ESTIMATION AND CONFIDENCE INTERVALS: Single Mean: Q1. An electrical firm manufacturing light bulbs that have a length of life that is normally distributed with a standard deviation of 30 hours. A sample of 50 bulbs were selected randomly and found to have an average of 750 hours. Let m be the population mean of life lengths of all bulbs manufactured by this firm. (1) Find a point estimate for m. (2) Construct a 94% confidence interval for m. Solution: Q2. Suppose that we are interested in making some statistical inferences about the mean, M, of a normal population with standard deviation S=2.0. Suppose that a random sample of size n=49 from this population gave a sample mean X=4.5. (1) The distribution of X is (A) N(0,1) (B) t(4,8) (2) A good point estimate of m is (A) 4.50 (B) 2.00 (3) The standard error of X is (A) 0.0816 (B) 2.0 (C) N(m, 0.2857) (D) N(m, 2.0) (E) N(m, 0.3333) (C) 2.50 (D) 7.00 (E) 1.125 (C) 0.0408 (D) 0.5714 (E) 0.2857 (D) (3.94,5.06) (E) (3.04,5.96) (4) A 95% confidence interval for m is (A) (3.44,5.56) (B) (3.34,5.66) (C) (3.54,5.46) (5) If the upper confidence limit of a confidence interval is 5.2, then the lower confidence limit is (A) 3.6 (B) 3.8 (C) 4.0 (B) 95.74% (C) 97.74% (D) 3.5 (6) The confidence level of the confidence interval (3.88, 5.12) is (A) 90.74% (D) 94.74% (E) 4.1 (E) 92.74% (7) If we use X to estimate m, then we are 95% confident that our estimation error will not exceed (B) E=0.59 (C) e=0.58 (D) e=0.56 (E) e=0.51 (A) e=0.50 (8) If we want to be 95% confident that the estimation error will not exceed e=0.1 when we use X to estimate m, then the sample size n must be equal to (A) 1529 Solution: (B) 1531 (C) 1537 (D) 1534 (E) 1530