Five new homes are built in a neighborhood with an average value of $168,500. Th

Question

Five new homes are built in a neighborhood with an average value of $168,500. The average value of homes in this area has been 159,750 with the standard deviation of $6,000. A one-sample z-test was conducted to determine if these new home values significantly differ from the norm in this area.

A. What is the name of the dependent variable?

B. Why would you conduct a z-test and not a t-test?

C. Would it be a one-tailed or a two-tailed test? Explain.
D. State the null and research hypotheses for this test in symbols and in words.

E. What critical value would you use for this test at alpha level of .05? Make sure to report the sign of this value (+, -, or +/-).

F. What would be the test statistic? Show your work.
G. Write your statistical decision and conclusions in full sentences (Step 4).

H. How would you interpret the result of the test? Write in full sentences. Make sure you answer the research question to the end, that is, report whether the new home values are higher or lower, if the result is significant, and explain how you determined this

Explanation / Answer

Given that,
population mean(u)=159750
standard deviation, =6000
sample mean, x =168500
number (n)=5
null, Ho: =159750
alternate, H1: !=159750
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 168500-159750/(6000/sqrt(5)
zo = 3.26093
| zo | = 3.26093
critical value
the value of |z | at los 5% is 1.96
we got |zo| =3.26093 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 3.26093 ) = 0.00111
hence value of p0.05 > 0.00111, here we reject Ho
ANSWERS
---------------

B. Z test
C.null, Ho: =159750
alternate, H1: !=159750
test statistic: 3.26093
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00111

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