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matrenka [14]
2 years ago
7

For each set of probabilities, determine whether the events A and B are independent or dependent.

Mathematics
1 answer:
lozanna [386]2 years ago
7 0

Answers:

  • (a) Independent
  • (b) Dependent
  • (c) Dependent
  • (d) Independent

========================================================

Explanation:

If events A and B are independent, then the two following equations must both be true

  • P(A | B) = P(A)
  • P(B | A) = P(B)

This is because the conditional probability P(A|B) means "P(A) when B has happened". If B were to happen, then P(A) must be the same as before. In other words, event B does not affect A, and vice versa.

For part (a), we have P(B) = 1/4 and P(B|A) = 1/4 showing that P(B|A) = P(B) is true, and therefore we can say the events are independent. We don't need the info that P(A) = 1/8.

------------------------

Unlike part (a), part (b) has the answer "dependent" because P(A) = 1/8 and P(A | B) = 1/3 differ in value. Event A starts off at probability 1/8, but then event B occurring means P(A) gets increased to 1/3. The prior knowledge about B changes the chances of A. The P(B) = 1/5 is unneeded.

------------------------

If A and B were independent, then,

P(A and B) = P(A)*P(B)

However,

P(A)*P(B) = (1/4)*(1/5) = 1/20

which is not the same as P(A and B) = 1/6. Therefore the two events are dependent.

------------------------

Refer back to part (a)

P(A) = 1/4 and P(A|B) = 1/4 are identical in value, so P(A|B) = P(A) which leads to the events being independent. Whether we know event B happened or not, it does not affect the outcome of event A. P(B) = 1/9 is unneeded.

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answer: 4
 
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3 years ago
Let the sample size of leg strengths to be 7 and the sample mean and sample standard deviation be 630 watts and 32 watts, respec
Colt1911 [192]

Answer:

a. There is_<u><em>sufficient</em></u> evidence that the leg

C. 0.010 < P-value < 0.025

b. Power of test = 1- β=0.2066

c. So the sample size is 88

Step-by-step explanation:

We formulate the null and alternative hypotheses as

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Critical value for a right tailed test with 6 df is 1.9432

Sample Standard deviation = s= 32

Sample size= n= 7

Sample Mean =x`= 630

Degrees of freedom = df = n-1= 7-1= 6

The test statistic used here is

Z = x- x`/ s/√n

Z= 630-600 / 32 / √7

Z= 2.4797= 2.48

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so it lies between 0.010 < P-value < 0.025

b) Power of test if true strength is 610 watts.

For  a right tailed test value of z is = ± 1.645

P (type II error) β= P (Z< Z∝-x- x`/ s/√n)

Z = x- x`/ s/√n

Z= 610-630 / 32 / √7

Z=0.826

P (type II error) β= P (Z< 1.645-0.826)

= P (Z> 0.818)

= 0.7933

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(c)

true mean = 610

hypothesis mean = 600

standard deviation= 32

power = β=0.9

Z∝= 1.645

Zβ= 1.282

Sample size needed

n=( (Z∝ +Zβ )*s/ SE)²

n=  ((1.645+1.282) 32/ 10)²

Putting the values  and solving we get 87.69

So the sample size is 88

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