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lozanna [386]
3 years ago
5

Which of the following is not true when using the confidence interval method for testing a claim about μ when σ is​ unknown? Cho

ose the correct answer below. A. The​ P-value method and the classical method are not equivalent to the confidence interval method in that they may yield different results. B. The​ P-value method, the traditional​ method, and the confidence interval method are equivalent and yield the same results. C. For a​ two-tailed hypothesis test with a 0.05 significance​ level, one must construct a​ 95% confidence interval. D. For a​ one-tailed hypothesis test with a 0.05 significance​ level, one must construct a​ 90% confidence interval.
Mathematics
1 answer:
sp2606 [1]3 years ago
3 0

Answer:

The​ P-value method and the classical method are not equivalent to the confidence interval method in that they may yield different results ( A )

Step-by-step explanation:

The False statement about using the confidence interval method when testing a claim about  μ when  σ is​ unknown is ; The​ P-value method and the classical method are not equivalent to the confidence interval method in that they may yield different results

This is because sometimes the values gotten from the p-value and confidence interval differs and this occurs mostly when the sample size is very small.

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Answer:

This question was worded weirdly however I am assuming this is algebra and you need the equation. So here- -2(q-3) or -2q+6

Step-by-step explanation:

Distribute the -2, to all the terms in the parenthesis of -2(q-3) you get 2q+6

If you didnt know how to get -2(q-3), -2(multiplied) by q-3, you use parenthesis to make sure you know it is multiplication in front of the parenthesis.

Hope I could help!

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If mean=30 and mode=15, then median​
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The median is the middle number
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4 years ago
A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with
Vikentia [17]

Answer:

(a) The PMF of <em>X</em> is: P(X=k)=(1-0.20)^{k-1}0.20;\  k=0, 1, 2, 3....

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let <em>X</em> = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable <em>X</em> follows a Geometric distribution.

The probability mass function of <em>X</em> is:

P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of <em>X</em> is:

P(X=k)=(1-0.20)^{k-1}0.20;\  k=0, 1, 2, 3....

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

              = 1 - P (X = 1) - P (X = 2)

              =1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

E(X)=\frac{1}{p}

Compute the expected number of opponents contested in a game as follows:

E(X)=\frac{1}{p}=\frac{1}{0.20}=5

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

              = 1 - P (X = 1) - P (X = 2) - P (X = 3)

              =1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2

Thus, the expected number of game plays until a player contests four or more opponents is 2.

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Answer:

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Step-by-step explanation:

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