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 19 people making inquiries at the first development is $148,000, with a standard deviation of $41,000. A corresponding sample of 25 people at the second development had a mean of $186,000, with a standard deviation of $27,000. Assume the population standard deviations are the same. 1. State the decision rule for .05 significance level: H0: μ1 = μ2; H1:μ1 ≠ μ2. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.) compute the value of the test statistics. at the 0.05 significance level can Fairfield conclude that the population means are different?

Answer 1

Answer:

The test statistic t = 3.61270

The tabulated value t₀.₀₅ =  3.61270 is greater than the tabulated value 2.4185 at 0.05 level of significance.

The null hypothesis is rejected at 0.05 level of significance.

the population means are not different

Step-by-step explanation:

Step(i):-

Given first sample size 'n₁' = 19

Given the mean annual family income for 19 people making inquiries at the first development is $148,000, with a standard deviation of $41,000

The mean of first sample 'x₁⁻ ' = $148,000

The standard deviation of first sample S₁ = $41,000

Given data  a corresponding sample of 25 people at the second development had a mean of $186,000, with a standard deviation of $27,000

The second sample size n₂ = 25

The mean of second sample 'x₂⁻ = $186,000

The standard deviation of first sample S₂ = $27,000

Step(ii) :-

Null hypothesis : H₀ : μ₁= μ₂

Alternative hypothesis :H₁: μ₁≠μ₂

Level of significance ∝ = 0.05

Degrees of freedom : ν = n₁+n₂-2 = 19+25-2 = 42

Test of hypothesis

                                        $t=\frac{x^{-} _{1}-x^{-} _{2} }{\sqrt{S^2(\frac{1}{n_{1} } +\frac{1}{n_{2} } } )}$

where

                                        $S^{2} = \frac{n_{1}S_{1} ^{2} +n_{2} S^{2} _{2} }{n_{1}+n_{2} -2 }$

on calculation , we get

                                       $S^{2} = \frac{19(41,000) ^{2} +25(27,000)^2 }{19+25-2 } = 1,194,380,952.381$

                                   

                                       $t=\frac{1,48,000 -1,86,000 }{\sqrt{1,194,380,952.381(\frac{1}{19 } +\frac{1}{25 } } )}$

on calculation , we get

                                      $t = \frac{-38,000}{10,518.4311} =-3.61270$

  Taking modulus

                              |t| = |-3.61270|

                               t = 3.61270

Step(iii):-

The degrees of freedom ν = n₁+n₂-2 = 19+25-2 = 42

The tabulated value t₀.₀₅ =  3.61270 is greater than the tabulated value 2.4185 at 0.05 level of significance.

Conclusion:-

The null hypothesis is rejected at 0.05 level of significance.

we accepted alternative hypothesis that is  H1:μ1 ≠ μ2

the population means are not different