Please Subscribe to this Channel for more solutions and lectures http://www.youtube.com/onlineteaching Chapter 9: Inferences from Two Samples 9.3 Two Means, Two Dependent Samples, Matched Pairs

1. Elementary Statistics
Chapter 9: Inferences
from Two Samples
9.3 Two Dependent
Samples (Matched Pairs)
1

2. Chapter 9: Inferences from Two Samples
9. 1 Two Proportions
9.2 Two Means: Independent Samples
9.3 Two Dependent Samples (Matched Pairs)
9.4 Two Variances or Standard Deviations
2
Objectives:
• Test the difference between two proportions.
• Test the difference between sample means, using the z test.
• Test the difference between two means for independent samples, using the t test.
• Test the difference between two means for dependent samples.
• Test the difference between two variances or standard deviations.

3. 1. The sample data are dependent (matched pairs).
2. The matched pairs are a simple random sample.
3. Either or both of these conditions are satisfied: The number of pairs of
sample data is large (n > 30) or the pairs of values have differences that are
from a population having a distribution that is approximately normal. These
methods are robust against departures for normality, so the normality
requirement is loose.
3
9.3 Two Means: Two Dependent Samples (Matched Pairs)
1. Verify that the sample data consist of dependent samples (or matched pairs)
2. Find the difference d for each pair of sample values.
3. Find 𝑑 & 𝑠𝑑
4. For hypothesis tests and confidence intervals, use the same t test procedures used for a single
population mean.
5. Because the hypothesis test and confidence interval in this section use the same distribution and
standard error, they are equivalent in the sense that they result in the same conclusions.

4. 4
Will a person’s cholesterol level change if the diet is supplemented by a certain
mineral? Six subjects were pretested, and then they took the mineral supplement for a
2-month period. The results follow in the table. (Cholesterol level is measured in mg /
dL) a) Can it be concluded that the cholesterol level has been changed at α = 0.10?
Assume the variable is approximately normally distributed. b) Find the 90% confidence
interval for the difference between the means.
Before (𝒙𝟏) 210 235 208 190 172 244
After (𝒙𝟐) 190 170 210 188 173 228
Difference d
Before (X1) After (X2) D = X1 – X2 D2
210
235
208
190
172
244
190
170
210
188
173
228
20
65
–2
2
–1
16
400
4225
4
4
1
256
Σ D = 100 Σ D2 = 4890
𝐷 =
𝐷
𝑛
𝑠𝐷 =
𝑛 𝐷2 − ( 𝐷)2
𝑛(𝑛 − 1)
=
6 ⋅ 4890 − 1002
6 ⋅ 5
= 25.3903
𝑑 =
𝑑
𝑛 𝑠𝐷 =
(𝑑 − 𝑑)2
𝑛 − 1
=
𝑛 𝑑2 − ( 𝑑)2
𝑛(𝑛 − 1)
TI Calculator:
How to enter data:
1. Stat
2. Edit
3. ClrList 𝑳𝟏 & 𝑳𝟐
4. Type in your data in
𝑳𝟏 & 𝑳𝟐
5. 𝑳𝟏 − 𝑳𝟐
6. Store in 𝑳𝟑
7. Enter
TI Calculator:
To get the Mean, SD, 5-
number summary
1. Stat
2. Calc
3. Select 1 for 1 variable
4. Type: L3 (second 3)
5. Calculate
=
100
6
= 16.667
Example 1

5. 5
Will a person’s cholesterol level change ….b) Find the 90%
confidence interval for the difference between the means.
Step 3: CV: α = 0.1, df= 6 − 1 = 5
Step 1:
H0: μD = 0 and H1: μD  0 (claim), 2TT
Before (𝒙𝟏) 210 235 208 190 172 244
After (𝒙𝟐) 190 170 210 188 173 228
Difference d 𝐷 = 16.667, 𝑠𝐷 = 25.3903
Step 2:𝑻𝑺:
𝑡 =
𝐷 − 𝜇𝐷
𝑠𝐷
𝑛
=
16.667 − 0
25.3903
6
= 1.6079
Step 4: Decision:
a. Do not reject H0
b. The claim is False
c. There is not sufficient
evidence to support the
claim that the mineral
changes a person’s
cholesterol level.
𝑡 =
𝑑 − 𝜇𝑑
𝑠𝑑
𝑛
TI Calculator:
Matched pair: T ‒ Test
1. Tests
2. T ‒ Test
3. Data
4. Enter 𝝁𝟎 =0, List:
𝑳𝟑 = 𝑳𝟏 − 𝑳𝟐, Freq:1
5. Choose RTT, LTT, or
2TT
6. Calculate
→ 𝑡 = ±2.015
𝑑 =
𝑑
𝑛
, 𝑑𝑓 = 𝑛 − 1
Example 1 Test

6. 6
Will a person’s cholesterol level change ….b) Find the 90%
confidence interval for the difference between the means.
Before (𝒙𝟏) 210 235 208 190 172 244
After (𝒙𝟐) 190 170 210 188 173 228
Difference d
𝐷 = 16.667,
𝑠𝐷 = 25.3903
→ 𝑏)𝐸 = 2.015 ⋅
25.3903
6
→ 16.667 ± 20.887
−4.22 < 𝜇𝐷 < 37.554
= 20.887
Since 0 is contained in the interval, the decision is to not reject the
null hypothesis H0: μD = 0.
𝑑 ± 𝐸
𝐸 = 𝑡𝛼
2
𝑠𝑑
𝑛
TI Calculator:
Matched pair: T ‒ Interval
1. Tests
2. T ‒ Interval
3. Data
4. Enter 𝝁𝟎 =0, List: 𝑳𝟑(𝒔𝒊𝒏𝒄𝒆 𝒊𝒕 𝒊𝒔 =
𝑳𝟏 − 𝑳𝟐), Freq:1
5. Calculate
CV: 𝑡 = ±2.015
Example 1 CI

