Question

city	Lat	Temp
Akron, OH	41.05	27
Albany-Schenectady-Troy, NY	42.4	23
Allentown, Bethlehem, PA-NJ	40.35	29
Atlanta, GA	33.45	45
Baltimore, MD	39.2	35
Birmingham, AL	33.31	45
Boston, MA	42.15	30
Bridgeport-Milford, CT	41.12	30
Buffalo, NY	42.54	24
Canton, OH	40.5	27
Chattanooga, TN-GA	35.01	42
Chicago, IL	41.49	26
Cincinnati, OH-KY-IN	39.08	34
Cleveland, OH	41.3	28
Columbus, OH	40	31
Dallas, TX	32.45	46
Dayton-Springfield, OH	39.54	30
Denver, CO	39.44	30
Detroit, MI	42.06	27
Flint, MI	43	24
Grand Rapids, MI	43	24
Greensboro-Winston-Salem-High Point, NC	36.04	40
Hartford, CT	41.45	27
Houston, TX	29.46	55
Indianapolis, IN	39.45	29
Kansas City, MO	39.05	31
Lancaster, PA	40.05	32
Los Angeles, Long Beach, CA	34	53
Louisville, KY-IN	38.15	35
Memphis, TN-AR-MS	35.07	42
Miami-Hialeah, FL	25.45	67
Milwaukee, WI	43.03	20
Minneapolis-St. Paul, MN-WI	44.58	12
Nashville, TN	36.1	40
New Haven-Meriden, CT	41.2	30
New Orleans, LA	30	54
New York, NY	40.4	33
Philadelphia, PA-NJ	40	32
Pittsburgh, PA	40.26	29
Portland, OR	45.31	38
Providence, RI	41.5	29
Reading, PA	40.2	33
Richmond-Petersburg, VA	37.35	39
Rochester, NY	43.15	25
St. Louis, MO-IL	38.39	32
San Diego, CA	32.43	55
San Francisco, CA	37.45	48
San Jose, CA	37.2	49
Seattle, WA	47.36	40
Springfield, MA	42.05	28
Syracuse, NY	43.05	24
Toledo, OH	41.4	26
Utica-Rome, NY	43.05	23
Washington, DC-MD-VA	38.5	37
Wichita, KS	37.42	32
Wilmington, DE-NJ-MD	39.45	33
Worcester, MA	42.16	24
York, PA	40	33
Youngstown-Warren, OH	41.05	28

Answer 1

You forgot to add question.

From your data, I think you want to fit regression equation for temp as dependent variable and lat as independent var.

If you want to do something else on this data please comment I will provide you answer accordingly.

Regression analysis: I used R software to solve this problem

R code:

> lat=scan("clipboard")
Read 59 items
> temp=scan("clipboard")
Read 59 items
> head(lat)
[1] 41.05 42.40 40.35 33.45 39.20 33.31
> head(temp)
[1] 27 23 29 45 35 45
> fit=lm(temp~lat)
> fit

Call:
lm(formula = temp ~ lat)

Coefficients:
(Intercept) lat
118.14 -2.15

> summary(fit)

Call:
lm(formula = temp ~ lat)

Residuals:
Min 1Q Median 3Q Max
-10.2978 -2.6353 -0.8719 0.3965 23.6789

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 118.139 6.743 17.52 <2e-16 ***
lat -2.150 0.171 -12.57 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 5.272 on 57 degrees of freedom
Multiple R-squared: 0.735, Adjusted R-squared: 0.7303
F-statistic: 158.1 on 1 and 57 DF, p-value: < 2.2e-16

Regression equation:

temp= 118.14 -2.15 lat

Adjusted R²= 0.7303 it means lat variable explain 73.03 % of variation in temp variable.