Question

The amount of time adults spend watching television is closely monitored by firms because this helps to determine advertising pricing for commercials. Complete parts (a) through (d). (b) According to a certain survey, adults spend 2.25 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 60 adults is obtained, describe the sampling distribution of x, the mean amount of time spent watching television on a weekday. and o=0 x is approximately normal with this = 2.25 (Round to six decimal places as needed.) (c) Determine the probability that a random sample of 60 adults results in a mean time watching television on a weekday of between 2 and 3 hours. The probability is . (Round to four decimal places as needed.) (d) One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 55 individuals who consider themselves to be avid Internet users results in a mean time of 1.88 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 1.88 hours or less from a population whose mean is presumed to be 2.25 hours. The likelihood is . (Round to four decimal places as needed.) Based on the result obtained, do you think avid Internet users watch less television? Yes No O

Answer 1

Ansbler : Given data = 2.25 1.93 n=60 6 He know that . In normal approximation 1x= el f să= 6 Here 42 = 2.25 and 6ł = 1.93=0,2491619 N V60 0.249 162 mi G = 0,249762 AnS : X is approximately els=2.25 and normal with so = 0. 249162 © Hege n = 60 find pl 2CX 23] = ? Test AP24X L3 -1 P{2{XL37 = pl 2-2.25 1. 936 Voo H 1.93 3-2.257 193 The Th Noo -0.25 ez c 0.75 20.249162 0.249162 = p [-1.0033 L2 (3.01007 PL 23.0100] – PLZ2-1.00337 = 0.99869 -0.15785 (from standard e normal table) P 2 L X 237 = 0.84084 x 0,8408 P/2 LTL3] = 0,8408 Ans: The probability is 10.8408)

Here, n=55 find PL < 1.88] - ? Don © -0.37 0.2602 : PEX (1.38] = P[p < 1.88 -- 2.251 1.93 155 P[2002] care l e p [2 (-1.4219] PrX 21.88] = 0.077522 ( from standard 0.0775 normal table) - PE C 1.88] = 0.0775 Ans: The likelihood is a 0.0775 Based on the avid internet result uers obtained, do you think watch less television Ans: No