Question
I need help constructing a conclusion to my slide presentation. I will include information I have gathered. Question: Residents of Melbourne are complaining about the
I need help constructing a conclusion to my slide presentation. I will include information I have gathered.
Question:
Residents of Melbourne are complaining about the high cost of housing within a 41-kilometer radius (about 25 miles) of the Central Business District, saying that even apartments are too expensive to purchase. Residents are pressuring city government leaders to address the need for affordable housing. At the same time, real-estate investors are downplaying the residents' concerns, saying that in fact, housing is affordable in Melbourne, and the real-estate investors are encouraging city-government leaders to focus their efforts on other issues of concern. Both the residents and real-estate investors presented their findings on apartment prices before the city government leaders.Your job is to determine the validity of each of their findings and to report your findings in the form of your own letter to city government officials. Write a conclusion that summarizes your findings and provides a recommendation to the city government officials regarding apartment prices in Melbourne.
The hypothesis to be tested is
H0 :The average apartment price is equal to $453,993.94. i.e., =453993.94
H1 :The average apartment price is higher than $453,993.94. i.e., >453993.94
Given that
Sample size n=300
Sample mean x= 612466.1462
Sample standard deviation s=247989.7244
Population mean or hypothesized mean =453993.94
As the population standard deviation is unknown, we have to use a t-test for one sample mean.
The test statistic (t) value is given by t=
The degrees of freedom=n-1=300-1=299.
The p-value for t=11.0683 with degrees of freedom is 0.00.
Decision: As the p-value(0.00) is less than the level of significance(0.05)(P-value < ), we reject the null hypothesis.
Conclusion: Hence we can conclude that the average apartment price is higher than $453,993.94
The p-value is 2.15*10-24=0.000.
Confidence interval for the mean
T interval
Given Sample Statistics
sample mean
612466.1462
Change
sample size
300
sample standard deviation
247989.7244
confidence level
Critical Value
0
Margin of Error
28176.21366
Confidence Interval .95
Lower Bound
584337.1568
Upper Bound
640595.1355
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