Simple Regression 2. A CEO of a large plastics manufacturing company would like to determine if she should be placing more money allotted in the budget next year for television advertising of a new...

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Simple Regression


2.
A CEO of a large plastics manufacturing company would like to determine if she should be placing more money allotted in the budget next year for television advertising of a new baby bottle marketed for controlling reflux and reducing gas. She wonders whether there is a strong relationship between the amount of money spent on television advertising for this new baby bottle called Gentle Bottle and the number of orders received. The manufacturing process of this baby bottle is very difficult and requires advanced quality control so the CEO would prefer to generate a stable number of orders. The cost of advertising is always an important consideration in the phase I roll-out of a new baby bottle. Data that have been collected over the past 20 months indicate the amount of money spent of television advertising and the number of orders received.
The use of linear regression is a critical tool for a manager's decision-making ability. Please carefully read the example below and try to answer the questions in terms of the problem context. The results are as follows:



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Simple Regression 2.  A CEO of a large plastics manufacturing company would like to determine if she should be placing more money allotted in the budget next year for television advertising of a new baby bottle marketed for controlling reflux and reducing gas. She wonders whether there is a strong relationship between the amount of money spent on television advertising for this new baby bottle called Gentle Bottle and the number of orders received. The manufacturing process of this baby bottle is very difficult and requires advanced quality control so the CEO would prefer to generate a stable number of orders. The cost of advertising is always an important consideration in the phase I roll-out of a new baby bottle. Data that have been collected over the past 20 months indicate the amount of money spent of television advertising and the number of orders received. The use of linear regression is a critical tool for a manager's decision-making ability. Please carefully read the example below and try to answer the questions in terms of the problem context. The results are as follows:     MonthAdvertising Cost Number of Orders1$77,430.002,902,000262,6203,800,000369,5803,299,000450,6702,430,000569,1803,367,000673,1404,011,000783,3704,888,000878,8804,935,000964,9905,555,0001077,2304,654,0001161,3805,598,0001262,9002,967,0001363,2702,899,0001489,1904,245,0001560,0304,934,0001679,2103,853,0001765,7705,625,0001884,5305,778,0001979,7605,999,0002082,6406,834,000 a. Set up a scatter diagram and calculate the associated correlation coefficient. Discuss how strong you think the relationship is between the amount of money spent on television advertising and the number of orders received. Please use the Correlation procedures within Excel under Tools > Data Analysis. The Scatterplot can more easily be generated using the Chart procedure. NOTE: If you do not have the Data Analysis option under...



Answered Same DayDec 23, 2021

Answer To: Simple Regression 2. A CEO of a large plastics manufacturing company would like to determine if she...

Robert answered on Dec 23 2021
117 Votes
2. a) Scatter Plot of Number of orders v/s Advertising cost is given below:
Correlation coefficient
between number of orders and advertising cost is 0.399 which can be
considered to be very low since it is even less than 0.5. Thus through this one can say that the
relationship between amount of money spent on advertising and number of orders is not very
strong.
b) On fitting a Regression with advertising cost as dependent variable and number of orders as
independent variables we get following results:
Regression Statistics
Multiple R 0.399413
R Square 0.159531
Adjusted R Square 0.112838
Standard Error 9594.676
Observations 20
ANOVA

df SS MS F
Significance
F
Regression 1 314525132.1 3.15E+08 3.416605 0.081041
Residual 18 1657040523 92057807

Total 19 1971565655

Coefficie
nts
Standard
Error t Stat
P-
value
Lower
95%
Upper
95%
Lower
95.0%
Upper
95.0%
Intercept
57216.9
6
8170.0238
12
7.0032
8
1.54E-
06
40052.3
8...
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