Question

Starting salaries of 60 college graduates who have taken a statistics course have a mean of 43,515. Suppose the distribution of this population is approximately normal and has a standard deviation of 10,863. Using the 98% confidence level, find both the following:

the margin error

the confidence interval for the mean ų:________ <ų<_______

Answer 1

Solution :

Given that,

Point estimate = sample mean =     = 43515

Population standard deviation =    =  10,863

Sample size = n =60

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z/2    * ( /n)

= 2.326 * (  10863 /  60)

= 3262
At 98% confidence interval mean
is,

- E < < + E

43515 - 3262 <   < 43515 + 3262

40253 <   < 46777