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

Answer:

Step-by-step 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:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense   No of   Probabilities(P)  Expe                $(O-E)^2$   $\dfrac{(O-E)^2}{E}$

             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

                                                                                           $\sum \dfrac{(O-E)^2}{E}= 2.0492$                                                                                                      

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

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 $H_o$