A random sample of 10 one-bedroom apartments from your local newspaper has these
ID: 3442162 • Letter: A
Question
A random sample of 10 one-bedroom apartments from your local newspaper has these monthly rents (dollars) Does these data give good reason to believe that the mean rent for all advertised apartments is greater than $500 per month? State hypotheses, find t statistic and its P-value, and state your conclusion. Find a 95% confidence interval for the mean monthly rent for one-bedroom apartments available for rent. 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.Explanation / Answer
a)
Set Up Hypothesis
Null, Rent for all apartment is equals to 500 H0: U=500
Alternate, Rent for all apartment is greater than 500 H1: U>500
Test Statistic
Population Mean(U)=500
Sample X(Mean)=531
Standard Deviation(S.D)=82.792
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =531-500/(82.792/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 ) = 0.13336
Hence Value of P0.05 < 0.13336,Here We Do not Reject Ho
We conclude that Rent for all apartment is equals to 500
b)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=531
Standard deviation( sd )=82.792
Sample Size(n)=10
Confidence Interval = [ 531 ± t a/2 ( 82.792/ 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 )=82.792
Sample Size(n)=10
Margin of Error = t a/2 * 82.792/ Sqrt ( 10)
= 3.2498 * (26.18)
= 85.08
It is larger with 95% confidence
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