SCI1020 Statistical Reasoning Question 4 (20 Marks) Is latitude a good predictor
ID: 3296627 • Letter: S
Question
SCI1020 Statistical Reasoning
Question 4 (20 Marks)
Is latitude a good predictor of the average winter temperature of a city in the USA? Consider the latitudes and the average temperatures in January for different cities of different latitude in USA. The data is given in the Excel file Temperature.xlsx (download from the SCI1020 Moodle page/Assignments.
a) Using appropriate graphs in Excel answer the following question. Is there a valid linear relationship between January temperature and latitude of the cities? Show and explain all Excel output that is your evidence for your answer.
Yes there is valid linear relationship between January Temperature and Latitude
Excel Analysis
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.869228922
R Square
0.755558919
Adjusted R Square
0.741978859
Standard Error
7.76199816
Observations
20
ANOVA
df
SS
MS
F
Significance F
Regression
1
3352.074922
3352.074922
55.63737686
6.54518E-07
Residual
18
1084.475078
60.24861544
Total
19
4436.55
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
127.9366563
12.36084161
10.3501574
5.24507E-09
101.9674917
153.9058209
Latitude
-2.408618899
0.322912433
-7.459046645
6.54518E-07
-3.087032747
-1.730205052
b) What is the best equation to relate these two variables? Do not use designations x and y ... use more descriptive symbols to show the variables.
Regression equation is January Temperature = 127.9367 - 2.4086*Latitude
c) What is the predicted January temperature for the city in the USA of Nashville which is at latitude 36N?
Predicted January temp for 36N latitude, just put latitude = 36 in part b] equation we get January temp
January Temp = 127.9367 - 2.4086*36 = 41.2263 that is the predicted January temperature for the city in USA of Nashville which Latitude is 36N = 41 degrees Fahrenheit
d) State the correlation value squared and state what this indicates.
Correlation value of squared, R squared = 0.7555 that is this regression model explained the 75.55% variation in January Temperature
e) What would change in the fitted equation and with correlation if you used degrees Celsius in place of degrees Fahrenheit for temperature?
After converts degree Fahrenheit to degree Celsius that is degree Celsius = (degree Fahrenheit - 32)*(5/9), we get
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.869228922
R Square
0.755558919
Adjusted R Square
0.741978859
Standard Error
4.3122212
Observations
20
ANOVA
df
SS
MS
F
Significance F
Regression
1
1034.591025
1034.591025
55.63737686
6.54518E-07
Residual
18
334.7145302
18.59525168
Total
19
1369.305556
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
53.29814238
6.867134228
7.761336914
3.76217E-07
38.87082873
67.72545603
Latitude
-1.338121611
0.179395796
-7.459046645
6.54518E-07
-1.715018193
-0.961225029
There is no change after using conversion of degrees Fahrenheit to degree Celsius
f) Find the equation of the regression line by hand by applying the formulae derived in the Method of Least Squares. You will need to use the value of correlation coefficient, r, from Excel, and other statistics in the data. [Your answer here should be the same equation that Excel has given you - check it is!]
(Only answer question F, the rest are already answered but the other information are required to answer question f)
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.869228922
R Square
0.755558919
Adjusted R Square
0.741978859
Standard Error
7.76199816
Observations
20
ANOVA
df
SS
MS
F
Significance F
Regression
1
3352.074922
3352.074922
55.63737686
6.54518E-07
Residual
18
1084.475078
60.24861544
Total
19
4436.55
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
127.9366563
12.36084161
10.3501574
5.24507E-09
101.9674917
153.9058209
Latitude
-2.408618899
0.322912433
-7.459046645
6.54518E-07
-3.087032747
-1.730205052
Explanation / Answer
(4)
(a)
In order to check the statistical significance of the slope, we check the corresponding p-value.
Given p-value = 6.54518E-07
This is very very small as compared to the generally chosen significance level of 0.05
So,
Yes the relationship between the latitude and the average winter temperatures is significant.
(f)
The regression equation is:
Y = aX + b
Here,
X = latitude
Y = Avg Winter temperatures
a = slope = r*(Sy/Sx)
Here, 'r' is Pearson coefficient, Sx is Standard deviation for X, Sy is Standard deviation for Y.
Putting values:
a = 0.86922*(Sy/Sx)
Next,
b = Y' - aX'
Here, Y' is the mean of Y values, and X' is the mean of X values.
You need to supply the five summary statistics, which will give us Sx, Sy, Y' and X', and then you can calculate 'a' and 'b' to construct the equation.
ANOVA output does not give the five summary statistics. Please provide the same.
Hope this helps !
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