In: Statistics and Probability
(Round all intermediate calculations to at least 4 decimal places.)
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 ventures. For the analysis, he uses a historical (population) standard deviation of $21,800 for condominiums and $19,600 for apartment buildings. (You may find it useful to reference the appropriate table: z table or t table)
Sample 1 represents condominiums and Sample 2 represents apartment buildings.
Condominiums | Apartment Buildings |
x¯1x¯1 = $247,600 | x¯2x¯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. Calculate the value of the test statistic. (Round your answer to 2 decimal places.)
c. Find the p-value.
p-value < 0.01
d-1. At the 10% significance level, what is the conclusion to the test?
d-2. At the 1% significance level, what is the conclusion to the test?
Here in this scenario our claim is that there is difference between the mean profitability condominiums and apartment buildings. To test this claim we selected the sample of size 35 each buildings.
Now to test this claim we have to use two sample z test because here the population standard deviations is known and sample size is large enough.
So using the given sample information we performed two sample z test at 0.01 level of significance as below,
The z critical value is calculated using Standerd normal z-table.
a. the hypotheses to test whether the mean profitability differs between condominiums and apartment buildings.
H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0
b. The value of the test statistic zo = 2.36. (Rounded to 2 decimal places.)
c. The p-value. = 2*P(z< 2.35).
P-value= 0.082
The p value is calculated using Standerd normal z-table.
d-1. At the 10% significance level, what is the conclusion to the test?
Since p-value is less than 0.10 level of significance so we Reject Ho null hypothesis and concluded that there is enough evidence to support claim that there is difference between mean profitability condominiums and apartment buildings.at 10% level of significance. Our result is significant.
d-2. At the 1% significance level, what is the conclusion to the test?
Since p value is greater than alpha level of significance so we fail to Reject null hypothesis and concluded that there is not enough evidence to support claim that there is difference between mean profitability condominiums and apartment buildings.
At 1% level of significance our result is not significant.
Thank you.