1. 1 Applied Calculus II Confidence Tests Slides subject to change

2. 2 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 X mean is approximately normal. σ mean = σ/sqrt N. Or, in other words, X mean distribution is approximately that of a normal distribution, with σ mean = σ/sqrt N.

3. 3 Example 1 Normal Population Random samples of a Normal population, N = 10. Population characteristics are μ = 5, σ = 2. 56344 34357 64645 57635 65746 25327 36927 74453 73336 85583 5.24.9 4.05.3 12345Mean

4. 4 Example 1 Normal Population Sample size N = 10. The central limit theorem says that the means of each sample of size N will have a normal distribution with standard deviation of σ/sqrt N. Sample size N = 10. The central limit theorem says that the means of each sample of size N will have a normal distribution with standard deviation of σ/sqrt N. σ is the standard deviation of the population. σ is the standard deviation of the population. The standard deviation of the means is σ mean = σ/sqrt N = 2/sqrt 10 = 0.63 The standard deviation of the means is σ mean = σ/sqrt N = 2/sqrt 10 = 0.63 In the five samples we took, the means were 5.2, 4.9, 4.9, 4.0, 5.3. In the five samples we took, the means were 5.2, 4.9, 4.9, 4.0, 5.3. Expect the “mean of the means” to be 5.0). Expect the “mean of the means” to be μ (= 5.0).

5. 5 Example 1 Normal Population σ mean = 2/sqrt 10 = 0.63 σ mean = 2/sqrt 10 = 0.63 Expect approximately 68% of the mean measurements within Expect approximately 68% of the mean measurements within X mean – σ mean and X mean + σ mean X mean – σ mean and X mean + σ mean Or, between 4.37 and 5.63.

6. 6 Samples of Normal Population 5.0 6.0 4.0 +σ mean −σ mean μ Mean of each sample 12345

7. 7 Statistical Inference Statistical inference provides methods for drawing conclusions about a population from sample data. One method: Normal Deviate (or z−) Test.

8. 8 Example: Normal Deviate Test A population of critters has mean body mass μ = 125 g and standard deviation σ = 30 g. A population of critters has mean body mass μ = 125 g and standard deviation σ = 30 g. You hypothesize that a diet treatment increases body mass. You hypothesize that a diet treatment increases body mass. You apply the treatment to 10 critters and find mean body mass of the 10 critters = 142 g. You apply the treatment to 10 critters and find mean body mass of the 10 critters = 142 g. Where does this mean lies within the normal curve of our reference population? Where does this mean lies within the normal curve of our reference population? Confidence of 95% if the mean lies above the top 5% of all sample means. Confidence of 95% if the mean lies above the top 5% of all sample means.

9. 9 Normal Deviate or z−Test Standard deviation of the sample means is σ x = 30/sqrt 10 = 9.49 Standard deviation of the sample means is σ x = 30/sqrt 10 = 9.49 z = (X avg − μ)/ σ x = (142 − 125)/ 9.49 = 1.79 z = (X avg − μ)/ σ x = (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 level − is in the 96.33 percentile. The diet treatment does increase body mass. The mean of this sample exceeds our confidence level − is in the 96.33 percentile. The diet treatment does increase body mass.

10. 10 Example A population of students score µ = 75 points on a standard exam with σ = 20 points. A population of students score µ = 75 points on a standard exam with σ = 20 points. You hypothesize that taking the exam early in the day affects performance on the exam. You hypothesize that taking the exam early in the day affects performance on the exam. You apply the treatment to 15 students and find mean score of sample = 66 points. You apply the treatment to 15 students and find mean score of sample = 66 points. We have 95% confidence in our hypothesis if the sample mean is below 5 th percentile or above the 95 th percentile. We have 95% confidence in our hypothesis if the sample mean is below 5 th percentile or above the 95 th percentile.

11. 11 Normal Deviate or z−Test Standard deviation of sample means is σ x = 20/sqrt 15 = 5.16 Standard deviation of sample means is σ x = 20/sqrt 15 = 5.16 z = (X avg − μ)/ σ x = (66 − 75)/ 5.16 = −1.74 z = (X avg − μ)/ σ x = (66 − 75)/ 5.16 = −1.74 Table A area = 0.0409 Table A area = 0.0409 This is only the percentile. The mean of this sample exceeds our confidence level − is in the 4.09 percentile. Taking the exam early in the day does affect performance on the exam This is only the percentile. The mean of this sample exceeds our confidence level − is in the 4.09 percentile. Taking the exam early in the day does affect performance on the exam

12. 12 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? σ mean The standard deviation of the sample means is σ mean = σ/sqrt 4 = 1.43. x mean Approximately 68% of the sample means 3.07 ≤ x mean ≤ 5.93. True? If not, would finding more social security numbers help?