1 An April 2004 article on Hear The Issuescom stated that Am

1.) An April 2004 article on Hear The Issues.com stated that Americans have an average of 2.24 televisions per household (source: Nielsen Media Research). If the standard deviation for the number of televisions in a U.S. household is 1.2 and a random sample of 80 American households is selected, the mean of this sample belongs to a sampling distribution.

2.)If the Confidence Level is 99%, the Confidence Coefficient for calculating the confidence interval for the mean is 2.33. T or F? Please show calculations as to why or why not.

Solution

CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=2.24
Standard deviation( sd )=1.2
Sample Size(n)=80
Confidence Interval = [ 2.24 ± Z a/2 ( 1.2/ Sqrt ( 80) ) ]
= [ 2.24 - 2.58 * (0.134) , 2.24 + 2.58 * (0.134) ]
= [ 1.894,2.586 ]
ANS: FALSE