Two-Sample Problems – Means
1. Comparing two (unpaired) populations
2. Assume: 2 SRSs, independent samples,
Normal populations
Make an inference for their difference: 1  2
Sample from population 1: n1 , x1 , s1
Sample from population 2: n2 , x2 , s2
1
S.E. – standard error in the two-sample process
2
S .E . 
2
s1
s2

n1
n2
Confidence Interval: Estimate ± margin of error
 ( x1  x2 )  t S .E.
*
Significance Test:
1 , 2  population means being tested
H 0 : 1  2  0 (or 1  2 ) df  min( n1 1, n2 1)
( x1  x2 )  0
Test Stat : t 

S .E .
x1  x2
2
2
s1 s2

n1 n2
2
Using the Calculator
Confidence Interval:
On calculator: STAT, TESTS, 0:2-SampTInt…
Given data, need to enter: Lists locations, C-Level
Given stats, need to enter, for each sample: x, s, n
and then C-Level
Select input (Data or Stats), enter appropriate info, then
Calculate
3
Using the Calculator
Significance Test:
On calculator: STAT, TESTS, 4:2-SampT –Test…
Given data, need to enter: Lists locations, Ha
Given stats, need to enter, for each sample : x, s, n
and then Ha
Select input (Data or Stats), enter appropriate info, then
Calculate or Draw
Output: Test stat, p-value
4
Ex 1. Is one model of camp stove any different at boiling
water than another at the 5% significance level?
Model 1: n1  10, x1  11.4, s1  2.5
Model 2: n2  12, x2  9.9, s2  3.0

H0 :

Ha :
 Test stat : t 
 p  value :

5
Ex 2. Is there evidence that children get more REM sleep
than adults at the 1% significance level?
Children: n1  11, x1  2.8, s1  0.5
Adults: n2  13, x2  2.1, s2  0.7

H0 :

Ha :
 Test stat : t 
 p  value :

6
Ex 3. Create a 98% C.I for estimating the mean difference
in petal lengths (in cm) for two species of iris.
Iris virginica: n1  35, x1  5.48, s1  0.55
Iris setosa: n2  38, x2  1.49, s2  0.21

 t - Interval :
 margin of error :

7
Ex 4. Is one species of iris any different at petal length
than another at the 2% significance level?
Iris virginica: n1  35, x1  5.48, s1  0.55
Iris setosa: n2  38, x2  1.49, s2  0.21
-4 -3 -2 -1 0 1 2 3 4

H0 :

Ha :
 Test stat : t 
 p  value :

8
Two-Sample Problems – Proportions
Make an inference for their difference:
p1  p2
Sample from population 1: n1 , x1 , pˆ1
Sample from population 2: n2 , x2 , pˆ 2
9
Using the Calculator
Confidence Interval: Estimate ± margin of error
*
ˆ
ˆ
 ( p1  p2 )  z S .E.
On calculator: STAT, TESTS, B:2-PropZInt…
Need to enter: n1 , x1 , n2 , x2 , C-Level
Enter appropriate info, then Calculate.
10
Using the Calculator
Significance Test:
p1 , p2  population proportion s being tested
H 0 : p1  p2  0 (or p1  p2 )
On calculator: STAT, TESTS, 6:2-PropZTest…
Need to enter: n1 , x1 , n2 , x2 , and then Ha
Enter appropriate info, then Calculate or Draw
Output: Test stat, p-value
11
Ex 5. Create a 95% C.I for the difference in proportions of
eggs hatched.
Nesting boxes apart/hidden: n1  478, x1  270(hatched )
Nesting boxes close/visible: n2  805, x2  270(hatched )

 z - Interval :
 margin of error :

12
Ex 6. Split 1100 potential voters into two groups, those
who get a reminder to register and those who do not.
Of the 600 who got reminders, 332 registered.
Of the 500 who got no reminders, 248 registered.
Is there evidence at the 1% significance level that the
proportion of potential voters who registered was greater
than in the group that received reminders?
Group 1:
n1  600, x1  332
Group 2:
n2  500, x2  248
13
Ex 6. (continued)

H0 :

Ha :
 Test stat : z 
 p  value :

14
Ex 7. “Can people be trusted?”
Among 250 18-25 year olds, 45 said “yes”.
Among 280 35-45 year olds, 72 said “yes”.
Does this indicate that the proportion of trusting people is
higher in the older population?
Use a significance level of α = .05.
Group 1: n1  250, x1  45
Group 2: n2  280, x2  72

