Answer:
C. Test for Goodness-of-fit.
Step-by-step explanation:
C. Test for Goodness-of-fit would be most appropriate for the given situation.
A. Test Of Homogeneity.
The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.
B. Test for Independence.
The chi square is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.
C. Test for Goodness-of-fit.
The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with their estimates and provided that one degree of freedom is deducted for each parameter estimated.
Question is not proper,Proper question is given below;
Nicholas buys 3/8 pound of cheese he put the same amount of cheese and three sandwiches how much cheese does Nicholas put on each sandwich
Answer:
Nichole can put 
of cheese on each of his sandwich.
Step-by-step explanation:
Total Amount of cheese he has = 
Number of cheese sandwiches = 3
We need to find Amount of cheese does Nicole put on each of his sandwich.
Solution:
Amount of cheese on each sandwich can be calculated by dividing Total Amount of cheese he has with Number of cheese sandwiches.
framing in equation form we get;
Amount of cheese on each sandwich = 
Hence Nichole can put of cheese on each of his sandwich.
Answer:
no
Step-by-step explanation:
Answer:
174
Step-by-step explanation:
Rental Fee ÷ Cost of Ticket = # of people needed
$4000÷$23=173.9....
Can't have .9 of a ticket, so round up to 174.
# of people needed is 174
Answer: To find the critical value, follow these steps.
Compute alpha (α): α = 1 - (confidence level / 100)
Find the critical probability (p*): p* = 1 - α/2.
To express the critical value as a z-score, find the z-score having a cumulative probability equal to the critical probability (p*).
Step-by-step explanation:
