Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order
ID: 3156795 • Letter: F
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 $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 .05 significance level, can Fairfield conclude that the population means are different? 1. State the decision rule for 0.05 significance level: H0: 1 = 2; H1:1 2. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) and Compute Value of test Statistic.
Explanation / Answer
1) degrees of freedom is 15+25-2=38; reject H0 if test statistic t exceeds 2.024 in absolute value
2) pool data to get common sp=sqrt[((15-1)(40000)^2 + (25-1)(30000)^2)/(15+25-2)]=34027.85
test statistic is t=(150000 - 180000)/[34027.85sqrt(1/15 + 1/25)]= -2.70
reject H0
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