Box Plots to Compare Medians of Home Prices
Determining if a mean price of a home with a pool is equal to the mean price of a home without will require the research team to calculate a score that will be compared with a critical value. A comparison of mean populations will allow the team to discover if the two variables vary significantly, which may answer many types of questions about homes with pools. For example, if homes with pools are significantly lower, would that mean the value is lower because of insurance costs? This example is one of many in which data can be enlightened to discover problems that may exist.
To answer the question if the median price of homes with pools is significantly equal to homes without pools requires the research team to identify the number of samples from both populations of data. Determining the number of samples will determine the statistical test that will used to test the null hypothesis. The number of samples used for the homes with pools is 36 and without pools the sample size is 67. Since both populations are n ≥ 30, a z- test will be used for solving the problem. To determine the answer to our research problem, a hypothesis test will provide the solution.
Step 1: Ho: µ1 = µ2
Ha: µ1 ≠ µ2
Step 2: The level of significance used for the research problem= Zα/2 0.05, which critical values for a two tailed test = -1.96 and 1.96. The decision rule states if z <-1.96 or z >1.96, then reject null hypothesis which states both means of populations are equal.
Step 3: The test statistic that will be used for the research study is a z-test:
Z= 201.49 -231.485 ÷ square root of 796.311÷36 + 2557.223 ÷ 67 = -29.995 ÷ 7.765= -3.863.
Step 4: Make the decision:
-3.863 < -1.96 or -3.863 > 1.96, which the first statement is true, so reject the null hypothesis.
Step: The research team has concluded that the means of the two populations are significantly unequal.
Descriptive Statistics: Address Box Plots
Please refer to the attachment for more data.