Answer:
A.
This assumption is that the distribution of polyunsaturated fat % of each if these four regimes must be with equal variances as well as uniform.
B.
The null hypothesis
H0: μ1 = μ2 = μ3 = μ4
The alternate hypothesis:
H1: at least 2 means are unequal
C.
First we calculate the grand mean
= 1/54[15(42.9)+17(43.1)+8(43.7)+14(43.9)]
= (643.5 +732.7 + 349.6 + 614.6)/54
= 2340.4/54
= 43.341
Sum of squared treatment
= [15(42.9-43.341)²+17(43.1-43.341)²+8(43.7-43.341)²+14(43.9-43.341)²]
= 9.3104
Mean square of treatment
= SST/I-1
= 9.3104/4-1
= 9.3104/3
= 3.1035
Error sum of squared
= (15-1)*(1.3)² + (17-1)*(1.5)² + (8-1)*(1.2)² + (14-1)*(1.2)²
= 88.46
Error mean square
MSE = 88.46/54-4
= 1.7692
Test statistic
= 3.1035/1.7692
= 1.75
Answer:
f = 5g/n^3 - 5h/n63
Step-by-step explanation:
Well, I believe one ordered pair that is a solution to the equation would be P(3,1). Since by plugging in the X and y values into the equation the left hand side is equal to the right hand side.
Y = X + 2/5
1 = 3+2/5
1 = 5/5
1 = 1.
w/5+11=41
subtract -11 -11
w/5=30
multiply to get rid of the divide w/5=30
the answer would be 150
A binomial probability density function should be used to represent the probability
<h3>How to determine the type of
probability density?</h3>
The given parameters are:
- Proportion that plays sport, p = 32%
- Number of students selected, p = 50
- The probability, P = (x ≤ 15)
The proportion that plays sport indicates that
68% of the students do not play sport
So, we have two events, which are
- Play sport
- Do not play sport
When there are two possible events, then the binomial probability density function should be used
Read more about binomial probability density at:
brainly.com/question/15246027
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