Question

b) According to a certain​ survey, adults spend 2.35hours per day watching television on a weekday. Assume that the standard deviation for​ "time spent watching television on a​ weekday" is 1.93 hours. If a random sample of 50 adults is​ obtained, describe the sampling distribution of x overbar​, the mean amount of time spent watching television on a weekday.

d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 45 individuals who consider themselves to be avid Internet users results in a mean time of 1.98 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.98 hours or less from a population whose mean is presumed to be 2.35 hours.

Answer 1

Solution:

Given:

Part b) Adults spend 2.35hours per day watching television on a weekday. That is: Mean =

The standard deviation for​ "time spent watching television on a​ weekday" is 1.93 , that is:

Sample size = n = 50

We have to describe the sampling distribution of , the mean amount of time spent watching television on a weekday.

Since sample size = n = 50 > 30, thus we can apply central limit theorem.

Thus sampling distribution of sample mean is approximately Normal with mean of sample means is

and standard deviation of sample means is

Part d) One consequence of the popularity of the Internet is that it is thought to reduce television watching.

Sample size = n= 45

Sample mean =

We have to find probability of obtaining a sample mean of 1.98 hours or less from a population whose mean is presumed to be 2.35 hours.

That is find:

Find z score:

where

Thus

Thus we get:

Look in z table for z = -1.2 and 0.09 and find corresponding area.

P( Z < -1.29 ) = 0.0985

Thus