Question

Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, it uses different media to reach potential buyers. The mean annual family income for 15 people making inquiries at the first development is $156,000, with a standard deviation of $44,000. A corresponding sample of 27 people at the second development had a mean of $182,000, with a standard deviation of $32,000. Assume the population standard deviations are the same. At the 0.05 significance level, can Fairfield conclude that the population means are different?

1. State the decision rule for 0.05 significance level: H₀: μ₁ = μ₂; H₁:μ₁ ≠ μ₂. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

2. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

3. At the 0.05 significance level, can Fairfield conclude that the population means are different?

Answer 1

=156000, =182000

s1​=44000, s2​=32000

n1​=15, n2​=27

= 0.05

Hypothesis is

Ho :

Ha:

a)

df = n1 + n2 -2 = 15 + 27 -2 = 40

now calculate critical value for two tailed test with df = 40 and = 0.05

to determine decesion rule.

we get critical value as

critical values are = ( - 2.021 , 2.021 )

Reject Ho if t is not between - 2.021 and 2.021

b)

Compute the value of the t test statistic

formula is

t test statistic = −2.203

c)

since (t test statistic = −2.203 ) < (critical value = −2.021 )

Reject the Null Hypothesis

Yes, there is enough significant evidence to conclude that the population means are different.