In: Statistics and Probability
A cereal manufacturer is planning to develop a new product. The product does not require adding milk because each piece has milk substitute inside of it. All the consumer needs to do is add water. The product is scheduled to be rolled out nationally next year. The marketing manager has to decide how to market the new product. There are three different marketing strategies that the firm is looking at. They want to emphasize all three points, but they also realize that they need to choose one as the primary hook before the national launch. The product is presently being sold in only the Columbus, Detroit and Pittsburgh markets. The marketing manager decides to launch three different advertising campaigns. In the Columbus market, convenience will be the main hook. In the Detroit market, quality will be the main hook. In the Pittsburgh market, the product's low price, due to the fact that no milk purchase is necessary, will be the main hook. Weekly sales were recorded for 20 weeks following the beginning of the campaigns. The data are below. (Answers must use excel). (A) State the null hypothesis and the alternative hypothesis. (B) Run the appropriate test to see if one campaign differs from the others. (C) What do you find? (D) If you find that there is a difference, in which pair(s) is it located? Bold and format with two digits after the decimal point the cells you looked at to make this decision. This will require running a test for each possible pair of data.
Convenience | Quality | Price |
529 | 804 | 630 |
658 | 630 | 531 |
793 | 774 | 443 |
514 | 717 | 526 |
663 | 679 | 602 |
719 | 604 | 502 |
711 | 620 | 619 |
606 | 697 | 689 |
461 | 706 | 675 |
529 | 615 | 512 |
498 | 492 | 691 |
663 | 719 | 733 |
604 | 787 | 628 |
495 | 699 | 776 |
485 | 572 | 561 |
557 | 523 | 572 |
353 | 584 | 469 |
557 | 634 | 581 |
542 | 580 | 639 |
614 | 624 | 532 |
a)
As we want to see if the weekly sale is different for at least one of the state from the others, so the appropriate test would be -
Null Hypothesis - H0: All the states have same population mean of weekly sales.
Alternate Hypothesis - H1: At least one of the state has different population mean of weekly sales from the others.
Symbolically, it can be written as -
Null Hypothesis - H0:
Alternate Hypothesis - H1: for at least one pair of (i, j), .
Where , and are the population mean of weekly sales for the states Columbus, Detroit and Pittsburgh respectively.
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b)
We would need to conduct an ANOVA test for this hypothesis. We can use Excel's 'Data Analysis' toolpak to access the 'Anova: Single Factor' option as shown -
Then, you enter the field values as shown -
Make sure to mark the check box for 'Labels in First Row' as we have labels in first row.
Output obtained from excel is as shown -
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Convenience | 20 | 11551 | 577.55 | 10775.00 | ||
Quality | 20 | 13060 | 653 | 7238.11 | ||
Price | 20 | 11911 | 595.55 | 7913.21 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 62114.7 | 2 | 31057.35 | 3.59 | 0.03 | 3.16 |
Within Groups | 492599.9 | 57 | 8642.10 | |||
Total | 554714.6 | 59 | ||||
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C)
As the p-value of test = 0.03, is less than significance level of = 0.05, so we reject the null hypothesis and conclude that there is enough evidence in the data to support the claim that the population mean of at least one state is different from the remaining two.
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D) To further check which two population means are different from each other, we need to perform a post-hoc test.
Tukey's HSD test can be conducted for difference of means using the test statistic critical value 'q' = 3.40-
Note that, you can get the critical value of 'd' from Tukey's critical value table corresponding to 3 treatments and error degrees of freedom = 57.
Following table shows comparison of differences of mean from the HSD value and conclusion -
Difference | Value | Critical Value | Decision |
|| | 75.45 | 70.6763 | Significant |
|| | 57.45 | 70.6763 | Not Significant |
|| | 18 | 70.6763 | Not Significant |
As the difference between the means of population 1 (Columbus - Convenience strategy) and population 2 (Detroit - Quality Strategy) is greater than the critical value of test statistic, so this difference is significant.
And as the average value of 'Quality' strategy = 653 is more than the average value of 'Convenience' strategy = 577.55, so we can say that the most effective strategy is the 'Quality' strategy that was used in Detroit.
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