1. INTEGRATED LEARNING CENTER
Screen
Lecturer’s desk
Cabinet
Cabinet
Table
Computer Storage Cabinet 4 3
Row A 19 18 5 17 16 15 14 13 12 11 10 9 8 7 6 2 1
Row B 3 23 22 6 5 4 21 20 19 7 18 17 16 15 14 13 12 11 10 9 8 2 1
Row C 24 4 3 23 22 5 21 20 6 19 7 18 17 16 15 14 13 12 11 10 9 8 1
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Row H 25 24 3 23 22 6 5 4 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 26 2 1
Row I 25 24 3 23 4 22 5 21 20 6 19 18 17 16 15 14 13 12 11 10 9 8 7 26 1 25 3 2
Row J 24 23 5 4 22 21 20 6 28 19 7 18 17 16 15 14 13 12 11 10 9 8 27 26 25 3 2 1
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INTEGRATED LEARNING CENTER
ILC 120
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desk

2. Introduction to Statistics for the Social Sciences SBS200, COMM200, GEOG200, PA200, POL200, or SOC200 Lecture Section 001, Fall, 2014 Room 120 Integrated Learning Center (ILC) 10: :50 Mondays, Wednesdays & Fridays.
Welcome

3. A note on doodling
Reminder

4. Schedule of readings Before next exam (November 21st)
Please read chapters 7 – 11 in Ha & Ha
Please read Chapters 2, 3, and 4 in Plous
Chapter 2: Cognitive Dissonance
Chapter 3: Memory and Hindsight Bias
Chapter 4: Context Dependence

5. By the end of lecture today 10/29/2014
Use this as your
study guide
By the end of lecture today 10/29/2014
Logic of hypothesis testing
Steps for hypothesis testing
Levels of significance (Levels of alpha)
Hypothesis testing with t-scores (one-sample)
Constructing brief, complete summary statements

6. Homework due
Assignment 16
One-sample z and t hypothesis tests
Due date extended: Friday, October 31st

7. Lab sessions
Labs continue
this week with Project 2

9. These would be helpful to know by heart – please memorize
Standard deviation and Variance
For Sample and Population
These would be helpful to know by heart – please memorize
these formula
Review

10. Standard deviation and Variance
For Sample and Population
Critical value gets smaller
Part 2:
When we move from a two-tailed test to a one-tailed test what happens to the critical z score (bigger or smaller?) - Draw a picture -
What affect does this have on the hypothesis test (easier or harder to reject the null?)
Gets easier to reject the null
Part 3:
What are the five steps for hypothesis testing?

11. Five steps to hypothesis testing
Step 1: Identify the research problem (hypothesis)
Describe the null and alternative hypotheses
Step 2: Decision rule: find “critical z” score
Alpha level? (α = .05 or .01)?
One versus two-tailed test
Step 3: Calculate observed z score
If we change from two-tailed to one tailed test
does critical z score get bigger or smaller?
Step 4: Compare “observed z” with “critical z”
If observed z > critical z then reject null
p < 0.05 and we have significant finding
Step 5: Conclusion - tie findings back in to research problem
Review

12. Five steps to hypothesis testing
Step 1: Identify the research problem (hypothesis)
How is a t score different than a z score?
Describe the null and alternative hypotheses
Step 2: Decision rule: find “critical z” score
Alpha level? (α = .05 or .01)?
One versus two-tailed test
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
Population versus sample standard deviation
Population versus sample standard deviation
Step 5: Conclusion - tie findings back in to research problem

13. 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

14. Hypothesis testing Step 1: Identify the research problem / hypothesis
Extra time
vs. no extra time
Independent
Variable?
Test Scores
Dependent Variable?
Did the extra time given to this sample of students affect
their chemistry test scores?
IV:
Nominal
IV: Nominal Ordinal Interval or Ratio?
DV:
Ratio
DV: Nominal Ordinal Interval or Ratio?
Two-tailed
test
One tail or two tail test?
Null: The extra time did not affect their chemistry test scores
Alternative: The extra time did affect their chemistry test scores

15. What is formula for degres of freedom?
Hypothesis testing
Step 2: Decision rule
n – 1
What is formula for degres of freedom?
= .05
n = 25
Degrees of freedom (df) = (n - 1)
= (25 - 1) = 24
two tail test

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

17. µ = 74 Hypothesis testing = = 868.16 = 6.01 24 . x (x - x) (x - x)2
Step 3: Calculations 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 = = = = = = = = = = = = = = =
0.1936
2.4336
2.0736
6.5536
0.3136
5.9536
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 =

18. µ = 74 Hypothesis testing = = = 76.44 - 74 1.20 2.03 .
Step 3: Calculations
µ = 74 Σx = N
1911 25 =
= 76.44
N = 25
s = 6.01 =
1.20
2.03
critical t
6.01 25
Observed t(24) = 2.03

