Suppose when will like to construct a 81% confidence interval for a sample mean when the population standard deviation is no determined the Z value that would be used to contract such a confidence interval round the solution to two places. Z=

Suppose one would like to contract a 99.9% confidence interval for a sample mean when the population standard deviation is known. Determine the Z value that would be used to contract such a confidence interval run the solution to two decimal places. Z=

what is the " lower limit " and the " upper limit"  to this problem ?   The mean diastolic blood pressure for a random sample of 150 people was 99 millimeters of mercury. If the standard deviation of individual blood pressure readings is known to be 12 millimeters of mercury, find a 95% confidence interval for the true mean diastolic blood pressure of all people. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

An article claims that babies spend twice as much time in REM (rapid eye movement) sleep than adults. This implies that babies have an average of 6 hours of REM sleep per night. A doctor would like to test the hypotheses = 6 versus  ≠ 6 where  = the true mean amount of REM sleep per night for babies. A 95% confidence interval for the true mean amount of REM sleep per night for babies is (5.45, 6.25) hours. Based on the confidence interval, what conclusion would you make for a test of these hypotheses? The 95% confidence interval  include  as a plausible value, so we  H0. We  convincing evidence that the true mean amount of REM sleep per night for babies  6 hours.

Can anyone get this right. I've tried multiple answers, but none of them are correct. One picture has the info needed and thr other one contains the actual question for the confidence level. The answer is NOT WHAT'S PICTURED OR 2.547, 6.547

Solve Question 2 Please show work

What is the confidence level if the confidence interval for population mean (μ) is (19.02, 20.98) assume sample of 100 data, sample mean 200, and standard deviation equals 5

I need A and B

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 41 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 9.1 and a standard deviation of 2.9. What is the 80% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to on 2 decimal place). < μμ <

A researcher collected sample data for 20 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 190.4, with a standard deviation of 9.7. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 99% confidence interval.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place Lower limit= Upper limit=