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 $50.67 million and the standard deviation is $11.07 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Advertising Expense ($ Million) Number of Companies 25 up to 35 7 35 up to 45 14 45 up to 55 29 55 up to 65 17 65 up to 75 8 Total 75 a. State the decision rule. Use the 0.01 significance level. (Round your answer to 2 decimal places.) H0: The population of advertising expenses follows a normal distribution. H1: The population of advertising expenses does not follow a normal distribution. b. Compute the value of chi-square. (Round your answer to 2 decimal places.) c. What is your decision regarding H0?
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 $50.67 million and the standard deviation is $11.07 million. Is it reasonable to conclude the sample data are from a population that follows a normal
| Advertising Expense ($ Million) |
Number of Companies |
||
| 25 up to 35 | 7 | ||
| 35 up to 45 | 14 | ||
| 45 up to 55 | 29 | ||
| 55 up to 65 | 17 | ||
| 65 up to 75 | 8 | ||
| Total | 75 | ||
a. State the decision rule. Use the 0.01 significance level. (Round your answer to 2 decimal places.)
H0: The population of advertising expenses follows a
H1: The population of advertising expenses does not follow a normal distribution.
b. Compute the value of chi-square. (Round your answer to 2 decimal places.)
c. What is your decision regarding H0?
The random variable is advertising expenses.
There are 5 companies of observations are given.
We have to test whether the sample data are from a population that follows a normal probability distribution or not.
This is Chi-square test for goodness-of-fit.
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