Answer:
There is enough evidence at the α-: 0.05 level of significance to support the claim that the true population mean market value of houses in the neighborhood where Rebecca works is greater than $250,000.
Step-by-step explanation:
We are given that Rebecca randomly selects 35 houses in the neighborhood and finds that the sample mean market value is $259,860 with a sample standard deviation of $24.922.
Let = <u><em>population mean market value of houses in the neighborhood.</em></u>
So, Null Hypothesis, :
= $250,000 {means that the population mean market value of houses in the neighborhood where she works is equal to $250,000}
Alternate Hypothesis, :
> $250,000 {means that the population mean market value of houses in the neighborhood where she works is greater than $250,000}
The test statistics that would be used here <u>One-sample t-test statistics</u> because we don't know about population standard deviation;
T.S. = ~
where, = sample mean market value = $259,860
s = sample standard deviation = $24,922
n = sample of houses = 35
So, <em><u>the test statistics</u></em> = ~
= 2.34
The value of t-test statistic is 2.34.
<u>Also, P-value of the test statistics is given by;</u>
P-value = P( > 2.34) = 0.0137
Since our P-value is less than the level of significance as 0.0137 < 0.05, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.
Therefore, we conclude that the population mean market value of houses in the neighborhood where she works is greater than $250,000.