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 19 people making inquiries at the first development is $160,000, with a standard deviation of $37,000. A corresponding sample of 27 people at the second development had a mean of $181,000, with a standard deviation of $32,000. Assume the population standard deviations are the same. At the 0.01 significance level, can Fairfield conclude that the population means are different?

1. State the decision rule for 0.01 significance level: H₀: μ₁ = μ₂; H₁:μ₁ ≠ μ₂. (Negative values should be indicated by a minus sign.Round your answers to 2 decimal places.)

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

3. At the 0.01 significance level, can Fairfield conclude that the population means are different?  

  Reject/Do no reject H0. Fairfield can/cannot conclude that the population means are different

Answer 1

Part a

We are given

α = 0.01

df = n1 + n2 – 2

n1 = 19

n2 = 27

df = 19 + 27 – 2 = 44

So, critical values by using t table are given as below:

Critical values = -2.6923 and 2.6923

Decision rule:

Reject null hypothesis H₀ if test statistic t is not lies between -2.69 and 2.69, otherwise do not reject the null hypothesis H₀.

Part b

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

n1 = 19

n2 = 27

X1bar = 160000

X2bar = 181000

S1 = 37000

S2 = 32000

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(19 – 1)* 37000^2 + (27 – 1)* 32000^2]/(19 + 27 – 2)

S_p² = 1165136363.6364

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (160000 – 181000) / sqrt[1165136363.6364*((1/19)+(1/27))]

t = -21000 / 10221.3583

t = -2.0545

Test statistic = -2.05

Part c

Here, test statistic value is lies between the critical values -2.69 and 2.69, so we do not reject the null hypothesis H₀

Fairfield cannot conclude that the population means are different.