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

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)"

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)(\sigma)/(√(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(670)/(√(10))=2186.53`

`2534+1.64(670)/(√(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)(\sigma)/(√(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(670)/(√(10))=2118.73`

`2534+1.96(670)/(√(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)(\sigma)/(√(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(670)/(√(10))=1987.37`

`2534+2.58(670)/(√(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)"