Question

1) The average annual salary based on a sample of 36 school administrators is $58,940, with a standard deviation of $18,490. (Please label your ans. a, b, c etc) a) Find the 95% confidence interval for the population mean annual salary of school administrators. b) What is the Margin of Error from part a? c) If the sample size is increased to 100, will the margin of error increase or decrease in part a?

Answer 1

1)

t critical value at 0.05 level with 35 df = 2.030

95% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

58940 - 2.030 * 18490 / sqrt( 36) < < 58940 + 2.030 * 18490 / sqrt( 36)

52684.22 < < 65195.78

95% CI is ( 52684.22 , 65195.78 )

2)

Margin of error =  t * S / sqrt(n)

=  2.030 * 18490 / sqrt( 36)

= 6255.78

3)

If sample size increased to 100 , margin of error decreases. (since quantity S/sqrt(n) will reduce)