Th weekly salaries of a sample 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
Not sure if this is correct?
=600+or-2.58(80)/10
=600+or-206.4/10
=600+or-20.64
I am not sure what "t" is supposed to be, but the interval that the salary will be in 99% of the time is 600 + or - 206.4
206.4 happens to be 2.58 times the standard deviation.
Your t may be what other texts usually call s, the deviation from the mean divided by sigma
To determine the confidence interval about the sample mean, you can use the formula:
Confidence Interval = Sample Mean ± (t * (Standard Deviation / √n))
In this case, the mean is $600, the standard deviation is $80, and the sample size, n, is 100. The critical value t for a 1% level of confidence with 99 degrees of freedom (n - 1) is 2.58.
Substituting the values into the formula:
Confidence Interval = 600 ± (2.58 * (80 / √100))
Confidence Interval = 600 ± (2.58 * (80 / 10))
Confidence Interval = 600 ± (2.58 * 8)
Confidence Interval = 600 ± 20.64
Thus, the interval about the sample mean that has a 1% level of confidence is:
(579.36, 620.64)