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Unformatted text preview: Stat 511 Homework 13 - Solutions Spring 2011 Due: 5pm, Tuesday Apr 26 1. The data in elnino.txt are records of sea surface temperature in four regions of the Pacific ocean from 1950 to current. The data labelled NINO1+2 is the average sea surface temper- ature for the extreme eastern equatorial Pacific. The actual temperature (in Nino1+2) is quite noisy because it includes seasonal variation. (Winter is cooler than summer, even in the tropics). The important column is labelled ANOM (just to the right of NINO1+2). This is the temperature anomaly (observed temperature - expected temperature for that month). Positive anomalies are times of El Nino conditions; negative anomalies are times of La Nina conditions. As an aside, El Nino/La Nina conditions have consequences throughout the west- ern hemisphere. For example, La Nina conditions are associated with June droughts in Iowa. Throughout, assume that observations are independent given the full model. This is not true for many full models, but I dont want to deal with that complication. (a) One crucial question is whether there is any trend in the temperature anomaly. Fit a linear regression, E ANOM = + 1 Date , where Date is Year + (Month-1)/12. Report the estimated slope and the p-value for the test of Ho: 1 = 0. 1 = 0 . 0046, p = 0.039 elnino <- read.table(elnino.txt,as.is=T,header=T) elnino$date <- elnino$YR + (elnino$MON-1)/12 summary(lm(ANOM~date,data=elnino)) (b) The trend may not be linear. Fit a loess curve, using f = 0.75. Use this to test whether there is lack of fit to a linear trend. Report your test statistic and p-value. Compare the linear span = 100 loess model to a linear span=0.75 loess model. F = 4.62, p = 0.0039. There is strong evidence that a straight line does not model the trend well. el1 <- loess(ANOM~date,data=elnino,degree=1, span=0.75) el0 <- loess(ANOM~date,data=elnino,degree=1, span=100) anova(el0,el1) You might also compare the linear span=100 loess model to a quadratic span = 0.75 model. That gives F = 6.55, p = 0.0078. Same conclusion....
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