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Question 1 Question 2 Question 3 Question 4 Math 144 tutorial 7 12 April, 2010 Question 5 Question 6 Question 1 Question 2 1. Let us define S02 = Question 3 1 n Pn i=1 (Xi Question 4 − X̄)2 . Show that E(S02 ) = n−1 2 σ . n Question 5 Question 6 Question 1 Question 2 1. Let us define S02 = Question 3 1 n Pn i=1 (Xi Question 4 − X̄)2 . Show that E(S02 ) = n−1 2 σ . n Question 5 Question 6 Question 1 Question 2 1. Let us define S02 = Question 3 1 n Pn i=1 (Xi Question 4 Question 5 − X̄)2 . Show that E(S02 ) = n−1 2 σ . n This result also indicate that S02 is a biased estimator of the σ 2 . Question 6 Question 1 Question 2 Question 3 Question 4 Question 5 2. If X is a binomial random variable, show that (a). p̂ = X/n√is an unbiased estimator of p; n/2 √ is a biased estimator of p; (b). p0 = X+ n+ n (c). Show that the estimator p0 becomes unbiased as n → ∞. Question 6 Question 1 Question 2 Question 3 Question 4 Question 5 2. If X is a binomial random variable, show that (a). p̂ = X/n√is an unbiased estimator of p; n/2 √ is a biased estimator of p; (b). p0 = X+ n+ n (c). Show that the estimator p0 becomes unbiased as n → ∞. Solution: (a). Since E[p̂] = E[X]/n = np/n = p, this shows that p̂ is an unbiased estimator of p. (b). E[p0 ] = √ E[X]+ n/2 √ n+ n = √ np+ n/2 √ n+ n 6= p. √ np + n/2 √ (c). lim n∞ n+ n p 1 √ √ = lim + lim n∞ 1 + n/n n∞ 2(1 + n) = p. Question 6 Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 3. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours. (a). Find a 96% confidence interval for population mean of all bulbs produced by this firm; (b). How large a sample is needed if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean? Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 3. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours. (a). Find a 96% confidence interval for population mean of all bulbs produced by this firm; (b). How large a sample is needed if we wish to be 96% confident that our sample mean will be within 10 hours of the true mean? Solution: The sample mean x̄ is 780, and the standard deviation of population is 40, using the Normal table, we can get z0.02 = 2.05. Hence the 96% confidence interval for µ is σ σ [x̄ − z0.02 × ( √ ), x̄ + z0.02 × ( √ )] = [765.029, 794.971]. 30 30 (b). We can be 96% confident that this error will less than zα/2 × ( √sn ), i.e, n = ( z0.02e×40 )2 = 67.24. Therefore, a sample with size 68 is needed if we wish be 96% confident that our sample mean will be within 10 hours of the true mean. Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 4. Many cardiac patients wear implanted pacemakers to control their hear-beat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 and an approximate normal distribution, (a). Find a 95% confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of 0.310 inch. (b). How large a sample is needed if we wish to be 95% confident that our sample mean will be within 0.0005 inch of true mean? Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 4. Many cardiac patients wear implanted pacemakers to control their hear-beat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 and an approximate normal distribution, (a). Find a 95% confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of 0.310 inch. (b). How large a sample is needed if we wish to be 95% confident that our sample mean will be within 0.0005 inch of true mean? Solution (a). From the problem, we obtain n = 75, x̄ = 0.310, σ = 0.0015, hence the 95% confidence interval for the mean of all modules is: 0.0015 0.0015 [0.310−z0.025 ×( √ ), 0.310+z0.025 ×( √ )] = [0.309661, 0.310339]. 75 75 ×0.0015 2 (b). n = ( z0.0250.0005 ) = 34.5744. That is, a sample with size 35 is needed if we wish to be 95% confident that our sample mean will be within 0.0005 inch of true mean. Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 5.The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a). Construct a 98% confidence interval for the mean height of all college students. (b). What can we assert with 98% confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters? Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 5.The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a). Construct a 98% confidence interval for the mean height of all college students. (b). What can we assert with 98% confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters? Solution: Because standard deviation of the population is unknown, so we should use the t distribution: (a). Because s = 6.9, x̄ = 174.5, n = 50, so 98% confidence interval for the mean height of all college students is: [174.5.t0.01,49 × ( √6.9 ), 174.5 + t0.01,49 ( √6.9 )] = [173.229, 175.771]. 50 50 (b). We can be 98% confident that this error will less than tα/2 ( √sn ), i.e, 6.9 e ≤ t0.01,49 × ( √ ) = 1.27148. 50 Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 6. A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a). Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b). What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year? Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 6. A random sample of 100 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,500 kilometers per year with a standard deviation of 3900 kilometers. Assume the distribution of measurements to be approximately normal. (a). Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b). What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year? Solution: (a). We use the t distribution, x̄ = 23500, s = 3900, n = 100, then the 99% confidence interval for population mean µ is: 3900 3900 ), 23500+t0.005,99 ×( √ )] = [22479.4, 24520.6]. [23500−t0.005,99 ×( √ 100 100 3900 (b). e ≤ t0.005,99 × ( √ ) = 1020.63. 100