Question

A sample of 25 values for monthly rent of two-bedroom apartments was 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 the population standard deviation was $165.

a. A group of students think the true mean rent for this type of apartment is over $650. Is there statistical evidence in the data to reject their claim?

b. A newspaper reports that the average rent for two-bedroom apartments last year was $500. Is there statistical evidence of a change in mean rent from last year to this?

c. Could you have made the decision in part b based only on a confidence interval? Explain.

Answer 1

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 mean rent 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 reject null hypothesis and conclude that there is significant difference in means.

c) Mean = $575
critical value= 1.96
sM = √(165^2/25) = 33

μ = M ± Z(sM)
μ = 575 ± 1.96*33
μ = 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 confidence interval significant therfore I would have made the same decision as in part b.

NOTE: ALPHA AND CONFIDENCE LEVEL WAS NOT GIVEN SO I ASSUMED HERE ALPHA=0.05 AND CONFIDENCE LEVEL 0.95