Question

Starting salaries of 110 college graduates who have taken a statistics course have a mean of $43,598. Suppose the distribution of this population is approximately normal and has a standard deviation of $8,635.
Using a 93% confidence level, find both of the following:

(a) The margin of error:  

(b) The confidence interval for the mean μ:

  <. ? <

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 43598  

Population standard deviation = = 8635

Sample size = n = 110

At 93% confidence level

= 1-0.93% =1-0.93 =0.07

/2 =0.07/ 2= 0.035

Z/2 = Z_{0.035 = 1.81}

Z/2 = 1.81
a) Margin of error = E = Z/2 * ( /n)

=1.81 *(8635 / 110 )

= 1490.2

b) At 93 % confidence interval estimate of the population mean is,

- E < < + E

43598 - 1490.2 <   < 43598 + 1490.2

42107.8 <   < 45088.2

(42107.8,45088.2 )