1. 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

2. A note on doodling

3. 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

4. Before next exam (April 7th)
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

5. Homework No homework due: Monday, March 27th
Just work on Project 3 to prepare for labs

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

9. 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

10. 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
= 74
H1:

11. 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

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

13. µ = 74 Hypothesis testing = = 868.16 = 6.01 24 x (x - x) (x - x)276 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 = = = = = = = = = = = = = = =
0.1936
2.4336
2.0736
6.5536
0.3136
5.9536
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 =

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

15. 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

16. 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 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

17. 26.08 < µ < 33.92 mean + z σ = 30 ± (1.96)(2)
95%
< µ < 33.92
mean + z σ = 30 ± (1.96)(2)
99%
< µ < 35.16
mean + z σ = 30 ± (2.58)(2)

18. 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

19. 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

20. 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

21. 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
2.667 6 √ 16

22. 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

23. 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

24. Thank you!
See you next time!!