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)
Set Up Hypothesis
Null,mean rent for all advertised apartments is equals to 500 H0: U=500
Alternate; the mean rent for all advertised apartments is greater than H1: U>500
Test Statistic
Population Mean(U)=500
Sample X(Mean)=531
Standard Deviation(S.D)=82.79
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =531-500/(82.79/Sqrt(9))
to =1.184
| to | =1.184
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |to| =1.184 & | t | =1.833
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Right Tail - Ha : ( P > 1.1841 ) = 0.13335
Hence Value of P0.05 < 0.13335,Here We Do not Reject Ho

We conclude that mean rent for all advertised apartments is equals to 500

b)
Mean(x)=531
Standard deviation( sd )=82.79
Sample Size(n)=10
Confidence Interval = [ 531 ± t a/2 ( 82.79/ Sqrt ( 10) ) ]
= [ 531 - 2.2622 * (26.18) , 531 + 2.2622 * (26.18) ]
= [ 471.77,590.23 ]

c)
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
Mean(x)=531
Standard deviation( sd )=82.79
Sample Size(n)=10
Margin of Error = t a/2 * 82.79/ Sqrt ( 10)
= 3.2498 * (26.18)
= 85.08
It is larger