A sample of size =n92 is drawn from a normal population whose standard deviation is =σ5.1. The sample mean is =x35.33.Part 1 of 2(a) Construct a 99.9% confidence interval for μ. Round the answer to at least two decimal places.A 99.9% confidence interval for the mean is 33.58<<μ37.08.Part: 1 / 21 of 2 Parts CompletePart 2 of 2(b) If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain.The confidence interval constructed in part (a) ▼(Choose one) be valid since the sample size ▼(Choose one) larg

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.

If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain.The confidence interval constructed in part (a) ▼(Choose one) be valid since the sample size ▼(Choose one) large.

A random sample of size n1 = 25, taken from a normal population with a standard deviation σ1 = 5, has a mean ¯x1 = 80. A second random sample of size n2 = 36, taken from a different normal population with a standard deviation σ2 = 3, has a mean ¯x2 = 75. Find a 94% confidence interval for μ1 − μ2.

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. Find a 98% confidence interval for the mean μ of a normal population withvariance 9 from the sample 32,42,41,35,48,55.b. Find a 98% confidence interval for the mean μ of a normal population from thesample 32,42,41,35,48,55.c. Using a sample size n = 15 and a sample variance s2 = 13 from a normalpopulation , test the hypothesis σ2 = σ02 = 10 against three kinds of alternatives,namely, a σ2 < σ02b σ2 > σ02c σ2 ≠ σ02by choosing a significance level 1%.