Question

You want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a random sample of 10 apartments advertised in the local newspaper is $2534. Assume that the standard deviation is $670. Find the 90%, 95%, and 99% confidence intervals for the mean monthly rent for this category of apartments. Look at the 95% confidence interval and say whether the following statement is true or false. "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area." Be sure to explain your answer.

Answer 1

Answer:

90% confidence interval: (2186.53;2881.47)

95% confidence interval: (2118.73;2949.27)

99% confidence interval: (1987.37;3080.63)

For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."

The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=2534$ represent the sample mean  

$\mu$ population mean (variable of interest)

$\sigma=670$ represent the population standard deviation

n=10 represent the sample size  

90% confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Now we have everything in order to replace into formula (1):

$2534-1.64\frac{670}{\sqrt{10}}=2186.53$    

$2534+1.64\frac{670}{\sqrt{10}}=2881.47$

So on this case the 90% confidence interval would be given by (2186.53;2881.47)

95% confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$2534-1.96\frac{670}{\sqrt{10}}=2118.73$    

$2534+1.96\frac{670}{\sqrt{10}}=2949.27$

So on this case the 95% confidence interval would be given by (2118.73;2949.27)

99% confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$2534-2.58\frac{670}{\sqrt{10}}=1987.37$    

$2534+2.58\frac{670}{\sqrt{10}}=3080.63$

So on this case the 99% confidence interval would be given by (1987.37;3080.63)

For the last part is not the best way say : "This interval describes the price of 95% of the rents of all the unfurnished one-bedroom apartments in the Boston area."

The best interpretation is this one: "We are 95% confident that the actual mean for the rents of unfurnished one-bedroom apartments in the Boston area is between (2118.73;2949.27)"