Question

In: Statistics and Probability

An entrepreneur owns some land that he wishes to develop. He identifies two development options: build...

An entrepreneur owns some land that he wishes to develop. He identifies two development options: build condominiums or build apartment buildings. Accordingly, he reviews public records and derives the following summary measures concerning annual profitability based on a random sample of 35 for each such local business venture. For the analysis, he uses a historical (population) standard deviation of $21,800 for condominiums and $19,600 for apartment buildings. Use Table 1. Sample 1 represents condominiums and Sample 2 represents apartment buildings. Condominiums Apartment Buildings x⎯⎯1 = $247,600 x⎯⎯2 = $235,900 n1 = 35 n2 = 35 a. Set up the hypotheses to test whether the mean profitability differs between condominiums and apartment buildings. H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0 H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0 H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0 b. Compute the value of the test statistic and the corresponding p-value. (Round "test statistic" value to 2 decimal places and "p-value" to 3 decimal places.) Test statistic p-value c-1. At the 10% significance level, what is the conclusion to the test? H0. At either the 10% significance levels, we conclude the mean profitability differs between condominiums and apartment buildings. c-2. At the 1% significance level, what is the conclusion to the test? H0. At either the 1% significance levels, we conclude the mean profitability differs between condominiums and apartment buildings.

Solutions

Expert Solution

Part a

Here, we have to use two sample z test for population means.

The required null and alternative hypotheses are given as below:

H0: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

Part b

The test statistic formula is given as below:

Z = (X1bar – X2bar) / sqrt[(σ12 / n1)+(σ22 / n2)]

We are given

X1bar = 247600

X2bar = 235900

σ1 = 21800

σ2 = 19600

n1 = 35

n2 = 35

Z = (247600 – 235900) / sqrt[(21800^2/35) + (19600^2/35)]

Z = 11700 / 4955.2281

Z = 2.3611

Test statistic = 2.36

P-value = 0.018

(by using z-table)

Part C1

We are given α = 0.10

P-value = 0.018 < α = 0.10

So, we reject the null hypothesis

At the 10% significance levels, we conclude the mean profitability differs between condominiums and apartment buildings.

Part C2

We are given α = 0.01

P-value = 0.018 > α = 0.10

So, we do not reject the null hypothesis

At the 1% significance levels, we can’t conclude the mean profitability differs between condominiums and apartment buildings.


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