An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the width of the 90% confidence interval?

Which of the following is the width of the confidence interval for the population mean?

Which confidence level would produce the widest interval when estimating the mean of a population from the mean and standard deviation of a sample of that population?A.26%B.54%C.11%D.38%

A student was asked to find a 98% confidence interval for widget width using data from a random sample of size n = 27. Which of the following is a correct interpretation of the interval 13.5 < μ < 22?Check all that are correct.There is a 98% chance that the mean of the population is between 13.5 and 22.With 98% confidence, the mean width of a randomly selected widget will be between 13.5 and 22.There is a 98% chance that the mean of a sample of 27 widgets will be between 13.5 and 22.With 98% confidence, the mean width of all widgets is between 13.5 and 22.The mean width of all widgets is between 13.5 and 22, 98% of the time. We know this is true because the mean of our sample is between 13.5 and 22.

A sample of size =n90 is drawn from a normal population whose standard deviation is =σ9.7. The sample mean is =x38.78.Part: 0 / 20 of 2 Parts CompletePart 1 of 2(a) Construct an 80% confidence interval for μ. Round the answer to at least two decimal places.An 80% confidence interval for the mean is <<μ.

You want to compute a 90% confidence interval for the mean of a population with unknown standard deviation. The sample size is 30. The correct distribution value for calculating the required confidence interval is:-Group of answer choices1.961.64491.6991.6971.311