Question

A company conducts surveys each year about the use of various media, such as television and video viewing. A representative sample of adult Americans was surveyed in 2010, and the mean number of minutes per week spent watching "time-shifted" television (watching television shows that were recorded and played back at a later time) for the people in this sample was 572 minutes. An independently selected representative sample of adults was surveyed in 2011, and the mean time spent watching time-shifted television per week for this sample was 648 minutes. Suppose that the sample size in each year was 1000 and that the sample standard deviations were 60 minutes for the 2010 sample and 80 minutes for the 2011 sample. Estimate the difference in the mean time spent watching time-shifted television in 2010 and the mean time spent in 2011 using a 99% confidence interval. (Use μ₂₀₁₀ − μ₂₀₁₁. Round your answers to two decimal places.)

Answer 1

x1            =	572.00	x2            =	648.00
n1           =	1000	n2           =	1000
σ1           =	60.00	σ2           =	80.00
std error σx1-x2=√(σ21/n1+σ22/n2)    =			3.162

Point estimate of differnce '=x1-x2 =			-76.000
for 99 % CI value of z=			2.576
margin of error E=z*std error =			8.145
lower bound=(x1-x2)-E =			-84.145
Upper bound=(x1-x2)+E =			-67.855
from above 99% confidence interval for population mean =(-84.15 <μ1-μ2 < -67.85)