Question

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

Answer 1

Explanation:

The table can be computed as:

Advertising Expenses ($ 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

Let's find the probabilities first:

`P(25 - 35) = P \Big((25-50.93)/(10.80)<z< (35-50.93)/(10.80)\Big) \\ \\ =P \Big((-25.93)/(10.80)<z< (-15.93)/(10.80)\Big) \\ \\ =P(-2.4009<z<-1.475) \\ \\ =(0.0694 -0.0082) \\ \\ =0.0612`

For 35 up to 45

`P(35 - 45) = P \Big((35-50.93)/(10.80)<z< (45-50.93)/(10.80)\Big)=P \Big((-15.93)/(10.80)<z< (-5.93)/(10.80)\Big) \\ \\ =P(-1.475<z<-0.5491) \\ \\ =(0.2912 -0.0694) \\ \\ =0.2218`

For 45 up to 55

`P(45 - 55) = P \Big((45-50.93)/(10.80)<z< (55-50.93)/(10.80)\Big)=P \Big((-5.93)/(10.80)<z< (4.07)/(10.80)\Big) \\ \\ =P(-0.5491<z<0.3769) \\ \\ =(0.6480 -0.2912) \\ \\ =0.3568`

For 55 up to 65

`P(55 - 65) = P \Big((55-50.93)/(10.80)<z< (65-50.93)/(10.80)\Big)=P \Big((4.07)/(10.80)<z< (14.07)/(10.80)\Big) \\ \\=P(0.3768<z<1.3028) \\ \\ =(0.9032-0.6480) \\ \\ =0.2552`

For 65 up to 75

`P(65 - 75) = P \Big((65-50.93)/(10.80)<z< (75-50.93)/(10.80)\Big)=P \Big((14.07)/(10.80)<z< (24.07)/(10.80)\Big) \\ \\ =P(1.3028<z<2.2287) \\ \\=(0.9871-0.9032) \\ \\ =0.0839`

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe
`(O-E)^2`
`((O-E)^2)/(E)`

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

`\sum ((O-E)^2)/(E)= 2.0492`

Using the Chi-square formula:

`X^2 = ((O-E)^2)/(E) \\ \\ Chi-square \ X^2 = 2.0942`

Null hypothesis:

`H_o: \text{The population of advertising expenses follows a normal distribution}`

Alternative hypothesis:

`H_a: \text{The population of advertising expenses does not follows a normal distribution}`

Assume that:

`\alpha = 0.02`

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from
`X^2 = 11.667`

Decision rule: To reject
`H_o \ if \ X^2` test statistics is greater than
`X^2` tabulated.

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