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

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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] · Answer 1 · 2026-02-26T06:50:35+03:00

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.

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