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SashulF [63]
3 years ago
11

In an experiment to see if light affected the amount of interaction, one group of students was put in a lighted room for an hour

, and another group was put in a room that was totally dark. Cameras recorded the n in each group. The independent variable in this experiment was?
Mathematics
2 answers:
Ludmilka [50]3 years ago
8 0
The Amount of light. 
dimulka [17.4K]3 years ago
4 0

Answer:

The amount of light is an independent variable.

Step-by-step explanation:

Consider the provided information.

The independent variable is the variable that the experimenter changes or controls to test the dependent variable impacts.

A dependent variable is the variable that depends on the independent variable and that the experimenter is testing and measuring.

For example:

In a study to determine whether how long you practice affects your game, the independent variable is the time you spent on practice while the dependent variable is the affects on your game.

Now consider the provided information.

In an experiment to see if light affected the amount of interaction, one group of students was put in a lighted room for an hour, and another group was put in a room that was totally dark. Cameras recorded the n in each group.

Here, the light is controlled by the experimenter but affect of interaction is dependents on light.

Thus, the amount of light is an independent variable.

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A coach is assessing the correlation between the number of hours spent practicing and the average number of points scored in a g
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Answer:

a) r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1  

We have a perfect linear relationship between the two variables

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Nowe we can find the means for x and y like this:  

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\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17  

And we can find the intercept using this:  

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So the line would be given by:  

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And the intercept 5 represent the minimum score expected for any game

Step-by-step explanation:

We have the following data:

Number of hours spent practicing (x) 0 0.5 1 1.5 2 2.5 3 3.5 4

Score in the game (y) 5 8 11 14 17 20 23 26 29

Part a

The correlation coefficient is given:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=9 \sum x = 18, \sum y = 153, \sum xy = 396, \sum x^2 =51, \sum y^2 =3141  

r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1  

We have a perfect linear relationship between the two variables

Part b

m=\frac{S_{xy}}{S_{xx}}  

Where:  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}  

With these we can find the sums:  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=51-\frac{18^2}{9}=15  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=396-\frac{18*153}{9}=90  

And the slope would be:  

m=\frac{90}{15}=6  

Nowe we can find the means for x and y like this:  

\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2  

\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17  

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b=\bar y -m \bar x=17-(6*2)=5  

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Part c

For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

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