Question

The Department of Housing and Urban Development (HUD) would like to test the hypothesis that the average size of a newly constructed house in 2010 is different from the average size of a newly constructed house in 2000. The following data summarizes the sample statistics for house sizes, in square feet, for both years. Assume that the population variances are equal.

2000 2010
Sample mean 2,180 2,390
Sample size 15 12
Sample standard deviation 300 320
If Population 1 is defined as homes built in 2000 and Population 2 is defined as homes built in 2010, which one of the following statements is true?

A. Because the
95
%
confidence interval includes zero, HUD cannot 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.

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.

C. Because the
95
%
confidence interval includes zero, HUD can conclude that the average size of a newly constructed house in 2010 is equal to the average size of a newly constructed house in 2000.

D. Because the
95
%
confidence interval does not include zero, HUD can conclude that the average size of a newly constructed house in 2010 is not different from the average size of a newly constructed house in 2000.

Answer 1

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.

Step-by-step explanation:

Here,

Null and alternative hypotheses are:

H0: u1 = u2

H1: u1 ≠ u2

Calculate test statistics:

`t = \frac{x'1 - x'2}{\sqrt{((n_1 - 1) (\sigma_1)^2 + (n_2 - 1)(\sigma_2)^2)/(n_1 + n_2 - 2) * ((1)/(n_1) + (1)/(n_2))}}`

`= \frac{2180 - 2390}{\sqrt{((14)(300)^2 + (11)(320)^2))/(15 +12 - 2) * ((1)/(15) + (1)/(12))}}`

`= \frac{-210}{\sqrt{((1260000) + (1126400))/(25) * (0.15)}}`

`t = -1.7549`

At 95% confidence interval, find t observed:

Significance level = 100% - 95% = 5% = 0.05

Degrees of freedom = 15 + 12 - 2 = 25

`t_o = t_\alpha_/_2_, _d_f = t_0_._0_5_/_2_, _2_5 = t_0_._0_2_5, _2_5 = 2.06`

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.