1. Introduction to Statistics for the Social Sciences SBS200 - Lecture Section 001, Fall 2018 Room 150 Harvill Building 10: :50 Mondays, Wednesdays & Fridays.
Welcome
10/31/18

2. ..
The
Green
Sheets

3. Before next exam (November 16th)
Schedule of readings
Before next exam (November 16th)
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

4. Lab sessions
This Week
Project 3

6. Confidence Interval of 95% Has and alpha of 5% α = .05
Critical z
-2.58
Critical z
2.58
Confidence Interval of 99%
Has and alpha of 1%
α = .01
99%
Critical z separates rare from common scores
Critical z
-1.96
Critical z
1.96
Confidence Interval of 95% Has and alpha of 5% α = .05
95%
Area associated with most extreme scores is called alpha
Critical z
-1.64
Critical z
1.64
Confidence Interval of 90%
Has and alpha of 10%
α = . 10
90%
Area in the tails is called alpha

7. Rejecting the null hypothesis
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 < 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

8. Reject the null hypothesis Support for alternative.
Reject the null hypothesis
95% ..
Relative to this distribution I am unusual maybe even an outlier X
95% X
Relative to this distribution I am utterly typical
Support for alternative
hypothesis

9. Deciding whether or not to reject the null hypothesis. 05 versus
Deciding whether or not to reject the null hypothesis .05 versus .01 alpha levels
What if our observed z = 2.0?
How would the critical z change?
Which situation
(alpha and tail) would make it easiest to reject the null?
α = 0.05
Significance level = .05
α = 0.01
Significance level = .01
why?
-1.96 or
+1.96
p < 0.05
Yes, Significant difference
Reject the null
Remember,
reject the null if the observed z is bigger than the critical z
-2.58 or
+2.58
Not a Significant difference
Do not Reject the null

10. One versus two tail test of significance 5% versus 1% alpha levels
What if our observed z = 2.45?
How would the critical z change?
Which situation
(alpha and tail) would make it easiest to reject the null?
One-tailed
Two-tailed
α = 0.05
Significance level = .05
α = 0.01
Significance level = .01
why?
-1.64 or
+1.64
-1.96 or
+1.96
Remember,
reject the null if the observed z is bigger than the critical z
Reject the null
Reject the null
-2.33 or
+2.33
-2.58 or
+2.58
Reject the null
Do not Reject the null

11. One versus two tail test of significance: Comparing different critical scores (but same alpha level – e.g. alpha = 5%)
1.64
95% 5%
95%
z score = -1.64
reject
null
2.5%
2.5%
95%
How would the critical z change?
Critical scores get smaller with one tailed test 5%
One tail test requires:
1. a unidirectional prediction (predict which group will have larger mean) and
2. that the results actually be in the predicted direction (predicted mean is larger)
So, in a one-tailed test the “region of rejection” refers to results in the predicted direction
If results are NOT in predicted direction, it is impossible to reject the null

12. Two tail tests One tail tests
In a one-tailed test do we use a negative or positive z score?
Doesn’t matter whether the z is positive or negative
If prediction was right, reject the null if observed score is larger than the critical score
Two tail tests
But, if prediction was wrong, it is impossible to reject the null anyway
One tail tests
Which type of test REQUIRES that you make a prediction about which group mean will be larger?
When we go from the regular two-tailed test to a one tail test, what happens to the critical z?
Only the one-tail test does
So, if prediction was right, it is easier to reject the null
The critical score get smaller
But, if prediction was wrong, it is impossible to reject the null
So, if prediction was right, it is easier to reject the null
But, if prediction was wrong, it is impossible to reject the null

13. Review of the homework assignment

14. 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 4 √ 16
= 4.0
.25
4.0
-1.96
1.96

15. 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

16. 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

17. two tail test
α= .05
(df) =15

18. 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 of is still bigger than critical t of 1.753
2.947
Yes
Because observed t of is not bigger than critical t of 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

19. 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

20. Value of observed statistic
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

21. Thank you!
See you next time!!