View
87
Download
0
Category
Tags:
Preview:
DESCRIPTION
Chapter 10: Inferential for Regression. http://jonfwilkins.blogspot.com/2011_08_01_archive.html. Model for Linear Regression. Linear Regression. b 0 = ȳ - b 1 x̄. ANOVA table for Linear Regression. Residual Plots. Example: Linear Regression 1. - PowerPoint PPT Presentation
Citation preview
1
Chapter 10: CI and HT Based on Two Samples or Treatments
https://www.cartoonstock.com/directory/s/statistics.asp
2
Notation
Mean Variance SDPopulation 1 μ1 1
Population 2 μ2 2
Population
Sample StatisticsSample size Mean Variance SD
Sample from Population 1
n1 x̄1 s1
Sample from Population 2
n2 x̄2 s2
3
Independent and Paired Samples
1. Two samples are independent if the process of selecting individuals or objects in sample 1 has no effect on, or no relation to, the selection of individuals or objects in sample 2. We call this 2 – sample independent.
2. A paired data set is the result of matching each individual or object in sample 1 with a similar individual or object in sample 2. We call this 2 – sample paired.
4
10.1/10.2: Comparing Two Population Means Using Independent Samples - Goals
• Be able to determine when you can perform 2-sample independent analyses.
• Perform a two-sample hypothesis test and summarize the results when the two samples are independent.
• Be able to construct a level C confidence interval for the difference between two means and interpret the results when the two samples are independent.
5
Conditions for Inference: 2 – sample independent
1. Each group is considered to be a sample from a distinct population.• We have an SRS from the population of
interest for each variable.2. The responses in each group are independent
of those in the other group.3. The statistic that we measure has a Normal
distribution.
6
Two-sample independent (Z): Hypothesis Test
• Step 1• Step 2
H0: µ1 - µ2 = Δz
• Step 3
• Step 4
7
Two-sample independent (Z): Confidence Interval
The 100 (1 – α)% confidence interval for µ1 - µ2 isestimator ± (critical value)(standard deviation of
the estimator)
8
Two-sample independent (t):
• Hypothesis test
• Confidence interval
9
Satterthwaite Approximation
df
s12
n1s22
n2
2
1n1 1
s12
n1
2
1n2 1
s22
n2
2
10
Two-sample Test (independent): Summary
Null hypothesis: H0: μ1 – μ2 = Δ0
Note: If we are determining if the two populations are equal, then Δ0 = 0
Alternative Hypothesis
P-Value
Upper-tailed Ha: μ1 – μ2 > Δ P(T ≥ tts)Lower-tailed Ha: μ1 – μ2 < Δ P(T ≤ tts)two-sided Ha: μ1 – μ2 ≠ Δ 2P(T ≥ |tts|)
11
Example: two-sample Independent tA group of 15 college seniors are selected to
participate in a manual dexterity skill test against a group of 20 industrial workers. Skills are assessed by scores obtained on a test taken by both groups. The data is shown in the following table:
a) Perform a significance test to determine if the skills are the same for college students and industrial works at a significance level of 0.05.
b) Calculate and interpret the 95% confidence interval.Group n x̅ sStudents 15 35.12 4.31Workers 20 37.32 3.83
12
Example: two-sample Independent t (cont)
The data does not provide support (P = 0.128) to the claim that there is a difference between the population mean tests for students and workers.
13
Two-sample independent (t):
• Hypothesis test
• Confidence interval
14
Example: two-sample Independent tA group of 15 college seniors are selected to
participate in a manual dexterity skill test against a group of 20 industrial workers. Skills are assessed by scores obtained on a test taken by both groups. The data is shown in the following table:
a) Perform a significance test to determine if the skills are the same for college students and industrial works at a significance level of 0.05.
b) Calculate and interpret the 95% confidence interval.Group n x̅ sStudents 15 35.12 4.31Workers 20 37.32 3.83
15
Example: two-sample Independent t (CI) (cont)
We are 95% confident that the difference between the population mean tests of students and workers is between -5.08 and 0.68.
P-value = 0.128, (-5.08, 0.68)
16
Robustness of the 2 sample t-procedure
• The t-procedure is very robust against normality. Let n = n1 + n2
–n < 15 : population distribution should be close to normal.–15 < n < 40: mild skewedness is acceptable–n > 40: procedure is usually valid.
