Question

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 77 manufacturing companies located in the Southwest. The mean expense is $50.26 million and the standard deviation is $11.24 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Advertising Expense ($ Million) 25 up to 35 35 up to 45 45 up to 55 55 up to 65 65 up to 75 Total Number of Companies 8 15 29 17 8 77 Click here for the Excel Data File State the decision rule. Use the 0.05 significance level. (Round your answer to 2 decimal places.) He: The population of advertising expenses follows a normal distribution. Hi: The population of advertising expenses does not follow a normal distribution. Reject HO if chi-square > Compute the value of chi-square. (Round your answer to 2 decimal places.) Chi-square value What is your decision regarding He? Do not reject HO. This data could be from a normal distribution.

Answer 1

Null and alternative hypothesis are                  
Ho : The population of advertising expenses follows a normal distribution                  
H1 : The population of advertising expenses does not follow a normal distribution                 
                  
Let X be the advertising expenses                  
X follows Normal Distribution with mean μ and standard deviation σ
Given
μ = $ 50.26 million         ....... Mean
σ = $ 11.24 million        ....... Standard Deviation            

We first find the expected probability for each of the Advertising Expense.

To find P(25 < X < 35)
P(25 < X < 35) = P(X < 35) - P(X < 25)
Using Excel Function "NORM.DIST", we get
                          = NORM.DIST(35, 50.26, 11.24, TRUE) - NORM.DIST(25, 50.26, 11.24, TRUE)
P(25 < X < 35) = 0.07498

Similarly

P(35 < X < 45) = 0.23262

P(45 < X < 55) = 0.34348

P(55 < X < 65) = 0.24176

P(65 < X < 75) = 0.081

Now multiply the expected probability with the total observations (77) to get the expected frequency

5 Advertising Observed Expected Expenses Freq (0) Freq (E) O-E (0 - E) 25 up to 35 8 5.77346 2.22654 35 up to 45 15 17.91174 -2.91174 45 up to 55 29 26.44796 2.55204 55 up to 65 17 18.61552 -1.61552 65 up to 75 8 6.237 1.763 8 (O-E)? / E 0.859 0.473 0.246 0.140 0.498 7 3 3 77 Sum = 2.22

Degrees of freedom = df = number of categories - 1 = 5 - 1 = 4                  
Degrees of Freedom = 4                  
                  
α = 0.05   that is 5% significance level              
Critical Chi square is found using Excel function CHISQ.INV.RT                  
χ2-critical = CHISQ.INV.RT(0.05, 4)                  
χ2-critical = 9.49     

           
Reject Ho if chi-square > 9.49                 
                  
                  
From the above table, calculate value of χ2 test statistic

           

2 r? = F (0-3)?

Fromt the above table the sum is 2.22

Chi-square value χ2 = 2.22                  
                  
2.22 < 9.49                  
that is calculated χ2 < χ2-critical                  
Hence, Do Not Reject Ho                  
                  
Do not Reject Ho. This data could be from a normal distribution                  

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