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
Subdivision	Square Footage
Greenwood	2,821	2,595	2,565	2,801	2,383	2,451	2,858	2,871	2,611
Pinewood	2,617	2,501	2,564	2,786	2,392	2,511	2,492	2,687	3,964

(b) Specify the decision rule with respect to the p-value.

Reject the null hypothesis if the p-value is (Click to select)greater thanless than 0.10.

(c) Find the test statistic t_calc. (A negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

t_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            

Answer 1

The t statistic to test whether the means are different can be calculated as follows:

t = (X bar1 − X bar2) /sqrt(2/n)*S_p

where S_p =sqrt((s_x^2+s_y^2)/2)

greenwood = c(2821,2595, 2565, 2801, 2383, 2451, 2858, 2871, 2611)

pinewood = c(2617,2501, 2564, 2786, 2392, 2511, 2492, 2687, 3964)

t.test(x = greenwood,y = pinewood,var.equal = TRUE,conf.level = 0.9)

Two Sample t-test

data: greenwood and pinewood
t = -0.36264, df = 16, p-value = 0.7216
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
-360.4905 236.4905
sample estimates:
mean of x mean of y
2661.778 2723.778

part b) pvalue = 0.7216 and alpha = 0.1

so we accept H_0 since pvalue>alpha (no singnificant )

c) test statistics t = -0.36264

d) unequal variance

t.test(x = greenwood,y = pinewood,var.equal = FALSE,conf.level = 0.9)

Welch Two Sample t-test

data: greenwood and pinewood
t = -0.36264, df = 10.263, p-value = 0.7242
alternative hypothesis: true difference in means is not equal to 0
90 percent confidence interval:
-371.0685 247.0685
sample estimates:
mean of x mean of y
2661.778 2723.778

p-value = 0.7242