The high temperatures (in degrees Fahrenheit) of a random sample of 15 small towns are: 98.2 99.7 96.8 96.7 98.8 98.7 98.1 97.7 99.3 97.8 99.5 99.6 96.9 97.1 98.5   Assume high temperatures are normally distributed. Based on this data, find the 99% confidence interval of the mean high temperature of towns. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place). 99% C.I. =

Tina catches a 14-pound bass.  She does not know the population mean or standard deviation.  So she takes a sample of five friends and they say the last bass they caught was 9, 12, 13, 10, and 10 pounds.  Find the t and calculate a 95% (α = .05) confidence interval.

The body temperatures in degrees Fahrenheit of a sample of adults in one small town are: 99.6 98.5 98.2 97.2 97.8 96.6 97.7 98.8 99.8 96.4 96.3 97.1 Assume body temperatures of adults are normally distributed. Based on this data, find the 80% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.80% C.I. =

A statistics instructor randomly selected four bags of oranges, each bag labeled 10 pounds, and weighed the bags. They weighed 10.3, 10.8, 10.4, and 10.6 pounds. Assume that the distribution of weights is Normal. Find a 95% confidence interval for the mean weight of all bags of oranges. Use technology for your calculations. Answer parts a and b below. a. Choose the correct interpretation of the confidence interval below and, if necessary, fill in the answer boxes to complete your choice. A. We are 95% confident the population mean is between and B. There is a 95% chance that all intervals will be between and C. We are 95% confident that the sample mean is between and D. The requirements for constructing a confidence interval are not satisfied. (Type integers or decimals rounded to the nearest thousandth as needed. Use ascending order.)

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 399 drivers and find that 285 claim to always buckle up. Construct a 83% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5]

Find 90% confidence interval

A local radio randomly selects 500 of its listeners. Listeners are offered to enter a competition where the price is a concert ticket. Among these 500 subscribers, 240 accept the promotional offer. The interval (45.76%, 50.23%) is a 95%-confidence interval for what quantity? a. the percentage of listeners in the sample who would accept the invitation to enter the competition b. the percentage of all listeners who would accept the invitation to enter the competition

Researchers studied the mean egg length (in millimeters) for a bird population. After taking a random sample of eggs, they obtained a 95% confidence interval of (45,60). What is the value of the margin of error? Choose the correct answer below. A. 15 mm B. 52.5 mm O c. 7.5 mm O D. 1.96

Find the 99% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).99% C.I. =

A local university administers a comprehensive examination to the candidates for B.S. degrees in Business Administration. Five examinations are selected at random and scored. The scores are shown below. Develop a 95% confidence interval estimate for the mean of the population. Assume the population is normally distributed. Grades 80 90 91 62 77 Select one: O a. 65.39 - 94.61 O b. 68.78 - 91.22 О с. O d. 67.52 - 92.48 69.48 - 90.52