hw1sol - STAT 4220 HW1 Solution 1 Problem 1 (a) Fitting the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: STAT 4220 HW1 Solution 1 Problem 1 (a) Fitting the model with all predictors yields the following output (see appendix for R code): > summary(mod1) Call: lm(formula = sr ~ ., data = savings) Residuals: Min 1Q Median 3Q Max-8.2422 -2.6857 -0.2488 2.4280 9.7509 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 28.5660865 7.3545161 3.884 0.000334 *** pop15-0.4611931 0.1446422-3.189 0.002603 ** pop75-1.6914977 1.0835989-1.561 0.125530 dpi-0.0003369 0.0009311-0.362 0.719173 ddpi 0.4096949 0.1961971 2.088 0.042471 *--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 3.803 on 45 degrees of freedom Multiple R-squared: 0.3385, Adjusted R-squared: 0.2797 F-statistic: 5.756 on 4 and 45 DF, p-value: 0.0007904 (b) The hypothesis is H : pop 15 = 0 vs. H 1 : pop 15 6 = 0. With p = . 002603 < . 01, we reject the null at the = . 01 level and conclude that pop15 is a significant predictor for the response....
View Full Document

Page1 / 3

hw1sol - STAT 4220 HW1 Solution 1 Problem 1 (a) Fitting the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online