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strojnjashka [21]
3 years ago
7

Classify the two given samples as independent or dependent. Sample 1: Pre-training blood pressure of 17 people Sample 2: Post-tr

aining blood pressure of 17 people a. dependent b. independent
Mathematics
1 answer:
Damm [24]3 years ago
7 0

Answer:

Sample  1 : independent

Sample  2 :   dependent

Step-by-step explanation:

From the question we are told that

   The  given date is  

                   Sample 1: Pre-training blood pressure of 17 people

                    Sample 2: Post-training blood pressure of 17 people

Now  Sample 1 is independent because  the Pre-training blood pressure of 17 people does not depend of Post-training blood pressure of 17 people

While  

        Sample 2  is  dependent because the Post-training blood pressure of 17 people is  dependent on the Pre-training blood pressure of 17 people

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Evaluate the expression m÷n for m=2 and n=14.
Sever21 [200]

Answer:

1/7

Step-by-step explanation:

Plug in 2 for m, and 14 for n in the expression:

m/n = 2/14

Simplify. Factor out common factors from both the numerator and denominator:

(2/14)/(2/2) = 1/7

1/7 is your answer.

~

3 0
3 years ago
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A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
weqwewe [10]

Answer:

99% confidence interval for the given specimen is [3.4125 , 3.4155].

Step-by-step explanation:

We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

Three weighing of a specimen on this scale give 3.412, 3.416, and 3.414 g.

Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

            \sigma = population standard deviation = 0.001 g

            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

                                             = [ 3.414-2.5758 \times {\frac{0.001}{\sqrt{3} } } , 3.414+2.5758 \times {\frac{0.001}{\sqrt{3} } } ]

                                             = [3.4125 , 3.4155]

Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

6 0
3 years ago
Adam drew two same size rectangles and divided in to the same number of equal parts. He shaded 1/3 of one rectangle and 1/4 of t
Eva8 [605]
It is at least seven or eight parts

5 0
3 years ago
Describe how to determine if a number is a solution to an equation.
marin [14]

Step-by-step explanation:

Determine whether a number is a solution to an equation.

Substitute the number for the variable in the equation.

Simplify the expressions on both sides of the equation.

Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.

Hoped that helped:P

5 0
2 years ago
Read 2 more answers
suppose you performed a regression analysis. you were given four observations of the target at [1.0, 1.5, 2.8, 3.7], and you pre
Arlecino [84]

Suppose you performed a regression analysis. The mse for this scenario is 0.105

Regression is a statistical method used in finance, making an investment, and different disciplines that attempt to determine the electricity and man or woman of the relationship between one established variable (commonly denoted through Y) and a sequence of different variables (called independent variables).

We are able to say that age and peak can be described through the usage of a linear regression version. because someone's peak will increase as age will increase, they have got a linear courting. Regression fashions are commonly used as statistical proof of claims regarding regular statistics.

"Regression" comes from "regress" which in turn comes from Latin "regresses" - to head returned (to something). In that feel, regression is the approach that permits "to head again" from messy, hard-to-interpret data, to a clearer and more significant version.

y ypred (y-ypred)^2

1 1.1 0.01

1.5 1.3 0.04

2.8 3.2 0.16

3.7 3.7 0

The error sum of the square is given by

ESS = (y- )

ESS=0.21

The mean square error is given by

ESS MSE = ESS/dfe

MSE = \frac{0.21}{2}

MSE = 0.105

Learn more about regression here brainly.com/question/26755306

#SPJ4

6 0
2 years ago
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