Drilling down beneath a lake in Alaska yields chemical evidence of past changes
ID: 3200798 • Letter: D
Question
Drilling down beneath a lake in Alaska yields chemical evidence of past changes in climate. Biological silicon, left by the skeletons of single-celled creatures called diatoms, measures the abundance of life in the lake. A rather complex variable based on the ratio of certain isotopes relative to ocean water gives an indirect measure of moisture, mostly from snow. As we drill down, we look farther into the past. Here are data from 2300 to 12,000 years ago:
(a) Make a scatterplot of silicon (response) against isotope (explanatory).
Ignoring the outlier, describe the direction, form, and strength of the relationship. The researchers say that this and relationships among other variables they measured are evidence for cyclic changes in climate that are linked to changes in the sun's activity.
weak positive associationmoderate positive association strong positive associationweak negative associationmoderate negative associationstrong negative association
Isotope(%) Silicon
(mg/g) Isotope
(%) Silicon
(mg/g) Isotope
(%) Silicon
(mg/g) 19.90 99 20.71 152 21.63 226 19.84 108 20.80 265 21.63 235 19.46 116 20.86 271 21.19 188 20.20 141 21.28 296 19.37 339 Isotope Silicon (mg/g) 100 150 200 250 300 350 9,5 20.0 20,5 21.0 21,5 22.0
Explanation / Answer
Answer:
(a).
Graph 2 (top right).
Ignoring the outlier, describe the direction, form, and strength of the relationship. The researchers say that this and relationships among other variables they measured are evidence for cyclic changes in climate that are linked to changes in the sun's activity.
moderate negative association
(b) Find the single outlier in the data. This point strongly influences the correlation. What is the correlation with this point? (Round your answer to two decimal places.)
outlier = (-19.37, 339)
Correlation = -0.33
What is the correlation without this point? (Round your answer to two decimal places.)
Correlation = -0.78
(c) Is the outlier also strongly influential for the regression line? Calculate the regression line with the outlier. (Round your slope to two decimal places, round your y-intercept to one decimal place.)
y = -486.0 33.49 x
Calculate the regression line without the outlier. (Round your slope to two decimal places, round your y-intercept to one decimal place.)
y = -1370.6 –75.5 x
Draw on your graph the two regression lines.
Graph 2 (top right)
Regression Analysis
r²
0.112
n
12
r
-0.334
k
1
Std. Error
79.562
Dep. Var.
Silicon
ANOVA table
Source
SS
df
MS
F
p-value
Regression
7,944.1459
1
7,944.1459
1.25
.2888
Residual
63,301.8541
10
6,330.1854
Total
71,246.0000
11
Regression output
confidence interval
variables
coefficients
std. error
t (df=10)
p-value
95% lower
95% upper
Intercept
-485.9716
615.4440
-0.790
.4481
-1,857.2662
885.3231
Isotope
-33.4899
29.8950
-1.120
.2888
-100.1002
33.1203
without outlier
Regression Analysis
r²
0.614
n
11
r
-0.784
k
1
Std. Error
46.778
Dep. Var.
Silicon
ANOVA table
Source
SS
df
MS
F
p-value
Regression
31,375.1196
1
31,375.1196
14.34
.0043
Residual
19,693.4259
9
2,188.1584
Total
51,068.5455
10
Regression output
confidence interval
variables
coefficients
std. error
t (df=9)
p-value
95% lower
95% upper
Intercept
-1,370.6426
412.5543
-3.322
.0089
-2,303.9054
-437.3799
Isotope
-75.4904
19.9360
-3.787
.0043
-120.5888
-30.3920
Regression Analysis
r²
0.112
n
12
r
-0.334
k
1
Std. Error
79.562
Dep. Var.
Silicon
ANOVA table
Source
SS
df
MS
F
p-value
Regression
7,944.1459
1
7,944.1459
1.25
.2888
Residual
63,301.8541
10
6,330.1854
Total
71,246.0000
11
Regression output
confidence interval
variables
coefficients
std. error
t (df=10)
p-value
95% lower
95% upper
Intercept
-485.9716
615.4440
-0.790
.4481
-1,857.2662
885.3231
Isotope
-33.4899
29.8950
-1.120
.2888
-100.1002
33.1203
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