Respuesta :
Answer:
B. Because the
95
%
confidence interval does not include zero, HUD can conclude that the average size of a newly constructed house in 2010 is different from the average size of a newly constructed house in 2000.
Explanation:
Here,
Null and alternative hypotheses are:
H0: u1 = u2
H1: u1 ≠ u2
Calculate test statistics:
[tex] t = \frac{x'1 - x'2}{\sqrt{\frac{(n_1 - 1) (\sigma_1)^2 + (n_2 - 1)(\sigma_2)^2}{n_1 + n_2 - 2} * (\frac{1}{n_1} + \frac{1}{n_2})}} [/tex]
[tex] = \frac{2180 - 2390}{\sqrt{\frac{(14)(300)^2 + (11)(320)^2)}{15 +12 - 2} * (\frac{1}{15} + \frac{1}{12})}} [/tex]
[tex] = \frac{-210}{\sqrt{\frac{(1260000) + (1126400)}{25} * (0.15)}} [/tex]
[tex] t = -1.7549 [/tex]
At 95% confidence interval, find t observed:
Significance level = 100% - 95% = 5% = 0.05
Degrees of freedom = 15 + 12 - 2 = 25
[tex] t_o = t_\alpha_/_2_, _d_f = t_0_._0_5_/_2_, _2_5 = t_0_._0_2_5, _2_5 = 2.06 [/tex]
T calculated = -1.76
T observed(critical) = -2.06
Since t calculated is bigger than t critical, reject null hypothesis H0.
Because the
95
%
confidence interval does not include zero, HUD can conclude that the average size of a newly constructed house in 2010 is different from the average size of a newly constructed house in 2000.