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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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