Question 1204624
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.