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Nikitich
1 month ago
9

One hundred teachers attended a seminar on mathematical problem solving. The attitudes of representative sample of 12 of the tea

chers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The twelve change scores are as follows...... 4; 7; -1; 1; 0; 5; -2; 2; -1; 6; 5; -3
What is the mean change score? (Round your answer to two decimal places.)
What is the standard deviation for this population? (Round your answer to two decimal places.)
What is the median change score? (Round your answer to one decimal place.)
Find the change score that is 2.2 standard deviations below the mean. (Round your answer to one decimal place.)
Mathematics
1 answer:
Svet_ta [12.7K]1 month ago
4 0

Answer:

a) 1.92

b) 3.25

c) 1.5

d) -5.23

Step-by-step explanation:

We are provided with the following in the question:

4, 7, -1, 1, 0, 5, -2, 2, -1, 6, 5, -3

a) mean change score

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{23}{12} = 1.92

b) standard deviation for the population

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n}}

where x_i represents data points, \bar{x} is the average, and n denotes the count of observations.

Sum of squared deviations = 126.92

\sigma = \sqrt{\frac{126.92}{12}} = 3.25

c) median change score

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Arranged data: -3, -2, -1, -1, 0, 1, 2, 4, 5, 5, 6, 7

Median equal to

\dfrac{6^{th} + 7^{th}}{2} = \dfrac{1+2}{2} = 1.5

d) change score that is 2.2 standard deviations lower than the mean.

x = \mu - 2.2(\sigma)\\x = 1.92-2.2(3.25)\\x = -5.23

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A) The product of x and y equals 196. Thus, A) x is equal to 196 divided by y.

B) The sum of x and y is 35. By substituting A) into B), we have
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Arrange the entries of matrix A in increasing order of their cofactors values
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To ascertain the cofactor of

A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]

First, eliminate the relevant row and column linked to the specified entries and calculate the determinant of the leftover 2\times 2 matrix while applying alternating signs.


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Ac_{11}=4\times 1- -1\times 2


Ac_{11}=4+ 2

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Ac_{12}=-(-7\times 1- -1\times -8)


Ac_{12}=-(-7- 8)

Ac_{12}=15




Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|


Ac_{21}=-(5\times 1- 3\times 2)


Ac_{21}=-(5-6)


Ac_{21}=1







A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|


Ac_{23}=-(7\times 2 -8\times 5)


Ac_{23}=-(14-40)


Ac_{23}=26




A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|


Ac_{31}=5\times -1 -4\times 3


Ac_{31}=-5-12


Ac_{31}=-17


A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|


Ac_{33}=7\times 4- -7\times 5


Ac_{33}=28+35


Ac_{33}=63


Therefore, the entries arranged in increasing order of their cofactors values are;

Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63



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