In: Statistics and Probability
General Instructions:
If nothing is specified, use LS = 5% and you may use MiniTab (or Excel) to perform the hypothesis test.
Question 2:
Twenty shoppers were given a standard shopping list of a limited number of items (say, twelve) for the upscale clothing requirements of one of their school-going children for the coming school year at an expensive private school. They were given the money by an independent firm called Market Research Inc. (MRI) to purchase the items for one child, from the standard list, from 4 different clothing chain stores. The dollar amount the shoppers spent on their clothing for each child, are given under the columns specifying which chain store they bought their clothing from. The MRI wanted only to compare the mean amount of money the various shoppers had spent in the 4 chain stores.
a. Clearly specify what hypothesis the MRI should test? Conduct the 4 step Hypothesis test at 1% level of significance (LS) and reach the appropriate conclusion regarding mean amounts spent in the 4 stores, by using the critical value of the statistic. You may use “SS” given by MiniTab.
b. Explain in words why, you can block the ‘Shopper’ variable. What is this method called?
c. Use the MiniTab approach to show how this can be demonstrated with appropriate analysis. Use 1% Level of Significance. (Hint: Think of ‘Shoppers’ and ‘Stores’ as the two factors.)
d. Which analysis is more appropriate? The one you did in Qu.#2 ‘c’ above or the one you did in Qu.#2 ‘a’? Explain with some specific numerical comparisons of relevant quality.
Data below..
Shopper | Store 1 | Store 2 | Store 3 | Store 4 | Explanations | |||||||
1 | 1021.05 | 1086.66 | 1050.84 | 1115.64 | ||||||||
2 | 689.31 | 738.72 | 718.29 | 1096.02 | ||||||||
3 | 605.97 | 662.58 | 631.53 | 854.01 | ||||||||
4 | 1121.04 | 1172.79 | 1145.79 | 1153.17 | Upscale School Clothing Prices at the 4 Chain Stores in $ | |||||||
5 | 611.1 | 623.43 | 637.56 | 1193.94 | ||||||||
6 | 1211.04 | 1247.04 | 1240.56 | 1152.45 | Column C1: Shopper (all shoppers have a standard list of Clothings to purchase) | |||||||
7 | 801 | 885.42 | 830.25 | 1118.61 | Column C2: Cost of Clothing on shopper's list at Store 1 | |||||||
8 | 409.77 | 469.89 | 437.67 | 587.52 | Column C3: Cost of Clothing on shopper's list at Store 2 | |||||||
9 | 985.05 | 1061.82 | 1009.53 | 1263.51 | Column C4: Cost of Clothing on shopper's list at Store 3 | |||||||
10 | 301.05 | 341.82 | 327.42 | 1171.26 | Column C5: Cost of Clothing on shopper's list at Store 4 | |||||||
11 | 1605.24 | 1625.13 | 1631.97 | 960.84 | ||||||||
12 | 250.2 | 347.4 | 275.85 | 1111.77 | Suggestion#1: Stack the Data in C2 to C5 in C9 and "ID" it in C10 for Qu.#1 | |||||||
13 | 558.72 | 625.68 | 584.73 | 676.08 | Suggestion#2: In C11 properly identify the "Which Shopper" for Qu.#2 | |||||||
14 | 232.11 | 280.17 | 260.55 | 777.51 | ||||||||
15 | 800.1 | 867.96 | 827.28 | 1313.01 | ||||||||
16 | 431.1 | 443.97 | 457.2 | 1053.54 | ||||||||
17 | 1004.13 | 1097.37 | 1029.96 | 945.09 | ||||||||
18 | 919.98 | 922.86 | 950.04 | 1309.77 | ||||||||
19 | 800.1 | 842.85 | 828.99 | 666.18 | ||||||||
20 | 830.97 | 977.67 | 858.78 | 1227.42 |
(a) H0: all store means are equal
Ha: atleast one store mean is different from others
this hypothesis testing require one way anova analysis and following information has been generated and we fail to accept H0( or reject H0) and conclude that atleast one strore is different from others , as p-value between group is less than alpha=0.05
Anova: Single Factor | ||||||
SUMMARY | ||||||
Groups | Count | Sum | Average | Variance | ||
Column 1 | 20 | 15189.03 | 759.4515 | 122819.8 | ||
Column 2 | 20 | 16321.23 | 816.0615 | 122052.4 | ||
Column 3 | 20 | 15734.79 | 786.7395 | 122955 | ||
Column 4 | 20 | 20747.34 | 1037.367 | 48141.69 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 969182 | 3 | 323060.7 | 3.106585 | 0.03138 | 2.724944 |
Within Groups | 7903408 | 76 | 103992.2 | |||
Total | 8872590 | 79 |
(b) this method is called completeley randomized design or one-way anova
(c) using both store and shoppers two factor, following information has been generated using ms-excel . It is found that both store and shoppers were significant at LS=0.05(alpha).
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
shoppers | 6058364 | 19 | 318861.2 | 9.850758 | 7.94E-12 | 1.771972 |
store | 969182 | 3 | 323060.7 | 9.980494 | 2.2E-05 | 2.766438 |
Error | 1845045 | 57 | 32369.21 | |||
Total | 8872590 | 79 |
(d) method c, is better because now store is highly signifiant as compared to in part b, aslo shoppers is also significant