Suppose that a researcher using data on class size (CS) and average test scores from 100 thirdgrade classes, estimates the following OLS regression. Test̂Score = 520.4 − 5.82 × CS n = 100, R 2 = 0.08, SER = 11.5
a. A class has 22 students. What is the regression’s prediction for this classroom’s average test score?
b. Last year, a class had 19 students, and this year it has 23 students. What is the regression’s prediction for the change in the classroom average test score?
c. The sample average class size across the 100 classrooms is 21.4. What is the sample average of the test scores across the 100 classrooms?
d. What is the sample standard deviation of test scores across the 100 classrooms?

Answers

Answer 1

This question is incomplete, the complete question is;

Suppose that a researcher using data on class size (CS) and average test scores from 100 third grade classes, estimates the following OLS regression. Test-Score = 520.4 - 5.82 × CS, n = 100, R² = 0.08, SER = 11.5

a. A class has 22 students. What is the regression's prediction for this classroom's average test score?

b. Last year, a class had 19 students, and this year it has 23 students. What is the regression's prediction for the change in the classroom average test score?

c. The sample average class size across the 100 classrooms is 21.4. What is the sample average of the test scores across the 100 classrooms?

d. What is the sample standard deviation of test scores across the 100 classrooms?

Answer:

a)

the regression's prediction for this classroom's average test score is 392.36

b)

the regression's prediction for the change in the classroom average test score is -23.28

c)

the sample average of the test scores across the 100 classrooms is 395.852

d)

the sample standard deviation of test scores across the 100 classrooms is 11.92887

Step-by-step explanation:

Given the data in the question;

Test-Score = 520.4 - 5.82 × CS, n = 100, R² = 0.08, SER = 11.5 -----1

the general formula for the average test score is as follows;

Test score = ^β₀ + ( ^β₁ × CS ) -------- 2

the general for change in test score ;

ΔTest Score = β[tex]_{ class-size[/tex] × ΔClass size  -------- 3

General formula for the sum of squared residuals SSR

SSR = ( n - 2 ) SER² ----- 4

General formula for total sum of squares TSS

TSS = SRR / 1 - R² -------- 5

General formula for sample standard deviation;

Sy = √(TSS / (n-1) )  ------ 6

now, from the given formula;

^β₀ = 520.4

aslo, β[tex]_{ class-size[/tex] = ^β₁ =  - 5.82

so

a) A class has 22 students. What is the regression's prediction for this classroom's average test score?

given that class size CS is 22, to get  the regression's prediction for this classroom's average test score, we make use of formula 2 above;

Test score = ^β₀ + ( ^β₁ × CS )

so we substitute

Test score = 520.4 + ( -5.82  × 22 )

Test score = 520.4 + ( - 128.04 )

Test score = 520.4 - 128.04

Test score = 392.36

Therefore,  the regression's prediction for this classroom's average test score is 392.36

b) Last year, a class had 19 students, and this year it has 23 students. What is the regression's prediction for the change in the classroom average test score.

we make use of formula 3 above

ΔTest Score = β[tex]_{ class-size[/tex] × ΔClass size

we substitute

ΔTest Score = -5.82 × ( 23 - 19 )

ΔTest Score = -5.82 × 4

ΔTest Score = -23.28

Therefore,  the regression's prediction for the change in the classroom average test score is -23.28

c) The sample average class size across the 100 classrooms is 21.4. What is the sample average of the test scores across the 100 classrooms?

we make use of formula 2 above;

Test score = ^β₀ + ( ^β₁ × CS )

we substitute

Test score = 520.4 + ( -5.82 × 21.4 )

Test score = 520.4 + ( -124.548 )

Test score = 520.4 - 124.548

Test score = 395.852

Therefore, the sample average of the test scores across the 100 classrooms is 395.852

d) What is the sample standard deviation of test scores across the 100 classrooms.

first we make use of formula 4 above; to calculate the sum of squared residuals SSR

SSR = ( n - 2 ) SER²

we substitute

SSR = ( 100 - 2 ) (11.5)²

SSR = 98 × 132.25

SSR = 12,960.5

Also, for total sum of squares TSS, we use formula 5

TSS = SRR / 1 - R²

we that R² = 0.08; from the given formula

so we substitute

TSS = 12,960.5 / 1 - 0.08

TSS = 12,960.5 / 0.92

TSS = 14087.5

so, the sample standard deviation will be;

from formula 6 above

Sy = √(TSS / (n-1) )

we substitute

Sy = √(14087.5 / (100-1) )

Sy = √(14087.5 / 99)

Sy = √142.297979

Sy = 11.92887

Therefore, the sample standard deviation of test scores across the 100 classrooms is 11.92887


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A.)

H0: μ ≤ 31

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Right tailed test

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Supplementary means the two angles add up to 180 degrees.  Now this turns into a system of equations problem.  let's call the two angles a and b.

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a = 2b - 45

Now this is actually super easy to do.  Normally you would have to do algebra to one of the two equations so there is one variable by itself.  the second equation already has that.

now look at the other equation.  instead of a though make a = 2b - 45

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2b - 45 + b = 180

Now use algebra to solve for b

2b - 45 + b = 180

b = 75

Now, go back to the first two equations.  put 75 in for b for either one of them and you will get the answer to a.

a + 75 = 180

a = 105

or

a = 2*75 - 45

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Now we can check.  a=105 and b = 75

do both add up to 180?  yep.  is a 45 less than twice b?  or using numbers, is 105 45 less than twice 75?  well twice 75 is 150.  and 105 is 45 less than that, so yep!

Let me know if there was anything you didn't get.  Also you could totally have started with the other equation first and got a = 180 - b or b = 180 - a and gotten the same answer.

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