Question

according to lear center local news archive the average amount of time that a half hour local tv news broadcast devotes to us foreign policy, including the war in iraq, is 38 seconds. (time, february 28, 2005). suppose a random sample of 40 such half-hour news broadcasts shows that an average of 38 seconds are devoted to us foreign policy with a standard deviation of 9 seconds. Find a 95% confidence interval for the mean time that all half-hour local TV news broadcasts devote to U.S. foreign policy.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 38

sample standard deviation = s = 9

sample size = n = 40

Degrees of freedom = df = n - 1 = 39

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,39 = 2.023

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

= 2.023* (9 / 40)

= 2.9

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

- E < < + E

38 - 2.9 < < 38 + 2.9

35.1 < < 40.9

A 95% confidence interval for the mean time that all half-hour local TV news broadcasts devote to U.S. foreign policy is (35.1 , 40.9)