A random sample of 89 observation produced a sample mean 25.8 and given population standard deviation 2.6.  (a) find a 95% confidence interval for μ. Round to 2 decimal places.

A random sample of 16 men was selected from the population of ISU students. The sample's average BMI value was 22.5 with a standard deviation of 4.8. Create a 99% confidence interval for the mean of the population of male student. I DFocus W

In a random sample of twelvetwelve ​people, the mean driving distance to work was 24.624.6 miles and the standard deviation was 4.54.5 miles. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 9999​% confidence interval for the population mean muμ. Interpret the results.

Assume that a sample is used to estimate a population mean μμ. Find the 95% confidence interval for a sample of size 63 with a mean of 49.8 and a standard deviation of 5.5. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.

A sample of n=17n=17 data values randomly selected from a normally distributed population has standard deviation s=6.8s=6.8. Construct a 95% confidence interval for the population standard deviation. Round your endpoints to one decimal place.

A sample of 14 male college students consumes an average of 14.3 gallons of beer per month, with a sample standard deviation of 5.2 gallons. You can safely assume that the population of monthly beer consumption is Normally distributed. What is the margin of error for a 90% Confidence Interval for the mean? (Leave your answer to two decimal places, ex: 18.12)

An SRS of 350 high school seniors gained an average of ?⎯⎯⎯=45points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation ?=26.deviation      Find a 90% confidence interval for ?� based on this sample. Give your answers to three decimal places. What is the lower bound?     What is the upper bound ?

A researcher collected sample data for 10 women ages 18 to 24. The sample had a mean serum cholesterol level (measured in mg/100 mL) of 188.4, with a standard deviation of 5.7. Assuming that serum cholesterol levels for women ages 18 to 24 are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. Then find the lower limit and upper limit of the  99%  confidence interval.

A sample of 120 students yielded a sample mean GPA of 3.35, end a sample standard deviation equal to 0.28. Calculate the 95% confidence interval for the population mean GPA.

A simple random sample of 44 female cottonmouths yeilded the number of young per liter. The sample mean calculated to 7.59.  Assume sigma = 2.4, obtain a 95% confidence interval for the mean number of young per litter of all female eastern cottonmouths.