Question

How much Tv do you watch each week? A random sample of 41 adults were asked this question and thee mean found to be 3.9 hours with standard deviation 1.6 hours.

a. Find the 99% confidence interval for the population mean number of hours of TV watched each week.

b. Is it thought that most adults watch 3 hours of TV each week. Do the data support this? Why of why not?

c. What is the margin of error in this situation (a value?)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 3.9

Population standard deviation =    = 1.6
Sample size = n =41

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* (1.6 / 41)

E= 0.6437

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

- E < < + E

3.9-0.6437 < < 3.9+ 0.6437

3.2563< < 4.5437

(3.2563, 4.5437)