Question

A random sample of 18 registered nurses in a large hospital showed that they worked on average 44.8 hours per week. The standard deviation of the sample was 2.4. Estimate the mean of the population with 95% confidence. Assume the variable is normally distributed. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place.

Answer 1

Solution :

Given that,

= 44.8

s =2.4

n =18

Degrees of freedom = df = n - 1 =18 - 1 = 17

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,17 = 2.110 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.110* ( 2.4/ 18)

= 1.2

The 95% confidence interval mean is,

- E < < + E

44.8 - 1.2 < < 44.8+ 1.2

43.6 < < 46.0