Question

In: Statistics and Probability

Since 2007, the American Psychological Association has supported an annual nationwide survey to examine stress across...

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.)

Solutions

Expert Solution

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 * ( / $\sqrt$ n)

= 1.96 * ( 2.3 / $\sqrt$ 360)

=0.238

- E < < + E

5.4-0.238 < < 5.4+0.238

5.162 ,5.68

< <

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 * ( / $\sqrt$ n)

= 2.576 * ( 2.3 / $\sqrt$ 360)

=0.312

- E < < + E

5.4-0.312< < 5.4+0.312

5.088< < 5.712

orchestra answered 11 months ago

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