Question

random election of 11 children tested and finds that their mean attention span is 31 minutes with a standard deviation of 8 minutes. assuming attention spans are normally distributed, find a 95% confidence interval for the mean attention span of children.   also calculate the upper and lower limit of the confidence interval

Answer 1

Solution :

Given that,

= 31

s = 8

n = 11

Degrees of freedom = df = n - 1 = 11 - 1 = 10

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,10 = 2.228

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

= 2.228 * (8 / 11)

= 5.37

Margin of error = 5.37

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

- E < < + E

31 - 5.37 < < 31 + 5.37

25.63 < < 35.37

The upper limit = 35.37

The lower limit = 25.63