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Softa [21]
3 years ago
6

Consider the hypothesis test H0: μ1= μ2 against H1: μ1.Suppose that the sample sizes aren1 = 15 and n2 =μ2. Assume thatσ21 = σ22

andthat the data are drawn from normal distributions. Use α =0.05.
(a) Test the hypothesis and find the P-value.

(b) Explain how the test could be conducted with a confidenceinterval.

(c) What is the power of the test in part (a) for a truedifference in means of 3?

(d) Assuming equal sample sizes, what sample size should beused to obtain β = 0.05 if the true difference in means is -2?Assume that α = 0.05.

Mathematics
1 answer:
torisob [31]3 years ago
8 0

Answer:

see attachments

Step-by-step explanation:

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A study conducted at a certain high school shows that 72% of its graduates enroll at a college. Find the probability that among
mixas84 [53]

Answer:

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

And we can use the probability mass function and we got:

P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615

And replacing we got:

P(X \geq 1) = 1-0.00615 = 0.99385

Step-by-step explanation:

Let X the random variable of interest "number of graduates who enroll in college", on this case we now that:  

X \sim Binom(n=4, p=0.72)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

We want to find the following probability:

P(X \geq 1)

And we can use the complement rule and we got:

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

And we can use the probability mass function and we got:

P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615

And replacing we got:

P(X \geq 1) = 1-0.00615 = 0.99385

5 0
3 years ago
PLEASE PLEASE PLEASE HELP ME. IM STUCK IN A SUMMER SCHOOL AND IF I DONT GET GOOD GRADES ILL GET IN BIG TROUBLE AND IM TOTALLY LO
Mrrafil [7]

Answer:

We know that in the box there are:

4 twix

3 kit-kat

Then the total number of candy in the box is:

4 +3 = 7

a)

Here we want to find the probability that we draw two twix.

All the candy has the same probability of being drawn from the box.

So, the probability of getting a twix in the first drawn, is equal to the quotient between the number of twix and the total number of candy in the box, this is:

p = 4/7

Now for the second draw, we do the same, but because we have already drawn one twix before, now the number of twix in the box is 3, and the total number of candy in the box is 6.

this time the probability is:

q = 3/6 = 1/2

The joint probability is the product of the individual probabilities, so here we have

P = p*q = (4/7)*(1/2) =  2/7

b) same reasoning than in the previous case:

For the first bar, the probability is:

p = 3/7

for the second bar, the probability is:

q = 2/6 = 1/3

The joint probability is:

P = p*q = (3/7)*(1/3) = 1/7

c) Suppose that first we draw a twix.

The probability we already know that is:

p = 4/7

Now we want another type, so we need to draw a kit-kat, the probability will be equal to the quotient between the remaining kit-kat bars (3) and the total number of candy in the box (6)

q = 3/6

The joint probability is:

P = p*q = (4/7)*(3/6) = 2/7

But, we also have the case where we first draw a kit-kat and after a twix, so we have a permutation of two, then the probability in this case is:

Probability = 2*P = 2*2/7 = 4/7

3 0
3 years ago
Please help me out with this so I can keep my kitten.
notka56 [123]

Answer:

y = \frac{32}{3}

Step-by-step explanation:

Firstly, move over the negative 3/4 fraction (don't forget to swap the operation i.e subtract to add):

\frac{y}{8} = \frac{7}{12}  + \frac{3}{4}

Now, to add the two fractions, simply multiply the numerator and denominator by 3:

\frac{3*3}{4*3} = \frac{9}{12}

Now add this to the other fraction:

\frac{9}{12} + \frac{7}{12} = \frac{16}{12}

This can be simplified down by dividing both the numerator and denominator by 4:

\frac{4}{3}

Which now simplifies the original equation to:

\frac{y}{8}  = \frac{4}{3}

Remove the y out of the fraction:

\frac{1}{8}y = \frac{4}{3}

Now multiply both sides by 8:

(\frac{1}{8}y) * 8 = (\frac{4}{3}) * 8

y = \frac{4*8}{3}

y = \frac{32}{3}

Hope this helps!

5 0
3 years ago
Read 2 more answers
IM GIVING BRAINLIEST!!!PLEASE HELP!!!I JUST KNOW ITS NOT A!!!
zmey [24]

Answer:

C. In y^2-y=6, 6 should have been subtracted on both sides first.

Step-by-step explanation:

It is a quadratic equation so you subtract the 6 and create

y^2-y-6=0

Then you factor it to:

(y-3)(y+2)=0

The solutions then are:

y-3=0, y=3 and y+2=0, y=-2

{-2,3}

4 0
3 years ago
Who is this?
Nastasia [14]

Answer

ALL OF THE ABOVE

8 0
2 years ago
Read 2 more answers
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