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IRINA_888 [86]
2 years ago
7

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua

tion. The mean GPA was 2.9 with a standard deviation of 0.4. Starting salaries had a mean of $47200 with a SD of $8500. The correlation between the two variables was r = 0.72. The association appeared to be linear in the scatterplot. a) Write an equation of the model that can predict salary based on GPA and find the slope and intercept. b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning? c) What proportion of the variation in salaries is explained by variation in GPA?
Mathematics
1 answer:
frez [133]2 years ago
8 0

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a)y = a+bx

b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} =  \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300

a=y-bx = 47200-(15300 \times 29) = 2830

So, slope =  15300

Intercept =  2830

So, equation : y = 2830+15300x

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

y = 2830+15300 \times 3.3 = 53320

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = r^2 = (0.72)^2 =0.5184

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2 years ago
2 divided by negative 1/5
oksian1 [2.3K]

Answer: 10

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4 0
2 years ago
A linear regression model is fitted to the data x y 37.0 65.0 36.4 67.2 35.8 70.3 34.3 71.9 33.7 73.8 32.1 75.7 31.5 77.9 with x
andrey2020 [161]

Answer:

b0= 144.59

b= -2.12

Se²= 1.02

99%CI E(Y/X=35): [68.78; 71.99]

Step-by-step explanation:

Hello!

I've arranged the given data:

X: 37.0, 36.4, 35.8, 34.3, 33.7, 32.1, 31.5

Y: 65.0, 67.2, 70.3, 71.9, 73.8, 75.7, 77.9

The equation of the linear regression model is:

Yi= β₀ + βXi + εi

Where

Yi is the dependent variable

Xi is the independent variable

εi represents the errors or residues

β₀ is the intercept of the line

β is the slope

The conditions to make a linear regression analysis are:

For each given value of X, there is a population of Y~N(μy;σy²)

Each value of Y is independent of the others.

The population variances of each population of Y are equal.

From these conditions the following characteristic is deduced:

εi~N(0;σ²)

The parameters of the regression are:

β₀, β, and σ²

If the conditions are met then you can estimate the regression line:

Yi= bo * bXi + ei.

And the point estimation of the parameters can be calculated using the formulas:

β₀ ⇒ b0= (∑y/n)-b(∑x/n)

β ⇒ b= [∑xy- ((∑x)(∑y))/n]/(∑x²-((∑x)²/n))

σ²⇒ Se²= 1/(n-2)*[∑y²-(∑y)²/n - b²(∑x²-(∑x)²/n)]

n= 7

∑y= 501.80

∑y²= 36097.88

∑x= 240.80

∑x²= 8310.44

∑xy= 17204.87

b0= 144.59

b= -2.12

Se²= 1.02

The estimated regression line is:

Yi= 144.59 -2.12Xi

You need to calculate a 99%CI E(Y/X=35), the formula is:

(b0 + bX0) ± t_{n-2;1-\alpha /2}*\sqrt{S_e^2(\frac{1}{n}+\frac{(X_0-X[bar])^2}{sumX^2-(\frac{(sumX)^2}{n} )} )}

(144.59 + (-2.12*35)) ± 4.032*\sqrt{1.02(\frac{1}{7}+\frac{(35-34.4)^2}{8310.44-(\frac{(240.80)^2}{7} )} )}

[68.78; 71.99]

With a 99% confidence level youd expect that the interval [68.78; 71.99] contains the true value of the average of Y when X= 35.

I hope it helps!

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