A sample of 25 values for mont
A sample of 25 values for monthly rent of two-bedroom apartmentswas selected from a recent edition of the Gainsville Sun.The distribution of the 25 values is symmetric, with no outliers.The mean of this sample was $575, and you may estimate that thepopulation standard deviation was $165.
a. A group of students think the true mean rent for this type ofapartment is over $650. Is there statistical evidence in the datato reject their claim?
b. A newspaper reports that the average rent for two-bedroomapartments last year was $500. Is there statistical evidence of achange in mean rent from last year to this?
c. Could you have made the decision in part b based only on aconfidence interval? Explain.
Answer:
NULL HYPOTHESIS H0: 
ALTERNATIVE HYPOTHESIS Ha: 
alpha=0.05
n=25
Z= 575-650/165/sqrt(25)
Z= -75/165/5
Z= -75/33
Z= -2.27
The P-Value is .016236.The result is significant at p <.05.
We reject the null hypothesis H0 and conclude that the true meanrent for this type of apartment is over $650.
b) NULL HYPOTHESIS H0: 
ALTERNATIVE HYPOTHESIS Ha:
alpha= 0.05
Z= 575-500/165/sqrt(25)
Z= 75/165/5
Z= 75/33
Z= 2.27
ALPHA= 0.05
The P-Value is .032473. The result is significant at p <.05.
CONCLUSION: Since P value is significant we therefore rejectnull hypothesis and conclude that there is significant differencein means.
c) Mean = $575critical value= 1.96sM = √(165^2/25) = 33μ = M ± Z(sM)μ = 575 ± 1.96*33μ = 575 ± 64.68
CI [510.32, 639.68].
You can be 95% confident that the population mean (μ)falls between 510.32 and 639.68.
Since Null hypothesis value not in interval hence confidenceinterval significant therfore I would have made the same decisionas in part b.
NOTE: ALPHA AND CONFIDENCE LEVEL WAS NOT GIVEN SO I ASSUMED HEREALPHA=0.05 AND CONFIDENCE LEVEL 0.95
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