The equation of the least squares regression line is y 2.0233x -2.3256 and the s

Question

The equation of the least squares regression line is y 2.0233x -2.3256 and the sum of squared residuals for this line is 0.7907. Answer parts (a)(d) x 3 457 8 (a) Predict the mean value of y #x-13. (Round to one decimal place as needed) (b) Construct a 95% confidence interval about the mean value of y if, 13 Lower Bound Upper Bound Round to one decimal place as needed) (c) Predict the value of y if x= 13. y(Round to one decimal place as needed) (d) Construct a 95% prediction interval about the value of y if x-13. Lower Bound Upper Bound Round to one decimal place as needed.) Enter your answer in each of the answer boxes

Explanation / Answer

y = c(4,6,7,12,14)
x =c(3,4,5,7,8)
df= as.data.frame(cbind(y,x))
model.lm= lm(y~x,df)
summary(model.lm)

Call:

lm(formula = y ~ x, data = df)

Residuals:

1 2 3 4 5

0.2558 0.2326 -0.7907 0.1628 0.1395

Coefficients:

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

(Intercept) -2.3256 0.7068 -3.29 0.046066 *  

x 2.0233 0.1238 16.34 0.000498 ***

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

Residual standard error: 0.5134 on 3 degrees of freedom

Multiple R-squared: 0.9889, Adjusted R-squared: 0.9852

F-statistic: 267.1 on 1 and 3 DF, p-value: 0.0004984

Part A)  predict the mean response of y value at x =13

y = model.lm$coefficients[1]+model.lm$coefficients[2]*13

y = 23.97674

Part B) construct the confidence interval of the

mean of y at x =13

predict(model.lm, newdata = new.dat, interval = 'confidence',level = 0.95)

Part C) predict the value of y at x =13

23.97674

Part d) construct a 95 percent prediction interval about the value of y if x =13.

predict(model.lm, newdata = new.dat, interval = 'prediction',level = 0.95)

Note: Notice the fitted value is the same as before, but the interval is wider. This is due to the additional term in the standard error of prediction. It should be noted prediction and confidence intervals are similar in that they are both predicting a response, however, they differ in what is being represented and interpreted. The best predictor of a random variable (assuming the variable is normally distributed) is the mean ?. The best predictor for an observation from a sample of x data points, x1,x2,?,xn and error ? is x¯. Since the prediction interval must take into account the variability of the estimators for ? and ?, the interval will be wider.

Upper Bound 27.05863 Lower Bound 20.89485

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