STAT 400 Discussion 11Spring 2015 1. A random sample of size n = 9 from a normal distribution is obtained: 4.4 3.7 5.1 4.3 4.7 3.7 3.5 4.6 4.7 Recall, from last discussion: x = 4.3 and s = 0.55. a) Construct a 95% (two-sided) confidence interval for the overall standard deviation. b) Construct a 90% one-sided confidence interval for that provides an upper bound for . c) Construct a 95% one-sided confidence interval for that provides a lower bound for . 2. An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating cost of 26.33 cents per mile and the sample standard deviation of 2.80 cents per mile. (Assume that operating costs are approximately normally distributed.) a) Construct a 95% confidence interval for the mean operating cost. b) Construct a 90% confidence interval for the variance of the operating costs. 3. Suppose the time spent on a particular STAT 400 homework follows a normal distribution with an overall standard deviation of 28 minutes and an unknown mean. a) Suppose a random sample of 49 students is obtained. Find the probability that the average time spent on the homework for students in the sample is within 5 minutes of the overall mean. b) A sample of 49 students has a sample mean of 234 minutes spent on the homework. Construct a 90% confidence interval for the overall mean time spent on the homework. c) What is the minimum sample size required if we want to estimate the overall mean time spent on the homework to within 5 minutes with 90% confidence?4. An economist states that 10% of Springfield's labor force is unemployed. A random sample of 400 people in the labor force is obtained, of whom 28 are unemployed. a) Construct a 95% confidence interval for the unemployment rate in Springfield. b) What is the minimum sample size required in order to estimate the unemployment rate in Springfield to within 2% with 95% confidence? (Use the economist’s guess.) c) What is the minimum sample size required in order to estimate the unemployment rate in Springfield to within 2% with 95% confidence? (Assume no information is available.) 5. Let the joint probability density function for ( X , Y ) be f ( x, y ) = 128 x 2 e – 4 y, 0 < x < y < , zero otherwise. a) Set up the integral(s) for the probability P ( X Y < 0.64 ). b) Find the marginal probability density function of X, f X ( x ). Include its support. c) Find the marginal probability density function of Y, f Y ( y ). Include its support.1. a) Construct a 95% (two-sided) confidence interval for the overall standard deviation. Confidence Interval for 2 : nn 112222122ss,. = 0.05. 2 = 0.025. 12 = 0.975. number of degrees of freedom = n 1 = 9 1 = 8. 22 = 17.54. 122 = 2.180. 180.23025.019,54.173025.019 ( 0.13797 ; 1.11009 ) Confidence Interval for : 1.11009 , 0.13797 = ( 0.3714 ; 1.0536 ) b) Construct a 90% one-sided confidence interval for that provides an upper bound for . 21 2 αχs 1 , 0n 290.0 21 χχα = 3.490. 490.33025.0 19 ,0 ( 0 , 0.8327 ) c) Construct a 95% one-sided confidence interval for that provides a lower bound for . , 1 2 2 αχs n 205.0 2 χχα = 15.51. ,51.153025.0 19 ( 0.395 , ∞ )2. An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating cost of 26.33 cents per mile and the sample standard deviation of 2.80 cents per mile. (Assume that operating costs are approximately normally distributed.) X = 26.33, s = 2.80, n = 16. a) Construct a 95% confidence interval for the mean operating cost. n – 1 = 15 degrees of freedom t 0.025 = 2.131 n st2X 162.80 131.233.26 26.33 1.49 ( 24.84 , 27.82 ) b) Construct a 90% confidence interval for the variance of the operating costs. n – 1 = 15 degrees of freedom 20.05 = 25.00, 20.95 = 7.261. nn 112222122ss,. 261.780.2116,00.2580.211622 ( 4.704 , 16.196 )3. Suppose the time spent on a particular STAT 400 homework follows a normal distribution with an overall standard deviation of 28 minutes and an unknown mean. a) Suppose a random sample of 49 students is obtained. Find the probability that the average time spent on the homework for students in the sample is within 5 minutes of the overall mean. P ( 5 X + 5 ) = ? n = 49 large (plus the distribution we sample from is normal). .nZX P ( 5 X + 5 ) = 4928)(Z4928)(P μ5μμ5μ = P ( 1.25 Z 1.25 ) = 0.8944 – 0.1056 = 0.7888. b) A sample of 49 students has a sample mean of 234 minutes spent on the homework. Construct a 90% confidence interval for the overall mean time spent on the homework. is known. The confidence interval : n2zX. = 0.10 2 = 0.05. 2z = 1.645. 4928645.1234 234 6.58 ( 227.42 ; 240.58 ) c) What is the minimum sample size required if we want to estimate the overall mean time spent on the homework to within 5 minutes with 90% confidence? 222 51.645 α 82εσzn = 84.86. Round up. n = 85.4. An economist states that 10% of Springfield's labor force is unemployed. A random sample of 400 people in the labor force is obtained, of whom 28 are unemployed. n = 400, y = 28, 40028pˆny= 0.07. a) Construct a 95% confidence interval for the unemployment rate in Springfield. The confidence interval : n )pˆ1(pˆzpˆ2. 95% confidence level = 0.05 2 = 0.025. 2z = 1.960. 400)930)(070( 9601 070.... 0.07 0.025 ( 0.045 ,
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