Ask question

Ask question

All categories

3 years ago

6

Two studies were done on the same set of data, where study I was a one-sided test and study II was a two-sided test. The p-value

of the test corresponding to study I was found to be 0.030. What is the p-value for study II?

1 answer:

mixer [17]3 years ago

4 0

Answer:

$0.060$

Step-by-step explanation:

In a two tailed test the probability of occurrence is the total area under the critical range of values on both the sides of the curve (negative side and positive side)

Thus, the probability values for a two tailed test as compared to a one tailed test is given by the under given relation -

$p-value = P(Z< -\frac{\alpha }{2} )+P(Z >\frac{\alpha}{2})$ \

Here $P\frac{\alpha}{2} = 0.030$

Substituting the given value in above equation, we get -

probability values for a two tailed test

= $0.030 + 0.030\\= 0.060$

You might be interested in

What is the slope of the line that passes through the points E(5, 1) and F(2-7)?<br> 12 Points)

xxTIMURxx [149]

Answer:

8/3

Step-by-step explanation:

the slope is 8/3 i am sure

5 0

3 years ago

Determine the distance between each pair of points. Then determine the coordinates of the midpoint M of the segment joining the

dalvyx [7]

The midpoint is at (3/10 , 3/2 , 4/10)

In the given statement is

To find the midpoint of the line segment joining the pair of points

F (3/5, 0, 4/5) and G(0,3,0)

Midpoint:

The coordinates of the midpoint of a line segment are the average of the coordinates of the end points.

m = (A +B)/2

If we are given three points and we wish to find the midpoint of those points, we need to use the midpoint formula m =( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2} , \frac{z_{1}+z_{2} }{2}$ )

where:

(x1 , y1 ,z1) are the coordinates of the first point:

(x2 , y2 , z2) are the coordinates of the second point:

Now, We are given the three points F (3/5, 0, 4/5) and G(0,3,0)

Solving for the midpoint , we have,

m = ( $\frac{3/5+0}{2} , \frac{0+3}{2} ,\frac{4/5+0}{2}$ )

m = (3/10 , 3/2 , 4/10)

Therefore, the midpoint is at (3/10 , 3/2 , 4/10)

Learn more about Midpoint at :

brainly.com/question/14547479

#SPJ4

5 0

1 year ago

Please help ill give brainliest

Sveta_85 [38]

Answer:

$\sf A . \sf \frac{3}{-7}$

$\sf E. \sf -( \frac{-3}{-7} )$

This above options makes the expression true.

The negatives above can be distributed up as well, which happened in A.

In E, when the negatives inside the bracket cancels out, the negative outside the brackets multiplies and makes the expression true.

In B the expression turns positive, thus false same happens with C and D

3 0

2 years ago

It takes 19 minutes to get from Yermin’s house to the beach. If she has traveled for 5 minutes already, how many more minutes wi

Bezzdna [24]

it will be 14 more minuted because she has completed 5/19 minutes it takes to reach the beach.. we can get this answer by taking the total number of minuted to get to the beach (19) and subtract the number of minutes she has already walked (5)..... our answer is that she has 14 minutes left till she reaches the beach.

8 0

3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

4 0

3 years ago