Question

This is observation from previous years about the impact of students working while they are enrolled in classes, due to students too much work, they are spending less time on their classes. First, the observer need to find out, on average, how many hours a week students are working. They know from previous studies that the standard deviation of this variable is about 5 hours. A survey of 200 students provides a sample mean of 7.10 hours worked. What is a 95% confidence interval based on this sample?

((NO HANDWRITING PLEASE))

Answer 1

Solution :

Given that,

= 7.10

= 5

n = 200

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 * (5 / 200)

= 0.69

At 95% confidence interval estimate of the population mean is,

- E < < + E

7.10 - 0.69 < < 7.10 + 0.69

6.41 < < 7.79

(6.41 , 7.79)