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1 Example: A SRS of 1000 high school students gains an average of x̄ = 22 points in their second attempt at the SAT mathematical exam. The change in score has a normal distribution with standard deviation σ = 50. (1) Find the 95% CI for µ. (2) Find the margin of error for 99% confidence. Solution: (1) The 95% CI for µ is σ 50 σ 50 [x̄ − z ∗ √ , x̄ + z ∗ √ ] = [22 − 1.96 √ , 22 + 1.96 √ ] n n 1000 1000 = [18.9, 25.1] (2) The the margin of error for 99% confidence is σ 50 = 4.073. z ∗ √ = 2.576 √ n 1000 Example: Acid rain, caused by the reaction of certain pollution with rainwater, appear to be a growing problem in the northeastern United States. Pure rain falling through clean air registers a pH value of 5.7 (pH is a measure of acidity:0 is acid; 14 is alkaline). We knew that water pH standard deviation is 0.5. Suppose water SRS from 15 rainfalls are analyzed for pH, and X̄ = 3.7. (a) Find a 99% CI for the mean pH in rainfall. (b) What assumption must be made for the confidence interval to be valid? 2 Solution (a) Since t∗ = 2.57, we have σ σ [x̄ − z ∗ √ , x̄ + z ∗ √ ] n n 0.5 0.5 = [3.7 − 2.57 √ 3.7 + 2.57 √ ] 15 15 = [3.368, 4.317]. (b) We must assume that water pH has a normal distribution. Choosing the sample size √ We know that σ/ n → 0 as n → ∞. Thus the margin error of any CI tends to zero as n → ∞. Thus, the bigger the sample size, the closer the CI to the estimated parameter. However, the bigger the sample size, the more money we need to spend. To save money, time, manpower, and so on, and same time to attend our required accuracy, we can find suitable sample size for us. Sample size for desired margin of error: The confidence interval for the mean of a normal population will have a specified margin of error m when the sample size is µ ∗ ¶2 z σ . n= m If we require that the margin of error within m, then, we have to choose n such that µ ∗ ¶2 z σ n≥ . m 3 Example: Some high school students will try their second attempt at the SAT mathematical exam. Suppose the change in score has a normal distribution with standard deviation σ = 50. How large a SRS of high school students would be needed to estimate the mean change µ in SAT score to within ±2 points with 95% confidence? Solution: According to the formula, we have µ ∗ ¶2 µ ¶2 z σ (1.96)(50) n= = = 2401. m 2 Example: We are concerned with the daily adult intake of selenium in a region of the United States. Suppose that the daily intakes have a normal distribution with standard deviation of 24.3 micrograms and you wish to estimate the mean daily intake of this region correct to within 5 micrograms, with probability equal to 0.9.How large should n be? Solution: According to the question, we have σ 24.3 1.645 √ = 1.645 √ ≤ 5. n n So 7.9947 ≤ √ n or n ≥ 63.9152.