Section 8.5 Day 3.

Section 8.5 Day 3

For a two-sided significance test, find the following values:
Significance level for critical values z* = ± 2.36 (b) Critical value(s) for significance level α = 0.20

Significance level for critical values z* = ± 2.36 2[normalcdf(-1EE99, -2.36)] ≈ 0.018 (b) Critical value(s) for significance level α = 0.20

(b) Critical value(s) for significance level α = 0.20 invNorm(0.10) = ± 1.28

For a one-sided significance test, find the following values:
Significance level for critical value z* = 2.36 (b) Critical value for significance level α = 0.20

Significance level for critical value z* = 2.36 normalcdf(2.36, 1EE99) ≈ 0.009 (b) Critical value for significance level α = 0.20

(b) Critical value for significance level α = 0.20 invNorm(.20) =

Difference of 2 Proportions
Recall, for surveys we need two random samples independently selected from two populations.

Difference of 2 Proportions
Recall, for surveys we need two random samples independently selected from two populations. For experiments we do not randomly select subjects; the treatments are randomly assigned to participants

What is the main benefit of performing an experiment?

What is the main benefit of performing an experiment?
Allows us to establish cause and effect.

Suppose we want to study the effects of second-hand smoke on children ages 2-4 years old.
Would we use an experiment?

Would we use an experiment?
No because we can not randomly assign the treatments of exposing/not exposing the children to second-hand smoke.

Recall from Chapter 4 If we can not randomly assign treatments to
the children, then we have an ___________ ______.

Recall from Chapter 4 If we can not randomly assign treatments to the children, then we have an observational study.

Recall from Chapter 4 If we can not randomly assign treatments to the children, then we have an observational study. In observational studies, the conditions of interest (or treatments) are already built into the subjects being studied.

What is a major concern with observational studies?

What is a major concern with observational studies?
Confounding makes it impossible to determine whether the treatment or something else caused the response.

Observational Studies
Do not allow you: to make cause-and-effect conclusions

Do not allow you: to make cause-and-effect conclusions to estimate population parameters in a rigorous way

Do not allow you: to make cause-and-effect conclusions to estimate population parameters in a rigorous way Can provide evidence of possible associations (correlations).

The question being asked here is “Could the result I see in the observed data have reasonably happened by chance?”

The question being asked here is “Could the result I see in the observed data have reasonably happened by chance?” If the answer is “no”, then there is evidence of an association that should be investigated further.

What condition(s) should we check when doing an observational study?

What condition(s) should we check when doing an observational study? Each of n1p1, n1(1 – p1), n2p2, and n2(1 – p2) is at least 5.

Page 551, P57 a) This is an observational study.
There was no random sampling done nor was there random assignment of treatments.

Page 551, P57 b) 2-PropZInt x1: 103 } abused group n1: 908
x2: 53 } not abused group n2: 667 C-level: .90 Calculate

Page 551, P57 b) 2-PropZInt x1: 103 n1: 908 x2: 53 n2: 667
C-level: .90 Calculate (.00956, .0584)

, P57 c) Because 0 is not in this interval, we conclude that the difference in proportions of the people in this study who were abused as children who later committed crimes and the people in this study who were not abused as children who later committed crimes can not be reasonably attributed to chance.

, P57 c) There may be and probably are many other factors that contributed to this difference so you can not conclude from this study alone that abuse of children causes them to be more likely to commit violent crime later in life.

Naming the test, checking conditions, and doing the computations should be straight-forward by now.

, E75a The conditions are met so find and interpret a 95% CI for the difference of two proportions.

, E75a Let n1 be those skiers who received a placebo and n2 be those who received vitamin C. 95% CI is (.01139, ) If you reverse what n1 and n2 represent, then 95% CI is (-.1869, ).

, E75a Let n1 be those skiers who received a placebo and n2 be those who received vitamin C. 95% CI is (.01139, ) If you reverse what n1 and n2 represent, then 95% CI is (-.1869, ). So must specify n1 and n2.

, E75a I’m 95% confident that if all the skiers in the experiment had been given the placebo and all of the skiers in the experiment had been given vitamin C, then the difference in proportions who get colds would have been in the interval (.01139, ).

, E75a I’m 95% confident that if all the skiers in the experiment had been given the placebo and all of the skiers in the experiment had been given vitamin C, then the difference in proportions who get colds would have been in the interval (.01139, ). 0 is not in the CI, so what does that mean?

, E75a I’m 95% confident that if all the skiers in the experiment had been given the placebo and all of the skiers in the experiment had been given vitamin C, then the difference in proportions who get colds would have been in the interval (.01139, ). Because 0 is not in this CI, we believe vitamin C reduces the proportion of people who get colds.

, E79 Name the test and write the hypotheses in words and symbols.

, E79 This is a one-sided significance test for the difference of two proportions because we are determining if there is significant evidence that aspirin use reduces the rate of heart attacks in this group.

, E79 Ho: p1 = p2 , If all of the men could have been given aspirin, the proportion, p1, who had a heart attack would have been equal to the proportion, p2, of the men who had a heart attack if all could have been given the placebo.

, E79 Ho: p1 = p2 , If all of the men could have been given aspirin, the proportion, p1, who had a heart attack would have been equal to the proportion, p2, of the men who had a heart attack if all could have been given the placebo. Ha: p1 < p2 (Is there significant evidence aspirin use reduces rate of heart attacks?)

, E81 Is this an experiment, a sample survey, or an observational study? Why?

, E81 Is this an experiment, a sample survey, or an observational study? Subjects not randomly selected Treatments not randomly assigned Condition of interest (dementia/free of dementia) already inherent in the subjects

Sample survey, experiment, or observational study?

A randomized clinical trial on Linus Pauling’s claim that vitamin C helps prevent the common cold was carried out in Canada among 818 volunteers, with results reported in The data showed that 335 of the 411 in the placebo group got colds over the winter in which the study was conducted, while 302 of the 407 in the vitamin C group got colds.

In one early study (1956) of the relationship between smoking and mortality, Canadian war pensioners were asked about their smoking habits and then followed to see if they were alive 6 years later. Of the 1067 nonsmokers, 950 were still alive. Of the 402 smokers, 348 were still alive.

Subjects not randomly selected Treatments not randomly assigned

In 2005, as NASA prepared to launch the New Horizons probe that will fly to Pluto on a 9-year mission, a poll asked 1000 randomly selected adult Americans about their attitude toward the U.S space agency. Sixty percent gave NASA a good to excellent rating. At the end of 1999, in a poll of about the same size of randomly selected adult Americans, only 53% gave this high a rating.

Two samples randomly and independently selected

Questions?

Section 8.5 Day 3.

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Section 8.5 Day 3.

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