Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2019 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays.

Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2019 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays & Fridays. March 22

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The Green Sheets

Before next exam (April 5th)
Schedule of readings Before next exam (April 5th) Please read chapters 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

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

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 (two-tailed α = .05) 1 1 6 – 5 = = .25 = 4.0 4 √ 16 .25 review -1.96 1.96 z = 4.0

Because observed z of 4 is still bigger than critical z of 1.64
Because the observed z (4.0 ) is bigger than critical z (1.96) These three will always match Yes Probability of Type I error is always equal to alpha Yes Yes 0.05 1.64 No Because observed z of 4 is still bigger than critical z of 1.64 2.58 No Because observed z of 4 is still bigger than critical z of 2.58 there is a difference there is not there is no difference there is 1.96 review 2.58

Two tailed test (α = .05) n – 1  16 – 1 = 15 Critical t(15) = 2.131
-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

Two tailed test (α = .05) n – 1  16 – 1 = 15 Critical t(15) = 2.131
-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 2.667 6 √ 16

These three will always match Yes Yes
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

Yes Yes Yes .05 1.753 No Because observed t(15) = 2.67 is still bigger than critical t(15) of 1.753 2.947

These three will always match
Yes Yes Yes .05 1.753 No Because observed t(15) = 2.67 is still bigger than critical t(15) of 1.753 2.947 Yes Because observed t(15) = is not bigger than critical t(15) of 2.947 No These three will always match No No consultant did improve morale when in fact she did not improve morale consultant did not improve morale when in fact she did improve morale 2.131 2.947

The average weight of bags of potatoes from this particular plant
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

The average job-satisfaction score was 89 for the employees who went
Finish with statistical summary t(15) = 2.67; p < 0.05 Or if it *were not* significant: t(15) = 1.07; 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 job-satisfaction score was 89 for the employees who went On the retreat, while the average score for population is 85. A t-test was completed and this difference was found to be statistically significant. We should hire the consultant. (t(15) = 2.67; p<0.05) Value of observed statistic df

Independent samples t-test
19

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

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 21

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 22

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) = 3.5 23

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 = 1.964 3.5 3.5 3 3 Observed t Observed t = Critical t = 2.776 1.964 is not larger than so, we do not reject the null hypothesis t(4) = 1.964; n.s. Conclusion: There appears to be no difference between the groups 24

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 25

Complete a t-test Mean= 21 Mean= 24 Participant 1 2 3 Big Meal 22 25
Small meal 19 23 21 26

Small meal 19 23 21 27

Small meal 19 23 21 If checked you’ll want to include the labels in your variable range If checked, you’ll want to include the labels in your variable range If checked you’ll want to include the labels in your variable range 28

Complete a t-test Finding Means Finding Means 29

Complete a t-test This is variance for each sample
(Remember, variance is just standard deviation squared) Please note: “Pooled variance” is just like the average of the two sample variances, so notice that the average of 3 and 4 is 3.5 30

Complete a t-test This is “n” for each sample
(Remember, “n” is just number of observations for each sample) This is “n” for each sample (Remember, “n” is just number of observations for each sample) Remember, “degrees of freedom” is just (n-1) for each sample. So for sample 1: n-1 =3-1 = 2 And for sample 2: n-1=2-1 = 2 Then, df = 2+2=4 df = “degrees of freedom” 31

Finding degrees of freedom
Complete a t-test Finding degrees of freedom 32

Complete a t-test Finding Observed t 33

Complete a t-test Finding Critical t 34

Finding Critical t 35

Finding p value (Is it less than .05?)
Complete a t-test Finding p value (Is it less than .05?) 36

Step 4: Make decision whether or not to reject null hypothesis
Complete a t-test Step 4: Make decision whether or not to reject null hypothesis Reject when: observed stat > critical stat is not bigger than 2.776 “p” is less than 0.05 (or whatever alpha is) p = is not less than 0.05 Step 5: Conclusion - tie findings back in to research problem There was no significant difference, there is no evidence that size of meal affected test scores 37

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 38

Graphing your t-test results 39

Graphing your t-test results 40

Graphing your t-test results Chart Layout 41

Fill out titles 42

Where are we? 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= 24 Mean= 21 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. 43

What if we ran more subjects?
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. This time he had two classes, both with nine 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 44

Notice: Additional participants don’t affect this part of the problem
Independent samples t-test Notice: Additional participants don’t affect this part of the problem Step 1: Identify the research problem Did the size of the meal affect the test scores? Step 2: Describe the null and alternative hypotheses Ho: The size of the meal has no effect on test scores H1: The size of the meal does have an effect on test scores One tail or two tail test? 45

