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Unformatted text preview: 2/7/2011 Lecture 5: February 7, 2011
• Today: – Review the classical approach and the p‐value approach to hypothesis testing: one sample z for Ford trucks
– The t‐distribution
– Two‐Sample Hypothesis Tests
• Homework 4 opens at 10:00 p.m. tonight:
– Page 318: Problems 8.18 and 8.20 – Confidence intervals using t
– Pages 363‐364: Problem 9.23, 9.28, and 9.33 – Hypothesis tests of one mean using t
– Homework 4 closes Wednesday, February 9, at 6:00 p.m.
– Unlimited tries until it closes! Feedback!
• Non‐Connect Practice problems:
– Page 318: Problems 8.19 and 8.21 – Confidence intervals using t
– Page 364: Problem: 9.35 – Hypothesis tests of one mean using t
– Answers are in the back of the book.
– If you need help with these, see a TA during office hours.
• Worked‐Out Problems – On my web site
– Page 318: Problem 8.17 – Confidence interval using t
– Page 364: Problem 9.31 – Hypothesis test of one mean using t
– These will be ready some time on Tuesday afternoon. Lecture 5: February 7, 2011
Readings • Student’s t distribution: – Confidence intervals – Chapter 8, pages 310‐318
– One‐sample hypothesis tests – Chapter 9, pages 359‐
364 • Two‐Sample Hypothesis Tests:
– End of today and next two classes
– Chapter 10, pages 390‐401 – Put on your seat belt!
– Type of problems to look at:
• pages 400‐401: 10.1 and 10.2 What we did last class:
• We did a one‐sample hypothesis test of the population mean.
• We were given the set‐up on the following slide. 1 2/7/2011 A One‐Sample Hypothesis Test of the Population Mean • Test the null hypothesis that the true mean is 20 mpg versus the alternative that the true mean is greater than 20 mpg.
• Assume that the standard deviation of the sample mean (or standard error) for each sampling distribution is: X / n 1.818
• You gather 25 observations and calculate
X=24 • Test at the 0.025 level of significance.
• Assume the underlying variable of interest is normally distributed. What we did last class (continued):
• Next, we followed the steps in the handout titled Revised Steps In Hypothesis Testing.
• The approach that we took with these five steps is sometimes called the classical approach to hypothesis testing.
• Again, here are the dynamics. Dynamics Of The Classical Approach To Hypothesis Testing
H0: µ0=20
Test statistic: X = 24 1‐α X = 1.818 α=0.025 Is far from μ0 = 20? X X µ0=20 Standardized test statistic: z = 2.20 X = 24 Reject H0 Fail to Reject H0
z.025=1.96 z=2.20
HA: µ1>26 β z
Reject H0 … because z>z.025 1‐β µ1>20 X 2 2/7/2011 The P‐Value Approach To Hypothesis Testing
H0 True
α=0.025
P‐value=0.0139
X µ0=20 X = 24 Reject H0 Fail to Reject H0
Z.025=1.96 • z z=2.20 Compare α and p‐value • Reject H0 • Because p‐value < α Usefulness of the p‐value approach
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• In our hypothesis test problem
What is the meaning of P(z>2.20) = 0.0139?
Here’s what it means and how it is used.
0
It gets at the issue: Is this value of far away from ?
X
If the p‐value is small, this means is far away from .
X
0
This is evidence against the null hypothesis. Interpretation Of A P‐Value
H0 True
α=0.025
P‐value=0.0139
X
µ0=20 X = 24 A p‐value is the observed level of significance for our X
For this example, the p‐value = 0.0139
Interpretation: It is the probability of observing a value ofX=24 or greater if the null hypothesis is true. It is calculated asP(X>24) with μ=20 and X =1.818. Meaning: If H0 is true, we would observe a value of or greater only about 1 time in 100. X=24
Translation: We have extremely strong evidence against the null hypothesis.
Decisions reached with both the p‐value approach and the classical approach are the same. 3 2/7/2011 New Topic Moving beyond the z‐distribution
(Chapter 8, pages 310‐312) (1) Strict definition of z (Chapter 8):
z X
X / n (a) z has one shape: standard normal
(b) z is centered at zero.
(c) z has standard deviation one. (d) As long as is given, i.e., known,
X can be converted to z.
(In fact, this is what we just did.) Moving beyond the z‐distribution
(continued) (2) Q: In what situations did we use z?
X
A: Making probability statements about .
Confidence intervals.
Hypothesis tests. (3) (a) In practical situations, we do not have . (b) We must use its estimate, which is the sample standard deviation s. (c) If s replaces in the z formula, then the formula is really no longer z! 4 2/7/2011 Moving beyond the z‐distribution
(continued)
This image cannot currently be display ed. (4) Q: What is: X 
s
n
A: t
(5) Q: Is there a relationship between t and z?
A: Yes.
The relationship depends on the magnitude of n.
Notice the t and z formulas. As n increases, s begins to resemble better.
t becomes z. Moving beyond the z‐distribution
(continued)
This image cannot currently be display ed. Here’s what we have so far: z X
X / n t = X 
s
n As n increases, s begins to resemble better, t becomes z. (6) Let’s apply Points (1)‐(4) from the previous two slides and develop a set of pictures for t. Moving beyond the z‐distribution (continued)
(6) Pictures of t, continued ...
(a) Shape: a number of shapes, depending on something called degrees of freedom (df): df=n‐1 (See p. 311.)
(b) Centered at zero; symmetrical about zero.
(c) t is flatter and more spread out than the z curve for small values of n; t becomes more like z as n → .
This image cannot currently be display ed. 5 2/7/2011 Moving beyond the z‐distribution (continued)
(7) Each curve reports tail area probabilities () and corresponding t‐values (t) (in Appendix D on page 766 of the text) for a variety of different sample sizes, i.e., for different n.
This image cannot currently be display ed. (8) If I am given n (or df) and , I can find t.
df t0.10 t0.05 t0.025 t0.01 t0.005 1 6.314 12.706
19 1.729 2.093 1000 1.282 1.646 1.962 2.330 2.581 z 1.282 1.645 1.960 2.326 2.576 Moving beyond the z‐distribution (continued)
This is the same as the bottom of the previous slide.
t0.05
t0.025
t0.01
t0.005
df
t0.10
1 6.314 12.706
19 1.729 2.093 1000 1.282 1.646 1.962 2.330 2.581 z 1.282 1.645 1.960 2.326 2.576
This image cannot currently be display ed. e.g., Given: n 2 .05 df t 20 .05
1001 .05
z .05 6 ...
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