Question

Since 2007, the American Psychological Association has supported an annual nationwide survey to examine stress across the United States. A total of 360 Millennials (18- to 33-year-olds) were asked to indicate their average stress level (on a 10-point scale) during a month. The mean score was 5.4. Assume that the population standard deviation is 2.3.

(a) Give the margin of error for a 95% confidence interval. (Round your answer to three decimal places.)

Find the 95% confidence interval for this sample. (Round your answers to three decimal places.)

(b) Give the margin of error for a 99% confidence interval. (Round your answer to three decimal places.)

Find the 99% confidence interval for this sample. (Round your answers to three decimal places.)

Answer 1

Solution :

Given that,

sample mean = = 5.4

Population standard deviation =    = 2.3

Sample size = n =360

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * ( 2.3 /   360)

=0.238

- E < < + E

5.4-0.238 < < 5.4+0.238

5.162 ,5.68

< <

b.

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * ( 2.3 /   360)

=0.312

- E < < + E

5.4-0.312< < 5.4+0.312

5.088< < 5.712