Fairfield Homes is developing two parcels near Pigeon Fork, Tennessee. In order

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 12 people making inquiries at the first development is $153,000, with a standard deviation of $42,000. A corresponding sample of 24 people at the second development had a mean of $171,000, 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?

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.)

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

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

Explanation / Answer

n1 = 12
x1bar = 153000
s1 = 42000

n2 = 24
xbar2 = 171000
s2 = 30000

alpha = 0.05

Compute the standard error (SE) of the sampling distribution.
SE = sqrt[ (s12/n1) + (s22/n2) ]

= sqrt((42000^2)/12 + (30000^2)/24)
= 15383

calculate df

. The degrees of freedom (DF) is:
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }

(((42000^2)/12 + (30000^2)/24))^2) / ( [ (((42000^2)/12)^2) / (12 - 1) ] + [ ((302000^2)/24)^2) / (24 - 1) ] )
= 0.053
consider df =1

e test statistic is a t statistic (t) defined by the following equation.
t = [ (x1 - x2) ] / SE

(153000 - 171000)/15383 = -1.17

calculate the t critical value for alpha =0.05 , df = 1 and t stat = -1.17

The P-Value is .450228.

The result is not significant at p < .05 , so we fail to reject the null hypothesis
and we conclude that

H0: 1 = 2

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