Based on a random sample of 1080 adults, the mean amount of sleep per night is 8.41 hours. Assuming the population standard deviation for amount of sleep per night is 3.8hours, construct and interpret a 90% confidence interval for the mean amount of sleep per night.
A 90% confidence interval is (nothing,nothing).
(Round to two decimal places as needed.)
Based on a random sample of 1080 adults, the mean amount of sleep per night is...
Based on a random sample of 1140 adults, the mean amount of sleep per night is 8.56 hours. Assuming the population standard deviation for amount of sleep per night is 1.3 hours, construct and interpret a 90% confidence interval for the mean amount of sleep per night. A 9090% confidence interval is (nothing,nothing). (Round to two decimal places as needed.) Interpret the confidence interval.
Based on a random sample of 1040 adults, the mean amount of sleep per night is 8.37 hours. Assuming the population standard deviation for amount of sleep per night is 2.7 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (DD (Round to two decimal places as needed.)
Based on a random sample of 1160 adults, the mean amount of sleep per night is 8.49 hours. Assuming the population standard deviation for amount of sleep per night is 3.6 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (_____,_______). (Round to two decimal places as needed.)
Based on a random sample of 1180 adults, the mean amount of sleep per night is 7.85 hours. Assuming the population standard deviation for amount of sleep per night is 1.4 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (DD Round to two decimal places as needed.) Interpret the confidence interval O A. O B. ° C. 0 D. We are 95% confident that the interval actually...
0 Based on a random sample of 1140 adults, the mean amount of sleep per night is 8.42 hours. Assuming the population standard deviation for amount of sleep per night is 2.9 hours, construct and interpret a 95% confidence interval for the mean amount of sleep per night. A 95% confidence interval is (2.) (Round to two decimal places as needed.) Interpret the confidence interval. O A. We are 95% confident that the interval actually does contain the true value...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.69 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.07 hours, with a standard deviation of 1.54 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...
In a random sample of 12 adults from the United States, the mean waste recycled per person per day was 1.7 pounds and the standard deviation was 0.3 pound. Assuming that the sample is approximately bell-shaped construct a 90% confidence interval estimate for the mean waste recycled per day by all U.S. adults. What is the Confidence Interval: ( nothing , nothing ) (round each interval limit to two decimal places) The local waste management says that the mean waste...
In a random sample of six microwave ovens, the mean repair cost was $65.00 and the standard deviation was $12.50. Assume the population is normally distributed and use a t-distribution to construct a 90% confidence interval for the population mean mu. What is the margin of error of mu? Interpret the results. The 90% confidence interval for the population mean mu is ( nothing, nothing). (Round to two decimal places as needed.) The margin of error is nothing. (Round to...
A. A random sample of 32 different juice drinks has a mean of 98 calories per serving and a standard deviation of 31.5 calories. Construct a 99% confidence interval of the population mean number of calories per serving, and interpret the 99% confidence interval in 1 sentence: B. A random sample of 50 standard hotel rooms in Philadelphia, PA, has a mean nightly cost of $189.99 and a standard deviation of $35.25. Construct a 95% confidence interval of the mean...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.52 hours, with a standard deviation of 2.36 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.46 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...