = .05) n – 1 =16 – 1 = 15 Critical t(15) = 2.131 89 - 85 2.667 6 √ 16

Because the observed z (2.67) is bigger than critical z (2.13) These three will always match Yes Yes Probability of Type I error is always equal to alpha Yes .05 1.753 No Because observed t (2.67) is still bigger than critical t (1.753) 2.947 Yes Because observed t (2.67) is not bigger than critical t(2.947) No These three will always match No No consultant did improve morale she did not consultant did not improve morale she did 2.131 2.947

Value of observed statistic Finish with statistical summary z = 4.0; p < 0.05 Or if it *were not* significant: z = 1.2 ; n.s. Start summary with two means (based on DV) for two levels of the IV Describe type of test (z-test versus t-test) with brief overview of results n.s. = “not significant” p<0.05 = “significant” The average weight of bags of potatoes from this particular plant is 6 pounds, while the average weight for population is 5 pounds. A z-test was completed and this difference was found to be statistically significant. We should fix the plant. (z = 4.0; p<0.05) Value of observed statistic

Five steps to hypothesis testing Step 1: Identify the research problem (hypothesis) Describe the null and alternative hypotheses Step 2: Decision rule Alpha level? (α = .05 or .01)? Critical statistic (e.g. z or t) value? Step 3: Calculations Step 4: Make decision whether or not to reject null hypothesis If observed z (or t) is bigger then critical z (or t) then reject null Step 5: Conclusion - tie findings back in to research problem

Hypothesis testing with t-tests The result is “statistically significant” if: the observed statistic is larger than the critical statistic observed stat > critical stat If we want to reject the null, we want our t (or z or r or F or x2) to be big!! the p value is less than 0.05 (which is our alpha) p < 0.05 If we want to reject the null, we want our “p” to be small!! we reject the null hypothesis then we have support for our alternative hypothesis

Independent samples t-test Are the two means significantly different from each other, or is the difference just due to chance? Independent samples t-test Donald is a consultant and leads training sessions. As part of his training sessions, he provides the students with breakfast. He has noticed that when he provides a full breakfast people seem to learn better than when he provides just a small meal (donuts and muffins). So, he put his hunch to the test. He had two classes, both with three people enrolled. The one group was given a big meal and the other group was given only a small meal. He then compared their test performance at the end of the day. Please test with an alpha = .05 Big Meal 22 25 Small meal 19 23 21 Mean= 21 Mean= 24 Got to figure this part out: We want to average from 2 samples - Call it “pooled” x1 – x2 t = 24 – 21 variability t = variability 27

α = .05 Independent samples t-test Step 1: Identify the research problem Did the size of the meal affect the learning / test scores? Step 2: Describe the null and alternative hypotheses Step 3: Decision rule α = .05 Two tailed test n1 = 3; n2 = 3 Degrees of freedom total (df total) = (n1 - 1) + (n2 – 1) = (3 - 1) + (3 – 1) = 4 Critical t(4) = 2.776 Step 4: Calculate observed t score 28

Notice: Simple Average = 3.5 Mean= 21 Mean= 24 Big Meal Deviation From mean -2 1 Small Meal Deviation From mean -2 2 Squared deviation 4 1 Squared Deviation 4 Big Meal 22 25 Small meal 19 23 21 Σ = 6 Σ = 8 6 3 Notice: s2 = 3.0 1 2 1 Notice: Simple Average = 3.5 8 4 Notice: s2 = 4.0 2 2 2 S2pooled = (n1 – 1) s12 + (n2 – 1) s22 n1 + n2 - 2 S2pooled = (3 – 1) (3) + (3 – 1) (4) 31 + 32 - 2 = 3.5 29

S2p = 3.5 Mean= 21 Mean= 24 Big Meal Deviation From mean -2 1 Small Meal Deviation From mean -2 2 Squared deviation 4 1 Squared Deviation 4 Participant 1 2 3 Big Meal 22 25 Small meal 19 23 21 Σ = 6 Σ = 8 = 24 – 21 1.5275 24 - 21 = 1.964 3.5 3.5 3 3 Observed t Observed t = 1.964 Critical t = 2.776 1.964 is not larger than 2.776 so, we do not reject the null hypothesis t(4) = 1.964; n.s. Conclusion: There appears to be no difference between the groups 30

Type of test with degrees of freedom Value of observed statistic We compared test scores for large and small meals. The mean test scores for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there appears to be no significant difference in test scores between the two types of meals, t(4) = 1.964; n.s. Type of test with degrees of freedom n.s. = “not significant” p<0.05 = “significant” Value of observed statistic Start summary with two means (based on DV) for two levels of the IV Finish with statistical summary t(4) = 1.96; ns Describe type of test (t-test versus anova) with brief overview of results Or if it *were* significant: t(9) = 3.93; p < 0.05 31

Homework Assignment Using Excel ?

Homework Assignment Using Excel

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