Question

A community health association is interested in estimating the average number of maternity days women stay in the local hospital. A random sample is taken of 36 women who had babies in the hospital during the past year. The following numbers of maternity days each woman was in the hospital are rounded to the nearest day. 3 3 4 3 2 5 3 1 4 3 4 2 3 5 3 2 4 3 2 2 1 6 3 4 3 3 5 2 3 3 3 5 4 3 5 4 Use these data and a population standard deviation of 1.12 to construct a 98% confidence interval to estimate the average maternity stay in the hospital for all women who have babies in this hospital.

Answer 1

Solution

The sample mean is

Mean = (x / n) )

=3+3+4+3+2+5+3+1+4+3+4+2+3+5+3+2+4+3+2+2+1+6+3+4+3+3+5+2+3+3+3+5+4+3+5+4/36

=118/36

=3.2778

Mean = 3.28

= 1.12

n = 40

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z/2 = Z 0.01 = 2.326

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

= 2.326 * (1.12 / 36 )

= 0.43

At 98% confidence interval estimate of the population mean is,

- E < < + E

3.28 - 0.43 < < 3.28 + 0.43

2.85 < < 3.71

(2.85 , 3.71)