Question

[a] We want to predict the price of houses from the size of the house (sarft). Please graph a scatterplot and see if the asso[d] Use the output below in order to write a conclusion for a 95% confidence interval for a prediction at a 2000 square feet

[a] We want to predict the price of houses from the size of the house (sarft). Please graph a scatterplot and see if the association is linear enough. (20 pts) There is a fairly linear association with very few outliers. Statkey Descriptive Statistics for Two Quantitative Variables Summary Statistics sih 013695 293546014 77.192 102713 445 650000 Sample Siee 0.788 140.211 11204 145 600000 550000 Scatterplot Controls a Show Regression Line 450000 300000 50000 1500 [b] Find the regression equation for predicting house price from sarft, (30 pts) Correlation between house price and size of house: r 0.788 Equation of a regression line for predicting food expenditure from income: y 11204.145 + 140211x Construct a 95% confidence interval for the slope and write a conclusion. (30 pts) [d] Use the output below in order to write a conclusion for a 95% confidence interval for a prediction at a 2000 square feet house. (5 pts)
[d] Use the output below in order to write a conclusion for a 95% confidence interval for a prediction at a 2000 square feet house. (5 pts) Predicted values: X value Pred. Y s.e.( Pred. y) 95% c.1. for mean 95% P.1. for new 2000 291626.16783.5336 (278140.88, 30511132) (164442.59, 418809.61) [e] Use the output below in order to write a conclusion for a 95% prediction interval for a prediction at a 2000 square feet house. Predicted values: x value Pred. Y s.e.(Pred. y) 95% CI. for mean 2000 2916261 6783.5336 (278140.88, 305111.32) (164442.59, 418809.61) (5 pts) 95%, P.1. for new [f] What is the difference between conclusions between part [d] and part [el above? (10 pts)
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Answer #1

c) 95% CI for slope is given by:

CI = Slope Coefficient +/- T-statistic * StdErrorOfSlope

= 140.211 +/- T88, 0.025 * Std Dev/\sqrt{n}

= 140.211 +/- 1.988 * 77.192 / \sqrt{88}

= (123.85, 156.57)

d) Confidence interval already given in the question is calculated from:

1(r- 7)2 n-2, (1-a/2) * yY n(n-1) *s2

=(278140.88, 305111.32)

e) Prediction interval is calculated as:

n- 1)*s prediction interval, Pl-y+-tn-2, (a-a/2) Sy

= (164442.59, 418809.61)

f) Prediction intervals account for the uncertainty in population mean, plus the data scatter. So, it is wider than the confidence interval.

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