7. 7
Data lists ages of actresses when they won Oscars in the category of Best Actress, along with the ages of
actors when they won Oscars in the category of Best Actor. The ages are matched according to the year
that the awards were presented. This is a small random selection of the available data so that we can better
illustrate the procedures of this section. Use the sample data with a 0.05 significance level to test the
claim that for the population of ages of Best Actresses and Best Actors, the differences have a mean less
than 0. Assume the variable is approximately normally distributed.
Example 2
H0: µd = 0, H1: µd < 0, claim, LTT
Calculate: 𝑑 = −10.2, 𝑠𝑑 = 14.7
Actress (years) 28 28 31 29 35
Actor (years) 62 37 36 38 29
Difference d −34 −9 −5 −9 6
CV: α = 0.05, 𝑑𝑓 = 5 − 1 = 4
→ 𝑡 = −2.132
=
−10.2
14.7/ 5
= −1.557
Decision:
a. Do not reject H0
b. The claim is False
c. There is not sufficient evidence to support the claim
that for the population of ages of Best Actresses and
Best Actors, the differences have a mean less than 0.
(There is not sufficient evidence to conclude that Best
Actresses are generally younger than Best Actors.)
Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is / is not
sufficient evidence to support the claim
that…
d.f. = 𝑛 − 1, 𝑑 =
𝑑
𝑛
𝑡 =
𝑑 − 𝜇𝑑
𝑠𝑑
𝑛
𝑠𝐷 =
(𝑑 − 𝑑)2
𝑛 − 1
=
𝑛 𝑑2 − ( 𝑑)2
𝑛(𝑛 − 1)
𝑑 ± 𝐸 𝐸 = 𝑡𝛼
2
𝑠𝑑
𝑛
TS: 𝑡 =
𝑑−𝜇𝑑
𝑠𝑑
𝑛

8. Data lists ages of actresses when they won Oscars in the category of Best Actress, along with the ages of
actors when they won Oscars in the category of Best Actor. The ages are matched according to the year
that the awards were presented. This is a small random selection of the available data so that we can better
illustrate the procedures of this section. Use the sample data with a 0.05 significance level to test the
claim that for the population of ages of Best Actresses and Best Actors, the differences have a mean less
than 0. Assume the variable is approximately normally distributed. Construct a 90% confidence interval
estimate of µd, which is the mean of the age differences
8
Example 2: P-Value & CI
Calculate: 𝑑 = −10.2, 𝑠𝑑 = 14.7
P-Value Method:
T-distribution: the row with df
= 4 TS: t = −1.552
corresponds to an “Area in
One Tail” that is greater than
0.05 → P-value > 0.05
Technology: P-value = 0.0973.
Actress (years) 28 28 31 29 35
Actor (years) 62 37 36 38 29
Difference d −34 −9 −5 −9 6
𝐸 = 2.132
14.7
5
= 14.015853
−10.2 − 14.0159 < 𝜇𝑑 < −10.2 + 14.0159
−24.2 < 𝜇𝐷 < 3.8
Since 0 is contained in the interval, the decision is to not reject the null
hypothesis H0: μD = 0. We have 90% confidence that the limits of −24.2
years and 3.8 years contain the true value of the mean of the age differences.
d.f. = 𝑛 − 1, 𝑑 =
𝑑
𝑛 𝑡 =
𝑑 − 𝜇𝑑
𝑠𝑑/ 𝑛
𝑑 ± 𝐸 𝐸 = 𝑡𝛼
2
𝑠𝑑
𝑛

9. 9
A sample of nine local banks shows their deposits (in billions of dollars) 3 years ago and their deposits (in
billions of dollars) today. At a 0.05, can it be concluded that the average in deposits for the banks is
greater today than it was 3 years ago (was the average deposits less 3 years ago)? Use α = 0.05.
Example 3
H0: μD = 0 and H1: μD < 0 , claim, LTT
Calculate: 𝑑 & 𝑠𝑑
𝑑 = −1.081, 𝑠𝑑 = 1.937
CV: α = 0.05, 𝑑𝑓 = 9 − 1 = 8
→ 𝑡 = −1.860
=
−1.081
1.937/ 9
= −1.674
Decision:
a. Do not reject H0
b. The claim is False
c. There is not enough evidence to show that the
deposits have increased over the last 3 years.
Step 1: H0 , H1, claim & Tails
Step 2: TS Calculate (TS)
Step 3: CV using α
Step 4: Make the decision to
a. Reject or not H0
b. The claim is true or false
c. Restate this decision: There is / is not
sufficient evidence to support the claim
that…
d.f. = 𝑛 − 1, 𝑑 =
𝑑
𝑛
𝑡 =
𝑑 − 𝜇𝑑
𝑠𝑑/ 𝑛
𝑠𝐷 =
(𝑑 − 𝑑)2
𝑛 − 1
=
𝑛 𝑑2 − ( 𝑑)2
𝑛(𝑛 − 1)
𝑑 ± 𝐸 𝐸 = 𝑡𝛼
2
𝑠𝑑
𝑛
TS: 𝑡 =
𝑑−𝜇𝑑
𝑠𝑑/ 𝑛