The weekly salaries of 100 recent graduates of a private women's college are normally distributed with a mean of $600 and a standard deviation of $80. Determine the interval about the sample mean that has a 1% level of confidence. Use t=2.58
I don't know.
600/sqrt100
60
80+-2.58*60
2334.8 and -74.8
You have reversed your values.
The standard error is the standard deviation — not the mean — divided by the square root of n.
You also want the mean ($600) [not the standard deviation] ± 2.58 times the standard error.
I hope this helps. Thanks for asking.