Question

According to a certain​ survey, adults spend 2.35 hours 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.

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

Given that, according to a survey, on average adults spend 2.35 hours per day watching television on a weekday and the standard deviation is given to be 1.93 hours. The sampling distribution of the sample mean when a sample of size 50 bis drawn is given as follows: 11 = 1 = 2.35 %3D Vn 1.93 V50 = 0.2729 %3D Therefore, the mean amount of time spend by the 50 average adults on a weekday is 2.35 hours with a standard deviation of 0.2729 hours. Now, again a random sample of 45 individuals who are keenly interested in internet is drawn and their mean time for watching television on a weekday is 1.98 hours. The population mean is assumed to be 2.35. Therefore, the probability of obtaining a sample mean less than or equal to 1.98 hours is computed as follows: X- 4 1.98– 2.35 2// = P[Z<-1.21] PX<1.98]-P 1.93/40 = NORM.S.DIST (-1.21,TRUE) = 0.1131 Therefore, the likelihood of obtaining a sample mean less than or equal to 1.98 hours is 0.1131.