Question

Nine homes are chosen at random from real estate listings in two suburban neighborhoods, and the square footage of each home is noted in the following table. Size of Homes in Two Subdivisions Square Footage 2,309 2,726 2,725 2,878 2,427 2666 2,737 2,885 2,544 Subdivision Pinewood 1 2,675 2,803 2.629 2,749 2,888 2,340 2,684 3,888 (a) Choose the appropriate hypothesis to test if there is a difference between the average sizes of homes in the two neighborhoods at the.10 significance level. Assume H1 is the mean of home sizes in Greenwood and y2 is the mean of home sizes in Pinewood. (b) Specify the decision rule with respect to the p-value. Reject the null hypothesis if the p-value is less than0.10 (c) Find the test statistic tcal- (A negative value should be indicated by a minus sign. Round your answer to 3 decimal places.) calc (d) Assume unequal variances to find the p-value. (Use the quick rule to determine degrees of freedom. Round your answer to 4 decimal places.) p-value (e) Make a decision. We do not reject te null hypothesis (f) State your conclusion. We cannot conclude that there is a difference between the sizes of homes in the two neighborhoods

Answer 1

For greenwood

$\dpi{100} \bar x1$ = 2655.2222, s1^2 = 37929

For Pinewood

$\dpi{100} \bar x2$ = 2836.3333, s2^2 = 182147

c) The test statistic t_calc = ($\dpi{100} \bar x1 - \bar x2$)/sqrt(s1^2/n1 + s2^2/n2)

                                      = (2655.2222 - 2836.3333)/sqrt(37929/9 + 182147/9)

                                      = -1.158

d) df = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

        = (37929/9 + 182147/9)^2/((37929/9)^2/8 + (182147/9)^2/8)

        = 11

P-value = 2 * P(T < -1.158)

             = 2 * 0.1357

             = 0.2714