This question consists of multiple parts The excel files are below Question 4 (2

Question

This question consists of multiple parts
The excel files are below
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 Temperaturexlsx (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 b) What is the best equation to relate these two variables? Do not use designations x and c What is the predicted January temperature for the city in the USA of Nashville which d) State the correlation value squared and state what this indicates. explain all Excel output that is your evidence for your answer y... use more descriptive symbols to show the variables. is at latitude 36N? e What would change in the fitted equation and with correlation if you used degrees Celsius in place of degrees Fahrenheit for temperature? Find the equation of the regression line by hand by applying the formulae derived in the Method of Least Squares. Y will need to use the value of correlation co- efficient, 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!] f)

Explanation / Answer

a] Using Excel graphs

Yes there is valid linear relationship between January Temp and Latitude

Excel Analysis

b]   regression equation is January Temp = 127.9367 - 2.4086*Latitude

c] 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 tenp for the city in USA of Nashville which Latitude is 36N = 41 deg Fah

d] Correlation value of squared, R squared = 0.7555 that is this regression model explained the 75.55% variation in January Temp.

e]

After converts deg Fah to deg Celsius that is deg Celsius = (deg Fah - 32)*(5/9), we get

there is no change after using conversion of deg Fah to deg Celsius

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.