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17 April, 01:40

A real estate company would like to see if the average sale price of townhouses in a particular school district has changed in the last 12 months. A study conducted 12 months ago indicated that the average sale price of condominiums in this locality was $280,000. Data on recent sales were as follows:

Sample size n 36

Mean $294,365

Std ... Deviation $22,898

Required:

a. Write down the null and alternative hypotheses for this problem.

b. Specify the rejection region for conducting a two-tail test at a significance level of 5%.

c. Based on the data in the table, would you reject or fail to reject the null hypothesis for a 5% significance level?

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  1. 17 April, 05:26
    0
    Step-by-step explanation:

    a) We would set up the hypothesis test.

    For the null hypothesis,

    H:0 µ = 280000

    For the alternative hypothesis,

    H1: µ ≠ 280000

    This is a 2 tailed test

    Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

    z = (x - µ) / (σ/√n)

    Where

    x = sample mean

    µ = population mean

    σ = population standard deviation

    n = number of samples

    From the information given,

    µ = 280000

    x = 294365

    σ = 22898

    n = 36

    z = (294365 - 280000) / (22898/√36) = 3.76

    b) Since α = 0.05, the critical value is determined from the normal distribution table.

    For the left, α/2 = 0.05/2 = 0.025

    The z score for an area to the left of 0.025 is - 1.96

    For the right, α/2 = 1 - 0.025 = 0.975

    The z score for an area to the right of 0.975 is 1.96

    In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

    Therefore, the rejection regions are area to the left of - 1.96 and to the right of 1.96 on the normal distribution curve.

    The calculated test statistic is 3.76 for the right tail and - 3.76 for the left tail

    c) Since - 3.76 1.96, we would reject the null hypothesis.
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