Question

An educational organization in California is interested in estimating the mean number of minutes per day that children between the age of 6 and 18 spend watching television per day. A previous study showed that the population standard deviation was 21.5 minutes. The organization selected a random sample of n = 251 children and recorded the number of minutes of TV that each person watched on a particular day. The mean time was 171.3 minutes. If the leaders of the organization wish to develop an interval estimate with 90 percent confidence, what is the upper limit of the confidence level?

Group of answer choices

169.07

173.53

172.04

189.07

Answer 1

Solution :

Given that,

Point estimate = sample mean = =  171.3

Population standard deviation =    = 21.5
Sample size = n =251

At 90% confidence level the z is

Z/2= Z0.1 = 1.645 ( Using z table )

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

= 1.645 * ( 21.5 /  251  )

E= 2.23

At 90% confidence interval estimate of the population mean
is,

+ E

171.3 + 2.23

173.53

upper limit=173.53