Sea surface temperatures across much of the tropics have been increasing since t

Question

Sea surface temperatures across much of the tropics have been increasing since the mid-1970s. At the same time, the growth of coral has been decreasing. Scientists examined data on mean sea surface temperatures (SST) in degrees Celsius and mean coral growth in millimeters (mm) per year over a several-year period at locations in the Red Sea. Here are the data. (Data set may be found here.) SST 29.68 29.87 30.16 30.22 30.48 30.65 30.90 Growth 2.63 2.58 2.68 2.60 2.48 2.38 2.26 (a) Do the data indicate that coral growth decreases linearly as SST increases? Is this change statistically significant? O There is strong evidence of an absence of any linear relationship between SST and Coral growth. O We can not conclude the evidence of linear relationship between SST and Coral growth. O There is strong evidence of a positive linear relationship between SST and Coral growth There is strong evidence of a negative linear relationship between SST and Coral growth. mm per year when SSTis 30 (b Use the data to predict with 95% confidence the mean coral growth 005 degrees Celsius. Round your ans es re deci mal places. to .010

Explanation / Answer

> x <- c(29.68,29.87,30.16,30.22,30.48,30.65,30.90)
> x
[1] 29.68 29.87 30.16 30.22 30.48 30.65 30.90
> y <- c(2.63,2.58,2.68,2.6,2.48,2.38,2.26)
> mod = lm(y ~x)
> summary (mod)

Call:
lm(formula = y ~ x)

Residuals:
       1        2        3        4        5        6        7
-0.06755 -0.05997 0.12792 0.06610 0.02490 -0.02358 -0.06782

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept) 11.69213    2.39615    4.88 0.00455 **
x           -0.30305    0.07913   -3.83 0.01225 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.08327 on 5 degrees of freedom
Multiple R-squared: 0.7458,    Adjusted R-squared: 0.6949
F-statistic: 14.67 on 1 and 5 DF, p-value: 0.01225

a) r = -0.7458 ,

option D) is correct

b)

predict(mod, data.frame(x= 30),interval="confidence")
    fit             lwr       upr
1    2.600569 2.501632 2.699506

( 2.502,2.700)

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