Question

An agent for a real estate company wanted to predict the monthly rent for apartments based on the size of the apartment. The data for a sample of 25 apartments is available below. Perform a t test for the slope to determine if a significant linear relationship between the size and the rent exists.

a. At the 0.05 level of​ significance, is there evidence of a linear relationship between the size of the apartment and the monthly​ rent?

b. Construct a​ 95% confidence interval estimate of the population​ slope, beta1.

-Find the test statistic. tSTATequals nothing ​(Round to two decimal places as​ needed.)

-Find the p value and round to three decimal places.

-Review the regression results and identify the bounds of the confidence​ interval, rounding to four decimal places.

Size_(sq._ft)   Rent_($)
860   1975
1450   2650
1085   2225
1232   2500
728   1950
1495   2675
1146   2650
716   1925
710   1850
966   2125
1100   2400
1285   2650
1975   3300
1379   2775
1175   2425
1225   2450
1255   2125
1249   2725
1140   2175
886   2125
1371   2625
1030   2650
765   2225
990   1800
1210   2750

Answer 1

Ʃx =	28423
Ʃy =	59725
Ʃxy =	70048050
Ʃx² =	34299715
Ʃy² =	145858125
Sample size, n =	25
x̅ = Ʃx/n = 28423/25 =	1136.92
y̅ = Ʃy/n = 59725/25 =	2389
SSxx = Ʃx² - (Ʃx)²/n = 34299715 - (28423)²/25 =	1985037.84
SSyy = Ʃy² - (Ʃy)²/n = 145858125 - (59725)²/25 =	3175100
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 70048050 - (28423)(59725)/25 =	2145503

a) Null and alternative hypothesis:

Ho: β₁ = 0 ; Ha: β₁ ≠ 0

α = 0.05

Slope, b = SSxy/SSxx = 2145503/1985037.84 = 1.080837

Sum of Square error, SSE = SSyy-SSxy²/SSxx = 3175100-(2145503)²/1985037.84 = 856160.26482

Standard error, se = √(SSE/(n-2)) = √(856160.26482/(25-2)) = 192.93615

Standard error for slope, se(b1) = se/√SSxx = 192.93615/√1985037.84 = 0.1369

Test statistic:

t = b1/se(b1) = 1.0808/0.1369 = 7.89

p-value = T.DIST.2T(ABS(7.8928), 23) = 0.000

Conclusion:

p-value < α Reject the null hypothesis.

b) Significance level, α = 0.05

Critical value, t_c = T.INV.2T(0.05, 23) = 2.0687

95% Confidence interval for slope:

Lower limit = b1 - tc*se(b1) = 1.0808 - 2.0687*0.1369 = 0.7976

Upper limit = b1 + tc*se(b1) = 1.0808 + 2.0687*0.1369 = 1.3641