15
Ex 7. (continued)
H0 :

Ha :
 Test stat : z 
 p  value :

16
Scatterplots & Correlation
Each individual in the population/sample will
have two characteristics looked at, instead of one.
Goal: able to make accurate predictions for one
variable in terms of another variable based on a
data set of paired values.
17
Variables
Explanatory (independent) variable, x, is used to
predict a response.
Response (dependent) variable, y, will be the
outcome from a study or experiment.
height vs. weight,
age vs. memory,
temperature vs. sales
18
Scatterplots
Plot of paired values helps to determine if a
relationship exists.
Ex: variables – height(in), weight (lb)
Height
Weight
72
171
65
150
68
180
70
180
72
185
66
165
190
170
150
65 66
68
70
72
19
Scatterplots - Features
Direction: negative, positive
Form: line, parabola, wave(sine)
Strength: how close to following a pattern
Direction:
Form:
Strength:
190
170
150
65 66
70
72
20
Scatterplots – Temp vs Oil used
Direction:
Form:
Strength:
45
35
25
20 30
70
90
21
Correlation
Correlation, r, measures the strength of the linear
relationship between two variables.
r > 0: positive direction
r < 0: negative direction
Close to +1:
Close to -1:
Close to 0:
22
.85, -.02, .13, -.79
23
Lines - Review
y = a + bx
a:
b:
3
y  2 x
2
3
2
1
1 2
3
4
24
Regression
Looking at a scatterplot, if form seems linear, then
use a linear model or regression line to describe
how a response variable y changes as an
explanatory variable changes.
Regression models are often used to predict the
value of a response variable for a given explanatory
variable.
25
Least-Squares Regression Line
The line that best fits the data:
yˆ  a  bx
where:
br
sy
sx
a  y  bx
26
Example
Fat and calories for 11 fast food chicken sandwiches
Fat: x  20.6, s x  9.8
r

