Question

After collecting a random sample of 101 individuals living in Arizona, you determine that the 95% confidence interval for the mean income is (60,224; 72,515). What was the standard deviation of income for the sample?

Hint: you'll need to find the critical value for t using a t-table

Round your answer to four decimal places.

Answer 1

sample mean = ( upper + lower)bound/2
= ( 72515 + 60224)/2
= 66369.5

margin of error = ( upper - lower)bound/2
= ( 72515 - 60224)/2
= 6145.5

MArgin of error = t *SE
t value at 95% = 1.984

6145.5 = 1.984 *Se

SE = 3097.5302

SE = s/sqrt(n)
3097.5302 = s/sqrt(101)

std.deviation = 308.2158