A computer company makes a new computer chip with an advertised processing power of 2.4GHZ and a standard deviation of 0.1GHZ. Computer chip processing power is normally distributed. The company tests 39 chips that have an average of 2.5 GHz with a standard deviation of 0.1GHZ. Find a 95% confidence interval for the average processing power of all computer chips produced.
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- A large multi-branch bank is affiliated with both the MasterCard and Visa credit cards. For a random sample of 100 MasterCard cardholders, it is observed that the average month-end account balance is R682.58 with a sample standard deviation of R307.05. For a random sample of 100 Visa cardholders, the average month-end account balance is R550.55 with a sample standard deviation of R265.29. The 90% confidence interval for the mean difference between the month-end account balances for the two types of credit card holders (rounded to three decimals) is:A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4, find: The t-value for a 95% confidence interval is:A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.64 hours, with a standard deviation of 2.43 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.09 hours, with a standard deviation of 1.63 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2.
- A study was conducted to estimate hospital costs for car accident victims who woreseat belts. Suppose that 46 randomly selected cases had an average of $15150and a standard deviation of $8500. If the average hospital costs for those that didnot wear seatbelts was $19000, construct a 90% confidence interval to see if theaverage costs are significantly different.A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4. The 95% confidence interval between the difference in rates between the two departments (round off to three decimals) is:A random sample of 15adultRainbow trout in a fish hatchery showed the average weight was 3.5pounds and a standard deviation of 0.8 pounds. Constructa 90% confidence interval for the average weight of the trout at the hatchery.
- A study of two kinds of photocopying equipmentshows that 61 failures of the first kind of equipment took on the average 80.7 minutes to repair with a stan-dard deviation of 19.4 minutes, whereas 61 failures of the second kind of equipment took on the average 88.1minutes to repair with a standard deviation of 18.8 minutes. Find a 99% confidence interval for the dif-ference between the true average amounts of time it takes to repair failures of the two kinds of photocopyingequipment.A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.61 hours, with a standard deviation of 2.39 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.43 hours, with a standard deviation of 1.61 hours. Construct and interpret a 90%confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2. Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. -The 90% confidence interval for μ1−μ2is the range from____ hours to____hours. (Round to two decimal places as needed.) -What is the interpretation of this confidence interval? A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference…A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.135.13 hours, with a standard deviation of 2.462.46 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.254.25 hours, with a standard deviation of 1.711.71 hours. Construct and interpret a 9090% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2. Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18. The 90%confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from hours to hours. (Round to two decimal places as needed.)
- A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.03 hours, with a standard deviation of 2.46 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.48 hours, with a standard deviation of 1.52 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2. Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. The 90%confidence interval for μ1−μ2 is the range from nothinghours to nothing hours. Also interpret the confidence interval. (Round to two decimal places as needed.)A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.995.99 hours, with a standard deviation of 2.292.29 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.414.41 hours, with a standard deviation of 1.661.66 hours. Construct and interpret a 9090% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2. Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18. The 9090% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from nothing hours to nothing hours. (Round to two decimal places as needed.)A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.015.01 hours, with a standard deviation of 2.492.49 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.044.04 hours, with a standard deviation of 1.891.89 hours. Construct and interpret a 9595% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2. Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from hours to hours.