A regression was done for 20 cities with the latitude (Latitude), a measurement of the distance of a location on the Earth from the equator in degrees and the average January temperatures (Temp) as the dependent variable measure in. The regression equation is a) The value of the slope is [Select] [Select] TEMP=48.4 -0.31 x LATITUDE b) The predicted average January temperatures for the 20 cities for a latitude of 45 degrees is [Select] ✓. If in fact the [Select] average January temperature for a latitude of 45 degrees is 40° F the residual is [Select] which is a(n) which means [Select] c) The coefficient of determination is R² = 66.2%. This value means [Select] The correlation r, is which means [Select] d) The y-intercept is [Select] the problem interprets to [Select] make sense in the context of the problem? [Select] which in the context of Dos this value

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A regression was done for 20 cities with the latitude (Latitude), a
measurement of the distance of a location on the Earth from the
equator in degrees and the average January temperatures (Temp)
as the dependent variable measure in. The regression equation
is
a) The value of the slope is [Select ]
[Select]
TEMP=48.4 -0.31 x LATITUDE
b) The predicted average January temperatures for the 20 cities
for a latitude of 45 degrees is [Select]
✓. If in fact the
[Select]
average January temperature for a latitude of 45 degrees is 40° F
, the residual is [Select]
which is a(n)
[Select]
which means
c) The coefficient of determination is R²: 66.2%. This value
means [Select]
The correlation r, is
which means [Select]
d) The y-intercept is [Select]
the problem interprets to [Select]
make sense in the context of the problem? [Select]
Formulas: r = ±√R²
Residual actual y - predicted y
0.7<|rl≤1 very strong
0.5<Irl≤0.7 strong
0.3<Irl≤0.5 moderate
0.1<Irl≤0.3 weak
Irl≤0.1 no correlation
which in the context of
Dos this value
Transcribed Image Text:A regression was done for 20 cities with the latitude (Latitude), a measurement of the distance of a location on the Earth from the equator in degrees and the average January temperatures (Temp) as the dependent variable measure in. The regression equation is a) The value of the slope is [Select ] [Select] TEMP=48.4 -0.31 x LATITUDE b) The predicted average January temperatures for the 20 cities for a latitude of 45 degrees is [Select] ✓. If in fact the [Select] average January temperature for a latitude of 45 degrees is 40° F , the residual is [Select] which is a(n) [Select] which means c) The coefficient of determination is R²: 66.2%. This value means [Select] The correlation r, is which means [Select] d) The y-intercept is [Select] the problem interprets to [Select] make sense in the context of the problem? [Select] Formulas: r = ±√R² Residual actual y - predicted y 0.7<|rl≤1 very strong 0.5<Irl≤0.7 strong 0.3<Irl≤0.5 moderate 0.1<Irl≤0.3 weak Irl≤0.1 no correlation which in the context of Dos this value
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