A random sample of 10 onebedroom apartments from your local

A random sample of 10 one-bedroom apartments from your local newspaper has these monthly rents (dollars)

                        500      650      600      505      450      550      515      495      650      395

Does these data give good reason to believe that the mean rent for all advertised apartments is greater than $500 per month?

(a)State hypotheses, find t statistic and its P-value, and state your conclusion.

(b)Find a 95% confidence interval for the mean monthly rent for one-bedroom apartments available for rent.

(c)If you chose 99% rather than 95% confidence, would your margin of error in the part (b) be larger or smaller? Verify your answer by doing the calculation.

Solution

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   500  
Ha:    u   >   500   [ANSWER, HYPOTHESES]
              
As we can see, this is a    right   tailed test.      
              
          
              
Getting the test statistic, as              
              
X = sample mean =    531          
uo = hypothesized mean =    500          
n = sample size =    10          
s = standard deviation =    82.7915723          
              
Thus, t = (X - uo) * sqrt(n) / s =    1.184065053 [T STATISTIC]          
              
Also, the p value is

df = n - 1 =    9              
              
p =    0.133357594   [ANSWER, p value]      
              
Comparing t and tcrit (or, p and significance level), we   FAIL TO REJECT THE NULL HYPOTHESIS.      

Thus, there is no significant evidence that the mean rent for all advertised apartments is greater than $500 per month. [CONCLUSION]

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b)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    500          
t(alpha/2) = critical t for the confidence interval =    2.262157163          
s = sample standard deviation =    82.7915723          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
              
Lower bound =    440.774477          
Upper bound =    559.225523          
              
Thus, the confidence interval is              
              
(   440.774477   ,   559.225523   ) [ANSWER]

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c)

It will be smaller, as the confidence level is larger.

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    500          
t(alpha/2) = critical t for the confidence interval =    3.249835542          
s = sample standard deviation =    82.7915723          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
              
Lower bound =    414.9160753          
Upper bound =    585.0839247          
              
Thus, the confidence interval is              
              
(   414.9160753   ,   585.0839247   ) [ANSWER]

As we can see, this interval is wider, hence, larger margin of error.