19. .
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

20. .
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 20

21. What if we had chosen a one-tail test?.
What if we had chosen a one-tail test?
Step 1: Identify the research problem / hypothesis
Did the extra time given to this sample of students improve
their chemistry test scores?
Null: The extra time did not improve their chemistry test scores
Alternative: The extra time did improve their chemistry test scores
Step 2: Decision rule
= .05
Degrees of freedom (df) = (n - 1)
= (25 - 1) = 24
Critical t (24) =
1.711

22. .
one tail test
α= .05
(df) = 24
Critical t(24)
= 1.711

23. µ = 74 (exactly same as two-tail test) Hypothesis testing = = 868.16
(x - x)
(x - x)2
Step 3: Calculations 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 = = = = = = = = = = = = = = =
0.1936
2.4336
2.0736
6.5536
0.3136
5.9536
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 =

24. has no effect on calculations stage.
Hypothesis testing
(exactly same as two-tail test)
One-tailed test
has no effect on calculations stage
Step 3: Calculations
µ = 74 Σx = N
1911 25 =
= 76.44
N = 25
s = 6.01 =
1.20
2.03
critical t
6.01 25

25. .
Hypothesis testing
Step 4: Make decision whether or not to reject null hypothesis
Observed t(24) = 2.03
Critical t(24) = 1.711
2.03 is not farther out on the curve than 1.711,
so, we do not reject the null hypothesis
Step 6: Conclusion: The extra time did have a significant
effect on the scores

26. .
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 one-tailed t-test was
completed and there appears to be significant difference in the
test scores for these two groups t(24) = 2.03; p < 0.05.
Type of test with degrees of freedom
n.s. = “not significant”
p<0.05 = “significant”
Value of observed statistic 26

27. Comparing z score distributions with t-score distributions
z-scores
Similarities include:
Using bell-shaped distributions to make
confidence interval estimations and decisions
in hypothesis testing
Use table to find areas under the curve
(different table, though
– areas often differ from z scores)
t-scores
Summary of 2 main differences:
We are now estimating standard deviation
from the sample
(We don’t know population standard deviation)
We have to deal with degrees of freedom

28. Comparing z score distributions with t-score distributions
Differences include:
We use t-distribution when we don’t know standard deviation
of population, and have to estimate it from our sample
Critical t
(just like critical z)
separates common from rare scores
Critical t used to define both common scores “confidence interval”
and rare scores “region of rejection

29. Comparing z score distributions with t-score distributions
Differences include:
We use t-distribution when we don’t know standard deviation
of population, and have to estimate it from our sample
2) The shape of the sampling distribution is very sensitive to small sample sizes (it actually changes shape depending on n)
Please notice: as sample sizes get smaller, the tails get thicker. As sample sizes get bigger tails get thinner and look more like the z-distribution

30. Comparing z score distributions with t-score distributions
Please note: Once sample sizes get big enough the t distribution (curve) starts to look exactly like the z distribution (curve) scores
Comparing z score distributions with t-score distributions
Differences include:
We use t-distribution when we don’t know standard deviation
of population, and have to estimate it from our sample
2) The shape of the sampling distribution is very sensitive to small sample sizes (it actually changes shape depending on n)
3) Because the shape changes, the relationship between the scores and proportions under the curve change
(So, we would have a different table for all the different possible n’s
but just the important ones are summarized in our t-table)

31. We use degrees of freedom (df) to approximate sample size
Interpreting t-table
We use degrees of freedom (df) to approximate sample size
Technically, we have a different
t-distribution for each sample size
This t-table summarizes the most
useful values for several distributions
This t-table presents useful values for distributions (organized by degrees of freedom)
Each curve is based on its own degrees of freedom (df) - based on sample size, and its own table tying together t-scores with area under the curve
n = 17
n = 5 .
Remember these
useful values for z-scores?
1.64
1.96
2.58

32. Area between two scores Area between two scores
Area beyond two scores
(out in tails)
Area beyond two scores
(out in tails)
Area
in each tail
(out in tails)
Area
in each tail
(out in tails) df

33. useful values for z-scores? .
Area between two scores
Area between two scores
Area beyond two scores
(out in tails)
Area beyond two scores
(out in tails)
Area
in each tail
(out in tails)
Area
in each tail
(out in tails) df
Notice with large sample size it is same values as z-score
Remember these
useful values for z-scores? .
1.96
2.58
1.64

34. For very small samples, t-values differ substantially from the normal.
Comparison of z and t
For very small samples, t-values differ substantially from the normal.
As degrees of freedom increase, the t-values approach the normal z-values.
For example, for n = 31, the degrees of freedom are:
What would the t-value be for a 90% confidence interval?
n - 1 = 31 – 1 = 30 df

35. Degrees of Freedom
Degrees of Freedom (d.f.) is a parameter based on the sample size that is used to determine the value of the t statistic.
Degrees of freedom tell how many observations are used to calculate s, less the number of intermediate estimates used in the calculation.

36. 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)?
One or two tailed test?
Balance between Type I versus Type II error
Critical statistic (e.g. z or t or F or r) 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

37. Questions from the homework assignment?

38. 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)

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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

45. 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

46. Thank you!
See you next time!!