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![Page 1: Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.6 inches. We also have a population of adult women.](https://reader035.fdocuments.us/reader035/viewer/2022062401/5a4d1b0b7f8b9ab05998af83/html5/thumbnails/1.jpg)
![Page 2: Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.6 inches. We also have a population of adult women.](https://reader035.fdocuments.us/reader035/viewer/2022062401/5a4d1b0b7f8b9ab05998af83/html5/thumbnails/2.jpg)
Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.6 inches. We also have a population of adult women with a mean height of 65 inches and standard deviation of 2.3 inches. Assume heights are normally distributed.
Describe the distribution of the difference in heights between males and females (male-female).Normal distribution withx-y =6 inches & x-y =3.471 inches
![Page 3: Suppose we have a population of adult men with a mean height of 71 inches and standard deviation of 2.6 inches. We also have a population of adult women.](https://reader035.fdocuments.us/reader035/viewer/2022062401/5a4d1b0b7f8b9ab05998af83/html5/thumbnails/3.jpg)
7165
Female Male
6
Difference = male - female
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yxyx
22
yxyx
We will be
interested in the
difference of means, so we will use this to
find standard
error.
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The goal of these inference procedures is to compare the responses to two treatmentstwo treatments or to compare the characteristics of two populationstwo populations.
We have INDEPENDENT samples from each treatment or population
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Have two SRS’stwo SRS’s from the populations or two randomly two randomly assignedassigned treatment groups
Samples are independentBoth populations are
normally distributed Have large sample sizes Graph BOTH sets of data
’’ss known/unknown
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Since in real-life, we will NOTNOT know both ’s, we will do t-procedures.
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Option 1: use the smaller of the two values n1 – 1 and n2 – 1 This will produce conservative This will produce conservative results – higher p-values & results – higher p-values & lower confidence.lower confidence.
Option 2: approximation used by technology
2
2
2
21
2
1
1
2
2
2
2
1
2
1
11
11
ns
nns
n
ns
ns
df
Calculator does this
automatically!
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statistic of SD valuecritical statisticCI
21xx *t
2
2
2
1
2
1
ns
ns
Called standard
error
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Used for two populations with the same variance
When you pool, you average the two-sample variances to estimate the common population variance.
DO NOT use on AP Exam!!!!!We do NOT know the variances of the population,
so ALWAYS tell the calculator NO for pooling!
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Two competing headache remedies claim to give fast-acting relief. An experiment was performed to compare the mean lengths of time required for bodily absorption of brand A and brand B. Assume the absorption time is normally distributed. Twelve people were randomly selected and given an oral dosage of brand A. Another 12 were randomly selected and given an equal dosage of brand B. The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow: mean SD n Brand A
20.1 8.7 12 Brand B18.9 7.5 12
Describe the shape & standard error for sampling distribution of the differences in the mean speed of absorption. (answer on next screen)
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Describe the sampling distribution of the differences in the mean speed of absorption.
Find a 95% confidence interval difference in mean lengths of time required for bodily absorption of each brand. (answer on next screen)
Normal distribution with S.E. = 3.316Normal distribution with S.E. = 3.316
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Assumptions:
Have 2 independent SRS from volunteers Given the absorption rate is normally distributed ’s unknown
)085.8,685.5(125.7
127.8080.29.181.20
22
53.21*2
22
1
21
21 dfns
nstxx
We are 95% confident that the true difference in mean lengths of time required for bodily absorption of each brand is between –5.685 minutes and 8.085 minutes.
State assumptions!
Formula & calculations
Conclusion in contextFrom calculator df = 21.53, use t* for df = 21 & 95% confidence
level
Think “Price is Right”!
Closest without going over
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Matched pairs – refer to “mean difference”“mean difference”
Two-Sample – refer to “difference of means”“difference of means”
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In a recent study on biofeedback, it was reported that meditation could alter the alpha & beta waves in the brain thus changing the rate at which the heart beats. This is important for relieving the effects of stress.
Let’s test this!
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H0: 1 - 2 = 0
Ha: 1 - 2 < 0Ha: 1 - 2 > 0Ha: 1 - 2 ≠ 0
H0: 1 = 2
Ha: 1< 2Ha: 1> 2
Ha: 1 ≠ 2
Be sure to define BOTHBOTH 1 and 2!
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statistic of SDparameter - statisticstatisticTest
t
2
2
2
1
2
1
2121
ns
nsxx
Since we usually assume H0 is true,
then this equals 0 – so we can usually
leave it out
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The length of time in minutes for the drugs to reach a specified level in the blood was recorded. The results follow:
mean SD n Brand A 20.1 8.7 12 Brand B 18.9 7.5 12Is there sufficient evidence that these drugs differ in the speed at which they enter the blood stream?
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Assump.: Have 2 independent SRS from volunteers Given the absorption rate is normally distributed ’s unknown
05.53.217210.
361.
125.7
127.8
9.181.2022
2
22
1
21
21
αdfvaluepns
ns
xxt
Since p-value > a, I fail to reject H0. There is not sufficient evidence to suggest that these drugs differ in the speed at which they enter the blood stream.
State assumptions!
Formula & calculations
Conclusion in context
H0: A= B
Ha:A= B
Where A is the true mean absorption time for Brand A & B is the true mean absorption time for Brand B
Hypotheses & define variables!
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Suppose that the sample mean of Brand B is 16.5, then is Brand B faster?
05.53.212896.
085.1
125.7
127.8
5.161.2022
2
22
1
21
21
αdfvaluepns
ns
xxt
No, I would still fail to reject the null hypothesis.
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Two-sample procedures are more more robustrobust than one-sample procedures
BESTBEST to have equal sample sizes! (but not necessary)
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11.2 HOMEWORK11.33-11.35 all11.37 – 11.39, 40, 43, 53, 54, 58, 62,
64Due Friday, March27Post Test Chapters 1 - 11 on Wed.,
April 1st