Help me ..........................
1 answer:
You might be interested in
Answer:
A a house in that market that sells for $300,000 is unusual.
Step-by-step explanation:
Let the random variable <em>X</em> denote the price of a house and the random variable <em>Y</em> denote the living area of a house.
The number of houses surveyed by the real estate agent is, <em>n</em> = 1057.
Assume that both the random variables, <em>X </em>and <em>Y</em> are approximately normally distributed.
That is,
To compute the probability of a Normal distribution we first need to convert the raw scores to <em>z</em>-scores.
A <em>z</em>-score higher than 1.96 and lower than -1.96 are considered unusual. The values having these <em>z</em>-scores are considered as outliers.
(1)
Compute the <em>z</em>-score for <em>X</em> = 300000 as follows:
(2)
Compute the <em>z</em>-score for <em>Y</em> = 3000 as follows:
The <em>z</em>-score for a house in that market that sells for $300,000 is more than 1.96.
This implies, that the price $300,000 is unusually high.
The complete statement is:
The house that sells for $300 comma 000 has a z-score of <u>2.70</u> and the house with 3000 sq ft has a z-score of <u>1.19</u>.
Answer:
(a) The <em>p</em>-value of the test statistic is 0.147.
(b) The <em>p</em>-value of the test statistic is 0.294.
(c) The <em>p</em>-value of the test statistic is 0.8531.
(d) None of the <em>p</em>-values give strong evidence against the null hypothesis.
Step-by-step explanation:
The <em>p</em>-value is well defined as the probability,[under the null hypothesis (H₀)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic.
We reject a hypothesis if the p-value of a statistic is lower than the level of significance <em>α</em>.
The null hypothesis for the test of population proportion is defined as:
<em>H₀</em>: <em>p</em> = 0.50
The value of <em>z</em>-test statistic is,
<em>z</em> = 1.05
(a)
The alternate hypothesis is defined as:
<em>Hₐ</em>: <em>p</em> > 0.50
Compute the <em>p</em>-value of the test statistic as follows:
*Use a <em>z</em>-table for the probability value.
Thus, the <em>p</em>-value of the test statistic is 0.147.
(b)
The alternate hypothesis is defined as:
<em>Hₐ</em>: <em>p</em> ≠ 0.50
Compute the <em>p</em>-value of the test statistic as follows:
*Use a <em>z</em>-table for the probability value.
Thus, the <em>p</em>-value of the test statistic is 0.294.
(c)
The alternate hypothesis is defined as:
<em>Hₐ</em>: <em>p</em> < 0.50
Compute the <em>p</em>-value of the test statistic as follows:
*Use a <em>z</em>-table for the probability value.
Thus, the <em>p</em>-value of the test statistic is 0.8531.
(d)
The decision rule of the test is:
If the <em>p</em>-value of the test is less than the significance level <em>α</em>, then the null hypothesis is rejected at <em>α</em>% level of significance.
And if the <em>p</em>-value of the test is more than the significance level <em>α</em>, then the null hypothesis is failed to be rejected.
The most commonly used level of significance are:
<em>α</em> = 0.01, 0.05 and 0.10
The <em>p</em>-value for all the three alternate hypothesis are:
<em>p-</em>values = 0.147, 0.294 and 0.8531.
All the <em>p</em>-values are quite large compared to the <em>α</em> values.
Thus, none of the <em>p</em>-values give strong evidence against the null hypothesis.
The null hypothesis was failed to be rejected.