• Best when n1 n2
• Best when distributions are similar.
10.3: Paired Data - Goals• Be able to determine when you should use 2-sample
paired analyses.• Be able to construct a level C confidence interval for a
matched pair and interpret the results.• Perform a matched pair t hypothesis test and
summarize the results.
17
18
Matched Pairs Procedures
• To compare the responses to the two treatments in a matched-pairs design, find the difference between the responses within each pair. Then apply the one-sample t procedures to these differences.
19
Conditions for Inference: 2 – sample paired
1. Each pair is considered to be a sample from a population of pairs.• We have an SRS from the population of
pairs.2. Each pair is independent of the other pairs.3. The difference of the each pair that we
measure has a Normal distribution with mean D and standard deviation σD.
20
Two-sample Matched Pair
21
Two-sample matched pair Test: SummaryNull hypothesis: H0: μD = 0
Note: If we are determining if the two populations are equal, then Δ0 = 0
AlternativeHypothesis
P-Value
One-sided: upper-tailed Ha: μD > 0 P(T ≥ tts)One-sided: lower-tailed Ha: μD < 0 P(T ≤ tts)two-sided Ha: μD ≠ 0 2P(T ≥ |tts|)
22
Example: Paired t test ProcedureIn an effort to determine whether sensitivity training for
nurses would improve the quality of nursing provided at an area hospital, the following study was conducted. Eight different nurses were selected and their nursing skills were given a score from 1 to 10. After this initial screening, a training program was administered, and then the same nurses were rated again. On the next slide is a table of their pre- and post-training scores.
a) Conduct a test to determine whether the training could on average improve the quality of nursing provided in the population at a 0.01 significance level.
b) Calculate and interpret the 99% lower confidence bound of the population mean difference in nursing scores?
23
Individual Pre-Training Post-Training Pre - Post1 2.56 4.54 -1.982 3.22 5.33 -2.113 3.45 4.32 -0.874 5.55 7.45 -1.905 5.63 7.00 -1.376 7.89 9.80 -1.917 7.66 7.33 0.338 6.20 6.80 -0.60mean 5.27 6.57 -1.30stdev 2.018 1.803 0.861
24
Example: Paired t test ProcedureIn an effort to determine whether sensitivity training for
nurses would improve the quality of nursing provided at an area hospital, the following study was conducted. Eight different nurses were selected and their nursing skills were given a score from 1 to 10. After this initial screening, a training program was administered, and then the same nurses were rated again. On the next slide is a table of their pre- and post-training scores.
a) Conduct a test to determine whether the training could on average improve the quality of nursing provided in the population at a 0.01 significance level.
b) Calculate and interpret the 99% lower confidence bound of the population mean difference in nursing scores?
25
Example: Paired t test Procedure (cont)
The data does provide strong support (P = 0.002) to the claim that the population average score did improve after training.
26
Example: Paired t test ProcedureIn an effort to determine whether sensitivity training for
nurses would improve the quality of nursing provided at an area hospital, the following study was conducted. Eight different nurses were selected and their nursing skills were given a score from 1 to 10. After this initial screening, a training program was administered, and then the same nurses were rated again. On the next slide is a table of their pre- and post-training scores.
a) Conduct a test to determine whether the training could on average improve the quality of nursing provided in the population at a 0.01 significance level.
b) Calculate and interpret the 99% upper confidence bound of the population mean difference in nursing scores?
27
Example: Paired t test Procedure (cont)
We are 99% confident that the difference in the scores between pre-training and post-training scores is less than -0.39.
P = 0.00185 < -0.39
28
Independent vs. Paired
1. If there is great heterogeneity between experimental units and a large correlation within experimental units then a paired experiment is preferable.
2. If the experimental units are relatively homogeneous and the correlation within pairs is not large, then unpaired experiments should be used
29
In Class: 2-sample Independent or Paired
On the worksheet, state which method is better; independent or paired and why. The following explanations are wrong: 1) there is no information for one of the methods, 2) the data is matched in the exercise, 3) the number of data points is different (or the same).
Recommended