Realtor Elaine Snyderman took a random sample of 12 homes in a prestigious suburb of
Chicago and found the average appraised market value to be $780,000, and the standard
deviation was $49,000. Test the hypothesis that for all homes in the area, the mean appraised
value is $825,000 against the alternative that it is less than $825,000. Use the 0.05 level of
significance

Question

09-11-2021
Mathematics

contestada

Realtor Elaine Snyderman took a random sample of 12 homes in a prestigious suburb of
Chicago and found the average appraised market value to be $780,000, and the standard
deviation was $49,000. Test the hypothesis that for all homes in the area, the mean appraised
value is $825,000 against the alternative that it is less than $825,000. Use the 0.05 level of
significance

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joaobezerra joaobezerra · Answer 1 · 2021-11-15T14:02:24+01:00

Using the t-distribution, it is found that since the p-value of the test is of 0.0044 < 0.05, it can be concluded that found the average appraised market value is less than $825,000.

The null hypothesis is:

[tex]H_0: \mu = 825000[/tex]

The alternative hypothesis is:

[tex]H_1: \mu < 825000[/tex]

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

[tex]\overline{x}[/tex] is the sample mean.

[tex]\mu[/tex] is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

For this problem, the values of the parameters are: [tex]\overline{x} = 780000, \mu = 825000, s = 49000, n = 12[/tex].

The value of the test statistic is:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{780000 - 825000}{\frac{49000}{\sqrt{12}}}[/tex]

[tex]t = -3.18[/tex]

The p-value of the test is found using a left-tailed test, as we are testing if the mean is less than a value, with t = -3.18 and 12 - 1 = 11 df.

Using a t-distribution calculator, it is of 0.0044.

Since the p-value of the test is of 0.0044 < 0.05, it can be concluded that found the average appraised market value is less than $825,000.

A similar problem is given at https://brainly.com/question/25369247