Math

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

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  1. 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.

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    posted by PsyDAG

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