Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order to test different advertising approaches, they use different media to reach potential buyers. The mean annual family income for 75 people making inquiries at the first development is $150,000, with a standard deviation of $40,000. A corresponding sample of 120 people at the second development had a mean of $180,000, with a standard deviation of $30,000. At the .05 significance level, can Fairfield conclude that the population means are different?© BrainMass Inc. brainmass.com June 3, 2020, 5:06 pm ad1c9bdddf
The test statistic is:
t= (180000-150000)/sqrt[40000^2/75+30000^2/120]=5.59 and is distributed as t ...
With a brief calculation of the test statistic, degrees of freedom, and critical value, this solution determine whether the null hypothesis should be rejected or not.