Example 5 A researcher wants to estimate the variance for the amount of comp

# Example 5 a researcher wants to estimate the variance

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Example 5 : A researcher wants to estimate the variance for the amount of comp time accumulated per week in the aerospace industry. He randomly samples 18 managers and obtains a sample mean of 13.5556 hours with a sample standard deviation of 7.8006 hours. Use this data to construct a 99% confidence interval for the population variance of comp time worked per week by managers in the aerospace industry. From the text, we have s = 7.8006, n = 18, n-1 = 17. With a 99% confidence level: α = 1 - .99 = .01 α/2 = .01/2 = .005 , 1 - α/2 = 1 - .005 = .995 Using Excel, we get the following: 2 = 2 = 28.9609 2 181.5697 P (28.9609 2 181.5697) = 99% Interpretation: We’re 99% confident that the population variance of comp time hours per week is between 28.9609 and 181.5697 hours. Note: In this problem we were given the sample standard deviation (s) which had to be squared in the numerator of the equation on both sides to get the sample variance. Left (Lower Limit) Right (Upper Limit) X 2 α/2, n-1 Excel Syntax X 2 1- α/2, n-1 Excel Syntax X 2 .005,17 35.7185 =CHIINV(0.005,17) X 2 .995, 17 5.6972 =CHIINV(0.995,17)
Module 12: Confidence Intervals Part 2 B. Confidence intervals to estimate the population variance. Example 6 : The business council of Greece is interested in estimating the variance of the hourly wages of production workers in Greece. They take a random sample of 25 production workers and obtain a sample mean wage of \$9.12 with a sample standard deviation of \$1.12. Use this information to construct a 95% confidence interval for the variance of the hourly wages of production workers in Greece. From the text, we have s = 1.12, n = 25, n-1 = 24. With a 95% confidence level: α = 1 - .95 = .05 α/2 = .05/2 = .025 , 1 - α/2 = 1 - .025 = .975 Using Excel, we get the following: 2 = 2 = .7648 2 2.4276 P (.7648 2 2.4276) = 95% Interpretation: We’re 95% confident that the population variance of hourly wages for production workers in Greece is between \$.7648 and \$2.4276. Note: In example 6, as in example 5 we were given the sample standard deviation (s) which had to be squared in the numerator of the equation on both sides to get the sample variance. Left (Lower Limit) Right (Upper Limit) X 2 α/2, n-1 Excel Syntax X 2 1- α/2, n-1 Excel Syntax X 2 .025,24 39.3641 =CHIINV(0.025,24) X 2 .975, 24 12.4012 =CHIINV(0.975,24)
Module 12: Confidence Intervals Part 2 Module 12 Exercises Exercise 1 : A random sample of 85 fast food fans is obtained. 40 of those sampled preferred McDonald's over Burger King. Use this information to construct a 99% confidence for the population proportion of fast-food fans who prefer McDonalds over Burger King. Exercise 2 : A random sample of 24 pieces of standard chalk from a factory which produces chalk resulted in a sample mean diameter of .375” with a sample variance of .001 inches. Use this data to construct a 99% confidence interval for the population variance of the chalk’s diameter. Exercise 3 : A random sample of 15 wine refrigerators of a certain model resulted in a sample mean temperature of 42 degrees with a sample standard deviation of 1.4142 degrees. Use this data to construct a 98% confidence interval for the population variance of this wine refrigerator model’s temperature variance.
Module 12: Confidence Intervals Part 2 Solutions to Module 12 Exercises Exercise 1 : A random sample of 850 fast food fans is obtained. 400 of those sampled preferred McDonald's over Burger King.

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