A sample of 15 consumers provided the following product ratings for three different products. Five consumers were randomly assig
1 answer:
- The value of the test statistic is equal to 9.98.
- There's sufficient statistical evidence at a 5% level of significance to state that there's a significant difference among the ratings for the products A, B, and C.
The Kruskal-Wallis test is also referred to as one-way ANOVA on ranks or Kruskal-Wallis H test and it can be defined as a rank-based nonparametric test that is typically used to determine whether or not there are statistically significant differences between two (2) or more samples of an independent variable on either a continuous or ordinal dependent variable.
This ultimately implies that, a Kruskal-Wallis test can be used to determine whether two (2) or more samples originate from the same distribution.
Based on the given sample, the hypothesis is given by:
- H₀: Ratings are the same for all the three products.
- Ha: There are differences among the ratings for the products.
For this exercise, we would use R-studio to perform the Kruskal-Wallis test as follows:
df₁ = read.table (header = TRUE, text ="
product Ratings
A 55
A 64
A 79
A 41
A 69
B 85
B 99
B 91
B 88
B 95
C 65
C 49
C 35
C 51
C 58
")
DF₁ kruskal.test(Ratings ~ product, data = df1)
Based on the Kruskal-Wallis test, we obtained the following results (outputs):
Kruskal-Wallis chi-squared = 9.98.
Degrees of freedom, DF = 2.
P-value = 0.006806.
Next, we would determine the test statistic as follows:
Test statistic = Kruskal-Wallis chi-squared = 9.98.
P-value = 0.007.
Therefore, the p-value (0.007) is less than α = 0.05. Based on this, we should reject the null hypothesis (H₀) and accept the alternate hypothesis (Ha).
In conclusion, we can infer and logically deduce that there's sufficient statistical evidence at a 5% level of significance to state that there's a significant difference among the ratings for the products A, B, and C.
Read more on null hypothesis here: brainly.com/question/14913351
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Answer:
Annual demand (U) = 90.000 bags
Cost of each bag = $1.50
Inventory carrying cost per unit(C) = $1.50 × 20% = 0 30
Ordering cost per unit (O) = $15
Part A)
EOQ = 3,000
Part B)
Maximum inventory = EOQ + Safety inventory on hand
Maximum inventory = 3000 + 1000
Maximum inventory = 4.000
Part C)
Average inventory = Maximum inventory + Minimum or Safety /2
Average inventory = 4,000 + 1,000 / 2
Average inventory =2,500
Part D)
How often company order = Annual demand / EOQ
How often company order = 90,000 / 3.000
How often company order = 30