Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 72 manufacturing companies located in the Southwest. The mean expense is $52.0 million and the standard deviation is $15.10 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 72 manufacturing companies located in the Southwest. The mean expense is $52.0 million and the standard deviation is $15.10 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Solution a)reject Ho if x2>9.49 b) expectd value=72/5=14.4 chi square value= sum of (expected value-actual value)^2/expected value =((3-14.4)^2+(15-14.4)^2+(29-14.4)^2+(17-14.4)^2+(8-14.4)^2)/14.4=27.1667.

1. Advertising expenses are a significant component of the cost of goods sold. Listed below is a
frequency distribution showing the advertising expenditures for 72 manufacturing companies
located in the Southwest. The mean expense is $52.0 million and the standard deviation is $15.10
million. Is it reasonable to conclude the sample data are from a population that follows a normal
probability distribution?
Advertising expenses are a significant component of the cost of goods sold. Listed below is a
frequency distribution showing the advertising expenditures for 72 manufacturing companies
located in the Southwest. The mean expense is $52.0 million and the standard deviation is
$15.10 million. Is it reasonable to conclude the sample data are from a population that follows a
normal probability distribution?
Solution
a)reject Ho if
x2>9.49
b)
expectd value=72/5=14.4
chi square value=
sum of (expected value-actual value)^2/expected value
=((3-14.4)^2+(15-14.4)^2+(29-14.4)^2+(17-14.4)^2+(8-14.4)^2)/14.4=27.1667