Question

Social Services wanted to determine the average number of hours a week that working parents need for daycare services. A random sample of 50 working parents gave a sample mean of 38 hours with a standard deviation of 6 hours. Make a 95% confidence interval for the average number of hours of daycare services a week that working parents need.

Answer 1

Solution :

Given that,

t _/2,df = 2.010

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

= 2.010 * (6 / 50)

Margin of error = E = 1.7

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

- E < < + E

38- 1.7 < < 38 + 1.7

36.3 < < 39.7

(36.3 , 39.7)