Notice: Two different ways to think about it
Hypothesis testing Step 3: Decision rule α = .05 n1 = 9; n2 = 9 Degrees of freedom total (df total) = (n1 - 1) + (n2 – 1) = (9 - 1) + (9 – 1) = 16 Degrees of freedom total (df total) = (n total - 2) = 18 – 2 = 16 two tailed test Notice: Two different ways to think about it Critical t(16) = 2.12 46

two tail test α= .05 (df) = 16 Critical t(16) = 2.12 47

8 8 Step 4: Calculate observed t-score 18 2.25 Notice: s2 = 2.25 24
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 = 18 Σ = 24 18 2.25 Notice: s2 = 2.25 1 8 1 Notice: Simple Average = 2.625 24 3.00 Notice: s2 = 3.0 2 2 8 48

Sp2 = 2.625 S21 = 2.25 S22 = 3.00 Step 4: Calculate observed t-score
Mean= 21 Mean= 24 Big Meal 22 25 Small meal 19 23 21 Sp2 = 2.625 S21 = 2.25 S22 = 3.00 S2pooled = (n1 – 1) s12 + (n2 – 1) s22 n1 + n2 - 2 S2pooled = (9 – 1) (2.25) + (9 – 1) (3) = 2.625 49

Sp2 = 2.625 S21 = 2.25 S22 = 3.00 Step 4: Calculate observed t-score
Mean= 21 Mean= 24 Big Meal 22 25 Small meal 19 23 21 Sp2 = 2.625 S21 = 2.25 S22 = 3.00 = 24 – 21 0.7638 = 2.625 2.625 9 9 50

Step 5: Make decision whether or not to reject null hypothesis
Summarizing your t-test results Step 5: Make decision whether or not to reject null hypothesis Observed t = Critical t = 3.928 is farther out on the curve than 2.120 so, we do reject the null hypothesis t(16) = 3.928; p < 0.05 51

We compared test scores for large and small meals. The mean test
Summarizing your t-test results Step 6: Conclusion We compared test scores for large and small meals. The mean test score for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there was a significant difference in test scores between the two types of meals t(16) = 3.928; p < 0.05 52

Let’s run more subjects using our excel!
53

Finding Means Finding Means 54

This is variance for each sample (Remember, variance is just standard deviation squared) Please note: “Pooled variance” is just like the average of the two sample variances, so notice that the average of 2.25 and 3 is 2.625 55

This is “n” for each sample (Remember, “n” is just number of observations for each sample) This is “n” for each sample (Remember, “n” is just number of observations for each sample) Remember, “degrees of freedom” is just (n-1) for each sample. So for sample 1: n-1 =9-1 = 8 And for sample 2: n-1=9-1 = 8 Then, df = 8+8=16 df = “degrees of freedom” 56

Finding degrees of freedom Finding degrees of freedom 57

Finding Observed t 58

Finding Critical t 59

Remember, if the “t Stat” is bigger than the “t Critical” then we “reject the null”, and conclude we have a significant effect Remember, if the “t Stat” is bigger than the “t Critical” then we “reject the null”, and conclude we have a significant effect 60

Finding p value (Is it less than .05?) 61

In this case, p = which is less than 0.05, so we “do reject the null” Remember, if the “p” is less than 0.05 then we “reject the null”, and conclude we have a significant effect 62

We compared test scores for large and small meals. The mean test score for the big meal was 24, and was 21 for the small meal. A t-test was calculated and there was a significant difference in test scores between the two types of meals t(16) = 3.928; p < 0.05 Let’s run more subjects using our excel! 63

What happened? We ran more subjects: Increased n
So, we decreased variability Easier to find effect significant even though effect size didn’t change This is the sample size This is the sample size Small sample Big sample 64

What happened? We ran more subjects: Increased n
So, we decreased variability Easier to find effect significant even though effect size didn’t change This is variance for each sample (Remember, variance is just standard deviation squared) This is variance for each sample (Remember, variance is just standard deviation squared) Small sample Big sample 65

Homework .

This is significant with alpha of 0.05
Type of instruction Exam score 50 40 2-tail 0.05 CAUTION This is significant with alpha of 0.05 BUT NOT WITH alpha of 0.01 2.66 2.02 38 p = yes The average exam score for those with instruction was 50, while the average exam score for those with no instruction was 40. A t-test was conducted and found that instruction significantly improved exam scores, t(38) = 2.66; p < 0.05

. Type of Staff Travel Expenses 142.5 130.29 2-tail 0.05 2.20 11 p = 0.153 no The average expenses for sales staff is 142.5, while the average expenses for the audit staff was A t-test was conducted and no significant difference was found, t(11) = 1.54; n.s.

If the observed t is less than one it will never be significant
. Location of lot Number of cars 86.24 92.04 2-tail 0.05 -0.88 2.01 51 p = 0.38 no Fun fact: If the observed t is less than one it will never be significant The average number of cars in the Ocean Drive Lot was 86.24, while the average number of cars in Rio Rancho Lot was A t-test was conducted and no significant difference between the number of cars parked in these two lots, t(51) = -.88; n.s.

Thank you! See you next time!!

Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2019 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays.

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Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2019 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays.

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Presentation on theme: "Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Spring 2019 Room 150 Harvill Building 9:00 - 9:50 Mondays, Wednesdays."— Presentation transcript:

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