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solong [7]
2 years ago
5

what is the critical f-value when the sample size for the numerator is six and the sample size for the denominator is four? use

a two-tailed test and the 0.10 significance level. (round your answer to 2 decimal places.)
Mathematics
1 answer:
lubasha [3.4K]2 years ago
3 0

The critical f-value when the sample size for the numerator and the  denominator is given using a two-tailed test will be 0.12 and 14.88.

We have,

The sample size for the numerator = 6,

i.e.

Numerator degree of freedom = 6 - 1 = 5,

And,

The sample size for the denominator = 4,

i.e.

Denominator degree of freedom = 4 - 1 = 3

And,

We have to use a two-tailed test,

So,

The significance level = \frac{0.10}{2} = 0.05

And,

Now,

The test follows the F-distribution with (5, 3) degrees of freedom,

Now,

Using the F-table at significance level 0.05 and check for given degrees of freedom,

We get,

Critical f-values : 0.12 and 14.88

Hence, we can say that the critical f-value when the sample size for the numerator and the  denominator is given using a two-tailed test will be 0.12 and 14.88.

Learn more about critical f-value click here:

brainly.com/question/17164520

#SPJ4

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Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that p = 0.905

You and 2 friends:

So 3 people in total, which means that n = 3

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

P(X \geq 1) = 1 - P(X = 0)

With what we found in b:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

p = 0.905*0.095*0.095 = 0.0082

0.0082 = 0.82% probability that only your order will be filled correctly

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3 years ago
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