The table below gives the number of absences and the overall grade in the class

Question

The table below gives the number of absences and the overall grade in the class for seven randomly selected students. Based on this data, consider the equation of the regression line, y ?? = b?_0 ? + b?_1 x, for using the number of absences to predict a student's overall grade in the class. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. Number of Absences 0 2 3 4 5 7 8 Grade 3.8 3.4 3 2.7 2.1 1.8 1.6 Step 1. Find the estimated slope. Round your answer to three decimal places. Answer: ____________________ Step 2. Find the estimated y-intercept. Round your answer to three decimal places. Answer: ____________________ Step 3. Find the estimated value of y when x = 7. Round your answer to three decimal places. Answer: ____________________ Step 4. According to the estimated linear model, if the value of the independent variable is increased by one unit, then the change in the dependent variable y ? is given by? A)?b_0 B)?b_1 C)?x D)?y Step 5. Determine the value of the dependent variable y ? at x = 0. A)?b_0 B)?b_1 C)?x D)?y Step 6. Find the value of the coefficient of determination. Round your answer to three decimal places. Answer: ____________________ Answer: ____________________

Explanation / Answer

let y be the grade and x be the number of absences.

y = bo+b1*x

y = 2.17827+0.11040*x

step 1) estimated slope b1 = 0.11040

step 2) y intercept bo = 2.17827

step 3) y(x =7)

y = 2.17827+.011040*7 =2.95107

step 4) according to the estimated linear model, if the value of the independent variable is increased by one unit then the change in dependent variable y is B) b_1

step5) y value at x = 0 is A) b_0

step 6) coefficient of determination that is R square = 0.1884.

Note: I have used R language to get the solution please go through the attached output.

x = c(0,2,3,4,5,7,8)
y =c( 3.8,3.4,3,2.7,2.1,1.8,1.6)
model.reg = lm(y~x)
summary(model.reg)

Call:

lm(formula = y ~ x)

Residuals:

Min 1Q Median 3Q Max

-1.17191 -0.36728 0.00931 0.32650 1.11133

Coefficients:

Estimate Std. Error t value Pr(>|t|)   

(Intercept) 2.17827 0.27482 7.926 1.24e-08 ***

x 0.11040 0.04331 2.549 0.0166 *  

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.5337 on 28 degrees of freedom

Multiple R-squared: 0.1884, Adjusted R-squared: 0.1594

F-statistic: 6.499 on 1 and 28 DF, p-value: 0.01656

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