Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 manufacturing companies located in the Southwest. Estimate the mean and the standard deviation of advertising expenses. (Round squared deviations to nearest whole number and final answers to 2 decimal places.) Advertising Expenditure ($ millions) Number of Companies 25 up to 35 5 35 up to 45 10 45 up to 55 21 55 up to 65 16 65 up to 75 8 Total 60
Mean
Standard deviation

Answers

Answer 1

The estimated mean of advertising expenses is $52 million, and the estimated standard deviation is $11.23 million.

To estimate the mean and standard deviation of advertising expenses, we need to calculate the midpoint of each class interval, the product of the midpoint and the frequency, and the squared deviation from the mean.

Here's a table to organize the calculations:

Class Interval (x)     Frequency (f)     Midpoint (x)       fx

25 up to 35                   5                       30                 150

35 up to 45                   10                      40                 400

45 up to 55                   21                      50                1,050

55 up to 65                   16                      60                 960

65 up to 75                   8                        70                  560

To calculate the mean, we need to find the sum of the products of the midpoint and the frequency, divided by the total frequency.

Mean = (fx) / Total Frequency

Mean = (150 + 400 + 1,050 + 960 + 560) / 60

Mean = 3,120 / 60

Mean = 52

Calculate the squared deviation from the mean for each class interval.

(x - mean)² × f

(30 - 52)² × 5 = 484 × 5 = 2,420

(40 - 52)² × 10 = 144 × 10 = 1,440

(50 - 52)² × 21 = 4 × 21 = 84

(60 - 52)² ×  16 = 64 × 16 = 1,024

(70 - 52)² ×  8 = 324 × 8 = 2,592

Calculate the sum of the squared deviations.

Sum of (x - mean)× f = 2,420 + 1,440 + 84 + 1,024 + 2,592

= 7,560

Standard Deviation =√(Sum of (x - mean)² × f / Total Frequency)

Standard Deviation = √(7,560 / 60)

Standard Deviation = √(126)

Standard Deviation = 11.23

Hence, the estimated mean of advertising expenses is approximately $52 million, and the estimated standard deviation is approximately $11.23 million.

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Complete question

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 manufacturing companies located in the Southwest. Estimate the mean and the standard deviation of advertising expenses. (Round squared deviations to nearest whole number and final answers to 2 decimal places.)

Advertising Expenditure ($ millions)        Number of Companies

25 up to 35                                                                      5

35 up to 45                                                                    10

45 up to 55                                                                    21

55 up to 65                                                                   16

65 up to 75                                                                   8

Total                                                                               60


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[tex]Sb_2 = 0.41[/tex]

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[tex]P(t < -t_0) = P(t > t_0) = 0.025[/tex]

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[tex]df = 30 -2[/tex]

[tex]df = 28[/tex]

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[tex]t = \frac{b_1 - 0}{Sb_1}[/tex]

[tex]t = \frac{2.815 - 0}{0.75}[/tex]

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Testing the second independent variable

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[tex]t = \frac{b_2 - 0}{Sb_2}[/tex]

[tex]t = \frac{-1.249 - 0}{0.41}[/tex]

[tex]t = \frac{-1.249 }{0.41}[/tex]

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