Question

Fairfield Homes is developing two parcels near Pigeon Forge, 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 $150,000, with a standard deviation of $40,000. A corresponding sample of 25 people at the second development had a mean of $180,000, with a standard deviation of $30,000. Assume the population standard deviations are the same. At the 0.05 significance level, can Fairfield conclude that the population means are different? a. State the decision rule for 0.05 significance level: He: M1 = U2; H1:41 € 42. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.) Reject HO ift is not between and b. Compute the value of the test statistic. (Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.) Value of the test statistic

Answer 1

Part a)

Reject null hypothesis if | t | > t(α/2, n1 + n2 - 2)
Critical value t(α/2, n1 + n1 - 2) = t(0.05 /2, 15 + 25 - 2) = 2.024

Reject H0 if t is not between -2.024 and 2.024

Part b)

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))
$S_{p} = ( \sqrt{( n1 - 1 )S_{1}^{2} + (n2 - 1 )S_{2}^{2} ) / ( n1 + n2 - 2 )}$
$S_{p} = (\sqrt{( 15 - 1 ) 40000^{2} + ( 25 - 1 ) 30000 ^{2} ) / ( 15 + 25 - 2 )}$

t = ( 150000 - 180000) / 34027.8524 √ ( ( 1 / 15) + (1 / 25 ))
t = -2.70

Part c)

| t | > t(α/2, n1 + n2 - 2) = 2.6994 > 2.024
Result :- Reject Null Hypothesis

Reject H0, fairfield is sufficient evidence to conclude that the population means are different.