Question

A toy manufacturer wants to see how long, on average, a new toy captures children's attention. He tests 10 children selected at random and finds that their mean attention span is 22 minutes with a standard deviation of 7 minutes. If we assume that attention spans are normally distributed, find a 99% confidence interval for the mean attention span of children playing with this new toy. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 22

sample standard deviation = s = 7

sample size = n = 10

Degrees of freedom = df = n - 1 = 10-1=9

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.05

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,9 = 3.250

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

= 3.250 * (7 / 10)

E = 7.194

The 99% confidence interval estimate of the population mean is,

- E < < + E

22 - 7.194 < < 22 + 7.194

14.8 < < 29.2

(14.8,29.2)

lower limit of the confidence interval is 14.8

upper limit of the confidence interval is 29.2