Answer:
Step-by-step explanation:
Hello!
The variable of study is
X: number of complaints per industry (Categorized: Bank, Cable, Car, Cell, Collection
Considering this is a categorical variable, and the hypothesis is that all categories have the same probability, you have to apply a Chi-Square Goodness to Fit test.
Observed frequencies per category
1) Bank: 26
2) Cable: 44
3) Car: 42
4) Cell: 60
5) Collection: 28
Total= 200
Statistical hypotheses:
H₀: P₁=P₂=P₃=P₄=P₅=1/5
H₁: At least one of the proposed proportions is different.
α: 0.01

For this test the formula for the expected frequencies is:

So the expected values for each category is:






This test is one tailed and so is its p-value, under a Chi-square with 4 degrees of freedom p-value: 0.021484.
The p-value is less than the significance level so the decision is to reject the null hypothesis.
c. The industry with most complaints is the cellular phone providers
I hope this helps!
Answer:
D
Step-by-step explanation:
-18/-9
= 2
:. -8/-4
= 2
The answer is -50 because -5 squared is 25 and that times 3 is 75 -5 cubed is -150 and -150 + 75 is 50