Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 75 manufacturing companies located in the Southwest. The mean expense is $50.93 million and the standard deviation is $10.80 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Advertising Expense ($ Million) Number of Companies 25 up to 35 4 35 up to 45 19 45 up to 55 27 55 up to 65 16 65 up to 75 9 Total 75

[tex]P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}<z< \dfrac{35-50.93}{10.80}\Big) \\ \\ =P \Big(\dfrac{-25.93}{10.80}<z< \dfrac{-15.93}{10.80}\Big) \\ \\ =P(-2.4009<z<-1.475) \\ \\ =(0.0694 -0.0082) \\ \\ =0.0612[/tex]

For 35 up to 45

[tex]P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}<z< \dfrac{45-50.93}{10.80}\Big)=P \Big(\dfrac{-15.93}{10.80}<z< \dfrac{-5.93}{10.80}\Big) \\ \\ =P(-1.475<z<-0.5491) \\ \\ =(0.2912 -0.0694) \\ \\ =0.2218[/tex]

For 45 up to 55

[tex]P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}<z< \dfrac{55-50.93}{10.80}\Big)=P \Big(\dfrac{-5.93}{10.80}<z< \dfrac{4.07}{10.80}\Big) \\ \\ =P(-0.5491<z<0.3769) \\ \\ =(0.6480 -0.2912) \\ \\ =0.3568[/tex]

For 55 up to 65

[tex]P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}<z< \dfrac{65-50.93}{10.80}\Big)=P \Big(\dfrac{4.07}{10.80}<z< \dfrac{14.07}{10.80}\Big) \\ \\=P(0.3768<z<1.3028) \\ \\ =(0.9032-0.6480) \\ \\ =0.2552[/tex]

For 65 up to 75

[tex]P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}<z< \dfrac{75-50.93}{10.80}\Big)=P \Big(\dfrac{14.07}{10.80}<z< \dfrac{24.07}{10.80}\Big) \\ \\ =P(1.3028<z<2.2287) \\ \\=(0.9871-0.9032) \\ \\ =0.0839[/tex]

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe [tex](O-E)^2[/tex] [tex]\dfrac{(O-E)^2}{E}[/tex]

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

[tex]\sum \dfrac{(O-E)^2}{E}= 2.0492[/tex]

Using the Chi-square formula:

[tex]X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942[/tex]

Null hypothesis:

[tex]H_o: \text{The population of advertising expenses follows a normal distribution}[/tex]

Alternative hypothesis:

[tex]H_a: \text{The population of advertising expenses does not follows a normal distribution}[/tex]

Assume that:

[tex]\alpha = 0.02[/tex]

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from [tex]X^2 = 11.667[/tex]

Decision rule: To reject [tex]H_o \ if \ X^2[/tex] test statistics is greater than [tex]X^2[/tex] tabulated.

Conclusion: Since [tex]X^2 = 2.0942[/tex] is less than critical value 11.667. Then we fail to reject [tex]H_o[/tex]