University of Utah February 28, 2018
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Law of Large Numbers Draw your samples from any population with finite mean µ. Then LLN says
Law of Large Numbers Draw your samples from any population with finite mean µ. Then LLN says Sample mean of observed values gets closer and closer to the mean µ of the population.
Law of Large Numbers Draw your samples from any population with finite mean µ. Then LLN says Sample mean of observed values gets closer and closer to the mean µ of the population. In other words x µ
Central Limit Theorem If any random sample of size n is selected from any population with mean µ and standard deviation is σ, then when n is large the distribution of the sample mean X is approximately normal. That is X N(µ, σ n ) (1)
Confidence Intervals for Normal population All confidence intervals we construct for mean will have a form similar to this: estimate ± margine error Estimate is X, where X is sample mean
Confidence Intervals for Normal population All confidence intervals we construct for mean will have a form similar to this: estimate ± margine error Estimate is X, where X is sample mean Margin error is z σ n where σ is standard deviation of population, n is sample size, and z is found from standard Normal distribution. In other words X z σ n < µ < X + z σ n
Example 1 Consider a population that is distributed N(µ, 20). Find a 90% confidence interval for the population mean when the sample mean of size 9 is 8.
Example 1 Consider a population that is distributed N(µ, 20). Find a 90% confidence interval for the population mean when the sample mean of size 9 is 8. Solution: First, what do we need to find?
Example 1 Consider a population that is distributed N(µ, 20). Find a 90% confidence interval for the population mean when the sample mean of size 9 is 8. Solution: First, what do we need to find? P(X z σ n < µ < X + z σ n ) =.90 What do we know?
Example 1 Consider a population that is distributed N(µ, 20). Find a 90% confidence interval for the population mean when the sample mean of size 9 is 8. Solution: First, what do we need to find? P(X z σ n < µ < X + z σ n ) =.90 What do we know? We know X = 8, n = 9, σ = 20 and from the table z = 1.65.
Example 1 Consider a population that is distributed N(µ, 20). Find a 90% confidence interval for the population mean when the sample mean of size 9 is 8. Solution: First, what do we need to find? P(X z σ n < µ < X + z σ n ) =.90 What do we know? We know X = 8, n = 9, σ = 20 and from the table z = 1.65. So 90% Confidence Interval for µ is ( 8 1.65 20, 8 + 1.65 20 ) 9 9
Example 2 (15.28 (kind of)) To estimate the mean score of µ of those who took the Medical College Admission Test on your campus, you will obtain the scores of an SRS of students. From published information, you know that the scores are approximately Normal with standard deviation about 6.4. You want your sample mean X to estimate µ with an error of no more than 1 point in either direction. a) What standard deviation must X have so that 99.7% of all samples give an X within 1 point of µ? (Use: 68-95-99.7 rule) b) How large SRS do you need in order to reduce the standard deviation of X to the value you found in part (a)?
Solution of Example 2 a)we know that X N(µ, 6.4/ n). We know by 68-95-99.7 rule that 99.7% of all observations fall within 3 standard devations, so we want that 3σ/ n = 1 i.e. P(X 3 σ n < µ < X + 3 σ n ) =.997 So the standard deviation of x must therefore be 1/3.
Solution of Example 2 a)we know that X N(µ, 6.4/ n). We know by 68-95-99.7 rule that 99.7% of all observations fall within 3 standard devations, so we want that 3σ/ n = 1 i.e. P(X 3 σ n < µ < X + 3 σ n ) =.997 So the standard deviation of x must therefore be 1/3. b) Since σ/ n = 1/3, so 6.4 = n/3 that is n = 19.2 so n=368.64. We need to take n as a whole number so n = 369.
Example 3 or 16.8 An NHANES reports gives the data for 654 women aged 20 to 29 years. The mean of BMI of these 654 women was x = 26.8. We treated these data as an SRS from a normally distributed population with standard deviation σ = 7.5
Example 3 or 16.8 An NHANES reports gives the data for 654 women aged 20 to 29 years. The mean of BMI of these 654 women was x = 26.8. We treated these data as an SRS from a normally distributed population with standard deviation σ = 7.5 a)give three confidence intervals for the mean BMI µ in this population, using 90%, 95%, and 99% confidence b) What are the margins of error for 90%, 95%, and 99%?
Example 3 or 16.8 An NHANES reports gives the data for 654 women aged 20 to 29 years. The mean of BMI of these 654 women was x = 26.8. We treated these data as an SRS from a normally distributed population with standard deviation σ = 7.5 a)give three confidence intervals for the mean BMI µ in this population, using 90%, 95%, and 99% confidence b) What are the margins of error for 90%, 95%, and 99%? Solution a) We know x = 26.8 and σ = 7.5 and n = 654 for all three cases. So z = 1.645 for 90% confidence level and z = 1.960 for 95% confidence level and z = 2.576 for 99% confidence level.
Example 3 solution 90% Confidence Interval is 26.8 ± 1.645(7.5/ 654), that is 26.32 to 27.28. Margin of error is 0.4824.
Example 3 solution 90% Confidence Interval is 26.8 ± 1.645(7.5/ 654), that is 26.32 to 27.28. Margin of error is 0.4824. 95% Confidence Interval is 26.8 ± 1.960(7.5/ 654), that is 26.23 to 27.37. Margin of error is 0.5748
Example 3 solution 90% Confidence Interval is 26.8 ± 1.645(7.5/ 654), that is 26.32 to 27.28. Margin of error is 0.4824. 95% Confidence Interval is 26.8 ± 1.960(7.5/ 654), that is 26.23 to 27.37. Margin of error is 0.5748 99% Confidence Interval is 26.8 ± 2.576(7.5/ 654), that is 26.04 to 27.56. Margin of error is 0.7555.
Example 4 or 16.19(kind of) A class survey in a large class for first-year students asked, About how many minutes do you study on a typical weeknight?. The mean response of the 463 students was x = 118 minutes. Suppose that we know that the study time follows a Normal distribution with standard deviation σ = 65 minutes in the population of all first-year students at the university. a) Use the survey result to give a 99% confidence interval for the mean of the study time of all first-year students. b) Assume there were 464 responses and one student claimed to study 60000 minutes per night. But then x = 247 minutes. Now what is the 99% confidence interval for the mean?
Solution for Example 4 Solution:a) Margin of error is 2.576 65 463. So Confidence interval is 118 ± 2.576 65 463 which is from 110.22 minutes to 125.78 minutes.
Solution for Example 4 Solution:a) Margin of error is 2.576 65 463. So Confidence interval is 118 ± 2.576 65 463 which is from 110.22 minutes to 125.78 minutes. b) Margin of error is 2.576 65 464. So Confidence interval is 247 ± 2.576 65 464 which is from 239.23 minutes to 254.77 minutes.