SOLUTION: Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,900 pounds and the standard deviation is 290 pounds. Assume that the po

Algebra ->  Probability-and-statistics -> SOLUTION: Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,900 pounds and the standard deviation is 290 pounds. Assume that the po      Log On


   

Question 1204624: Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 5,900 pounds and the standard deviation is 290 pounds. Assume that the population follows the normal distribution. Fifty trucks are randomly selected and weighed.
Within what limits will 99 percent of the sample means occur? (Round your z-value to 2 decimal places and final answers to 1 decimal place.)
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population mean is assumed to be 5900 pounds with a standard deviation of 290 pounds.
sample size is 50.
two tailed confidence interval is 99% = .99.
that gives you an alpha at each end of the confidence interval of .05.
critical z-score for two tailed confidence interval of 99% is equal to plus or minus 2.58.
z-score formula is z = (x-m)/s
z is the z-score
x is the sample mean
m is the population mean
s is the standard error.
standard error = standard deviation / sqrt(sample size) = 290/sqrt(50) = 41.0122
at the low end of the confidence interval, -2.58 = (x-5900)/41.0122.
solve for x to get x = -2.58 * 41.0122 + 5900 = 5794.2.
at the high end of the confidence interval, 2.58 = (x-5900)/41.0122.
solve for x to get x = 2.58 * 41.0122 + 5900 = 6005.8
your 99% confidence interval is 5794.2 to 6005.8
all rounding not specified was done to 4 decimal digits.