Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2017 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays & Fridays. Welcome http://www.youtube.com/watch?v=oSQJP40PcGI http://www.youtube.com/watch?v=oSQJP40PcGI

A note on doodling

By the end of lecture today 3/24/17 Hypothesis testing with z scores Hypothesis testing with t-tests Interpreting Alpha levels p values Type I and Type II Errors

Before next exam (April 7th) 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 No homework due: Monday, March 27th Just work on Project 3 to prepare for labs

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

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 16

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

Thank you! See you next time!!