MATH SOLVE

4 months ago

Q:
# The diameter of bushings turned out by a manufacturing process is a normally distributed random variable with a mean of 4.035 mm and a standard deviation of 0.005 mm. A sample of 25 bushings is taken once an hour. (a) Within what interval should 95 percent of the bushing diameters fall? (Round your answers to 4 decimal places.) The 95% confidence interval is from to (b) Within what interval should 95 percent of the sample means fall? (Round your answers to 5 decimal places.) The 95% confidence interval is from to (c-1) What conclusion would you reach if you saw a sample mean of 4.020? The sample came from a population that a population mean equal to 4.035. (c-2) What conclusion would you reach if you saw a sample mean of 4.055? The sample came from a population that a population mean equal to 4.035.

Accepted Solution

A:

Complete question is;The diameter of bushings turned out by a manufacturing process is a normally distributed random variable with a mean of 4.035 mm and a standard deviation of 0.005 mm. A sample of 25 bushings is taken once an hour.a. Within what range should 95 percent of the bushing diameters fall?b. Within what range should 95 percent of the sample means fall?c. What conclusion would you reach if you saw a sample mean of 4.020?Answer:A) (4.0252, 4.0448).B) (4.0331, 4.0369).C) If i saw a sample mean of 4.020, i would conclude that the sample came from a population that did not have a population mean equal to 4.035Step-by-step explanation:A) We know that an approximate 95% confidence interval for the item in question has a range of diameter given by the equation; Range = x¯ ± 1.96σNow, since we have a mean of 4.035 mm and a standard deviation of 0.005 mm, we can solve;Range of bushing diameter = 4.035 ± (1.96 × 0.005) = 4.035 ± (0.0098)= 4.0252 or 4.0448 and it can be written as;(4.0252, 4.0448).B) We also know that In the large-sample case, a 95% confidence interval estimate for the population mean is given by the formula;CI = x¯ ± (1.96s/√n)) Thus, if we solve, we get;CI = 4.035 ± (1.96 × 0.005 )/√25CI = 4.035 ± 0.0019or (4.0331, 4.0369).c. If i saw a sample mean of 4.020, i would conclude that the sample came from a population that did not have a population mean equal to 4.035 because it doesn't fall into the range of confidence intervals we got earlier.