A company conducts surveys each year about the use of variou

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 574 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 642 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 2010 2011. Round your answers to two decimal places.)

Solution

CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=574
Standard deviation( sd1 )=60
Sample Size(n1)=1000
Mean(x2)=642
Standard deviation( sd2 )=80
Sample Size(n1)=1000
CI = [ ( 574-642) ±t a/2 * Sqrt( 3600/1000+6400/1000)]
= [ (-68) ± t a/2 * Sqrt( 10) ]
= [ (-68) ± 2.581 * Sqrt( 10) ]
= [-76.16 , -59.84]