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Applied Calculus IIApplied Calculus II

Confidence TestsConfidence Tests

Slides subject to changeSlides subject to change

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Central Limit Theorem Draw a simple random sample of size N from any

population with mean μ and standard deviation σ.

When N is large, the distribution of the sample mean Xmean is approximately normal.

Or, in other words, Xmean distribution is approximately that of a normal distribution, with σσmeanmean = = σσ/sqrt /sqrt N.N.

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Example 1 Normal Population Random samples of a Normal population, N = 10. Population characteristics are μ = 5, σ = 2.

5 6 3 4 4

3 4 3 5 7

6 4 6 4 5

5 7 6 3 5

6 5 7 4 6

2 5 3 2 7

3 6 9 2 7

7 4 4 5 3

7 3 3 3 6

8 5 5 8 3

5.2 4.9 4.9 4.0 5.3

11 22 33 44 55

MeanMean

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Example 1 Normal Population Sample size Sample size NN = 10. The central limit theorem says = 10. The central limit theorem says

that the means of each sample of size that the means of each sample of size NN will have a will have a normal distribution with standard deviation of normal distribution with standard deviation of σσ/sqrt /sqrt N. N.

σσ is the standard deviation of the population. is the standard deviation of the population.

The standard deviation of the means is The standard deviation of the means is σσmeanmean = =

σσ/sqrt /sqrt NN = 2/sqrt 10 = 0.63 = 2/sqrt 10 = 0.63

In the five samples we took, the means were 5.2, 4.9, In the five samples we took, the means were 5.2, 4.9, 4.9, 4.0, 5.3. 4.9, 4.0, 5.3.

Expect the “mean of the means” to be Expect the “mean of the means” to be μ (= 5.0).5.0).

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Example 1 Normal Population σσmeanmean = = 2/sqrt 10 = 0.632/sqrt 10 = 0.63

Expect approximately 68% of the mean Expect approximately 68% of the mean measurements withinmeasurements within

XXmeanmean – – σσmeanmean and and XXmeanmean + + σσmeanmean

Or, between 4.37 and 5.63.

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Samples of Normal Population

5.05.0

6.06.0

4.04.0

++σσmeanmean

−−σσmeanmean

μμ

Mean of each sampleMean of each sample

11 22 33 44 55

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Statistical Inference Statistical inference provides methods for

drawing conclusions about a population from sample data.

One method: Normal Deviate (or z−) Test.

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Example: Normal Deviate TestExample: Normal Deviate Test A population of critters has mean body mass A population of critters has mean body mass μμ

= 125 g and standard deviation = 125 g and standard deviation σσ = 30 g. = 30 g. You hypothesize that a diet treatment increases You hypothesize that a diet treatment increases

body mass.body mass. You apply the treatment to 10 critters and find You apply the treatment to 10 critters and find

mean body mass of the 10 critters = 142 g.mean body mass of the 10 critters = 142 g. Where does this mean lies within the normal Where does this mean lies within the normal

curve of our reference population?curve of our reference population? Confidence of 95% if the mean lies above the top Confidence of 95% if the mean lies above the top

5% of all sample means.5% of all sample means.

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Normal Deviate or z−Normal Deviate or z−TestTest Standard deviation of the sample means is Standard deviation of the sample means is σσxx= =

30/sqrt 10 = 9.4930/sqrt 10 = 9.49

zz = (X = (Xavgavg − − μμ)/)/ σ σxx = (142 − 125)/ 9.49 = 1.79 = (142 − 125)/ 9.49 = 1.79

Use Table A, “C” column. Area = 0.0367.Use Table A, “C” column. Area = 0.0367.

The mean of this sample exceeds our confidence The mean of this sample exceeds our confidence level − is in the 96.33 percentile. The diet treatment level − is in the 96.33 percentile. The diet treatment doesdoes increase body mass. increase body mass.

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ExampleExample A population of students score µ = 75 points on A population of students score µ = 75 points on

a standard exam with σ = 20 points.a standard exam with σ = 20 points. You hypothesize that taking the exam early in You hypothesize that taking the exam early in

the day affects performance on the exam.the day affects performance on the exam. You apply the treatment to 15 students and find You apply the treatment to 15 students and find

mean score of sample = 66 points.mean score of sample = 66 points. We have 95% confidence in our hypothesis if We have 95% confidence in our hypothesis if

the sample mean is below 5the sample mean is below 5thth percentile or percentile or above the 95above the 95thth percentile. percentile.

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Normal Deviate or z−Normal Deviate or z−TestTest Standard deviation of sample means is Standard deviation of sample means is σσxx= =

20/sqrt 15 = 5.1620/sqrt 15 = 5.16 zz = ( = (XXavgavg − − μμ)/)/ σ σxx = (66 − 75)/ 5.16 = −1.74 = (66 − 75)/ 5.16 = −1.74

Table A area = 0.0409Table A area = 0.0409

This is only the percentile. The mean of this This is only the percentile. The mean of this sample exceeds our confidence level − is in the sample exceeds our confidence level − is in the 4.09 percentile. Taking the exam early in the day 4.09 percentile. Taking the exam early in the day doesdoes affect performance on the exam affect performance on the exam

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Exercise Analyze the last 4 digits in each social security number.

Find the average of each sample, N = 4. The population is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. For this population μ = 4.50, σ = 2.87. The mean of the sample means should be close to 4.50.

True? The standard deviation of the sample means is σσmeanmean =

σ/sqrt 4 = 1.43. Approximately 68% of the sample means 3.07 ≤ xxmeanmean

≤ 5.93. True? If not, would finding more social security numbers help?