.
947
Calories: y  472.7, s y  144.2
27
Example
Fat and calories for 11 fast food chicken sandwiches
Fat: x  20.6, s x  9.8
r  .947
Calories: y  472.7, s y  144.2
Calories
Fat
28
Example-continued
yˆ  185.65  13.93x
What is the slope and what does it mean?
What is the intercept and what does it mean?
How many calories would you predict a sandwich
with 40 grams of fat has?
29
Why “Least-squares”?
The least-squares lines is the line that minimizes
the sum of the squared residuals.
Residual: difference between actual and predicted
x
y
ŷ
y  yˆ
1
10
14
-4
3
25
24
1
…
…
…
…
27
18
9
1
3
30
Scatterplots – Residuals
To double-check the appropriateness of using a
linear regression model, plot residuals against the
explanatory variable.
No unusual patterns means good linear relationship.
31
Other things to look for
Squared correlation, r2, give the percent of
variation explained by the regression line.
Chicken data:
r  .947
32
Other things to look for
Influential observations:
Prediction vs. Causation:
x and y are linked (associated) somehow but
we don’t say “x causes y to occur”. Other forces may
be causing the relationship (lurking variables).
33
Extrapolation: using the regression for a prediction outside
of the range of values for the explanatory variables.
age
weight
20
180
230
y = 1.6126x + 148.49
R² = 0.7157
220
25
190
32
190
36
200
36
225
40
215
47
220
weight
210
200
y
190
Linear (y)
180
170
160
0
10
20
30
40
50
age
34
On calculator
Set up: 2nd 0(catalog), x-1(D), scroll down to
“Diagnostic On”, Enter, Enter
Scatterplots: 2nd Y=(Stat Plot), 1, On, Select Type
And list locations for x values and y values
Then, ZOOM, 9(Zoom Stat)
Regression: STAT, CALC, 8: LinReg (a + bx), enter,
List location for x, list location for y, enter
Graph: Y=, enter line into Y1
35
Examples:
Cat Chick Dog
Duck
Goat
Lion
Bird
Pig
Bun
ny
Squir
rel
x 63
22
63
28
151 108 18 115 31
44
Incubation,
days
y 11
7.5
11
10
12
9
Lifespan,
years
x 2
5
1
age, years
2
5
y 16 11 17 10
4
5
10
1
8
1
10
4
2
7
6
12 11 20 19 10 16 11 20
resale,
thousands $
36
Contingency Tables
Making comparisons between two categorical variables
• Contingency tables summarize all outcomes
– Row variable: one row for each possible value
– Column variable: one column for each possible value
– Each cell (i,j) describes number of individuals with those
values for the respective variables.
Age\Income
<15
15-30
>30
Total
<21
5
3
1
9
21-25
4
9
6
19
>25
2
2
8
12
Total
11
14
15
40
37
• Info from the table
– # who are over 25 and make under $15,000:
– % who are over 25 and make under $15,000:
– % who are over 25:
– % of the over 25 who make under $15,000:
Age\Income
<15
15-30
>30
Total
<21
5
3
1
9
21-25
4
9
6
19
>25
2
2
8
12
Total
11
14
15
40
38
Age\Income
Marginal Distributions
<15
15-30
>30
Total
<21
5
3
1
9
21-25
4
9
6
19
>25
2
2
8
12
Total
11
14
15
40
– Look to margins of tables for individual variable’s distribution
– Marginal distribution for age:
Age
Freq.
<21
9
21-25
19
>25
12
Total
40
Rel. Freq
– Marginal distribution for income:
Income <15
Freq.
11
15-30
>30
Total
14
15
40
Rel. Freq.
39
Conditional Distributions
– Look at one variable’s distribution given another
– How does income vary over the different age groups?
– Consider each age group as a separate population and compute
relative frequencies:
Age\Income
Age\Income
<15
15-30
>30
Total
<21
5
3
1
9
21-25
4
9
6
19
>25
2
2
8
12
<15
15-30
>30
Total
<21
21-25
>25
40
Independence Revisited
Two variables are independent if knowledge of one
does not affect the chances of the other.
In terms of contingency tables, this means that the
conditional distribution of one variable is (almost) the
same for all values of the other variable.
In the age/income example, the conditionals are not
even close. These variables are not independent.
There is some association between age and income.
41
Test for Independence
Is there an association between two variable?
– H0: The variables are
( The two variables are
– Ha: The variables
(The two variables are
)
)
Assuming independence:
– Expected number in each cell (i, j):
(% of value i for variable 1)x(% of j value for variable 2)x
(sample size) =
42
Example of Computing Expected Values
Rh\Blood A B AB
+ 176 28 22
O
198
Total
424
- 30 12 4
Total 206 40 26
30
228
76
500
Expected number in cell (A, +):
Rh\Blood
+
A
B
Total
206
40
AB
O
22.048 193.344
Total
424
3.952
34.656
76
26
228
500
43
Chi-square statistic
To measure the difference between the observed
table and the expected table, we use the chisquare test statistic:
 
2
observed count  expected count 
2
expected count
where the summation occurs for each cell in the table.
1. Skewed right
2. df = (r – 1)(c – 1)
3. Right-tailed test
44
Test for Independence – Steps
 State variables being tested
 State hypotheses: H0, the null hypothesis, vars independent
Ha, the alternative, vars not independent
 Compute test statistic: if the null hypothesis is true, where
does the sample fall? Test stat = X2-score
 Compute p-value: what is the probability of seeing a test stat
as extreme (or more extreme) as that?
 Conclusion: small p-values lead to strong evidence against H0.
45
ST – on the calculator
On calculator: STAT, TESTS, C:X2 –Test
Observed: [A]
Expected: [B]
Enter observed info into matrix A, then perform test with
Calculate or Draw.
To enter observed info into matrix A:
2nd, x-1 (Matrix), EDIT, 1: A, change dimensions,
enter info in each cell.
Output: Test stat, p-value, df
46
Ex . Test whether type and rh factor are independent at a 5%
significance level.

H0 :

Ha :
 Test stat :  
2
 p  value :
 conclusion :
47
Ex . Test whether age and stance on marijuana legalization are
associated.
stance\age 18-29 30-49
50Total
for
against
172
52
313
103
258
119
743
274
Total
224
416
377
1017

H0 :

Ha :
 Test stat :  2 
 p  value :
 conclusion :
48
Additional Examples
personality\college Health
extrovert
68
introvert
94
Job grade\marital status
1
2
3
Science
56
81
Lib Arts
62
45
Single
58
222
Married
874
3450
Divorced
15
60
74
1204
93
City size\practice status Government Judicial
<250,000
250-500,000
>500,000
30
79
22
Educator
47
66
44
102
34
Private
258
Salaried
36
651
90
127
23
49