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Free_Kalibri [48]
2 years ago
15

If testing the claim that sigma subscript 1 superscript 2 baseline not equals sigma subscript 2 superscript 2σ21≠σ22​, what do w

e know about the two samples if the test statistic is fequals=​1?
Mathematics
1 answer:
12345 [234]2 years ago
6 0

Answer:

The statistic for this system of hypothesis is given by:

F=\frac{s^2_1}{s^2_2}

If the statistic is equal to 1 then that means s^2_1 = s^2_2 and we don't have enough evidence to conclude that the two population variances and deviations are different.

Step-by-step explanation:

System of hypothesis

We want to test if the variation for a group1 is equal to another one 2, so the system of hypothesis are:

H0: \sigma^2_1 = \sigma^2_2

H1: \sigma^2_1 \neq \sigma^2_2

Calculate the statistic

The statistic for this system of hypothesis is given by:

F=\frac{s^2_1}{s^2_2}

If the statistic is equal to 1 then that means s^2_1 = s^2_2 and we don't have enough evidence to conclude that the two population variances and deviations are different.

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the answer should be 780π

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Explain how to write and evaluate an algebraic expression that represents the situation. The blueberries were divided equally in
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The variable is the number of blueberries, the operation is division, and the constant is 6. Write the division expression as the fraction, b/6 . To evaluate, substitute 210 in for the variable, b, and simplify. 210 divided by 6 is 35. So each bowl has 35 blueberries.





4 0
2 years ago
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Suppose that Kevin can choose to get home from work by car or bus. When he chooses to get home by car, he arrives home after 7 p
astraxan [27]

Answer:

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = 0.838

Step-by-step explanation:

Let the probability that Kevin arrives home after 7 pm be P(L)

Probability that Kevin uses the bus = P(B)

Probability that Kevin uses the car = P(C)

Probability of arriving home after 7 pm if the car was taken = P(L|C) = 4% = 0.04

Probability of arriving home after 7 pm if the bus was taken = P(L|B) = 15% = 0.15

The bus is cheaper, So, he uses the bus 58% of the time.

P(B) = 58% = 0.58

P(C) = P(B') = 1 - P(B) = 1 - 0.58 = 0.42

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm = P(B|L)

The conditional probability P(A|B) is given mathematically as

P(A|B) = P(A n B) ÷ P(B)

Hence, the required probability, P(B|L) is given as

P(B|L) = P(B n L) ÷ P(L)

But we do not have any of P(B n L) and P(L)

Although, we can obtain these probabilities from the already given probabilities

P(L|C) = 0.04

P(L|B) = 0.15

P(B) = 0.58

P(C) = 0.42

P(L|C) = P(L n C) ÷ P(C)

P(L n C) = P(L|C) × P(C) = 0.04 × 0.42 = 0.0168

P(L|B) = P(L n B) ÷ P(B)

P(L n B) = P(L|B) × P(B) = 0.15 × 0.58 = 0.087

P(L) = P(L n C) + P(L n B) = 0.0168 + 0.087 = 0.1038 (Since the bus and the car are the two only options)

The approximate probability that Kevin chose to get home from work by bus, given that he arrived home after 7 pm

= P(B|L) = P(B n L) ÷ P(L)

P(B n L) = P(L n B) = 0.087

P(L) = 0.1038

P(B|L) = (0.087/0.1038) = 0.838150289 = 0.838

Hope this Helps!!!

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Step-by-step explanation:

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