# 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