Question

Refer to Exercise \(13.5\).

a. Find and interpret a \(95%\) confidence interval for the mean monthly rent of newly completed apartments in the Midwest.

b. Find and interpret a \(95%\) confidence interval for the difference between the mean monthly rents of newly completed apartments in the Northwest and South.

c. What assumptions are you making in solving parts (a) and (b)?

Short answer

Part a.

Part b.

Part c.

• The sample should be independent.
• Data should be originating from the normal population (the samples are small then the central limit theorem will not apply).

• The population standard deviation in the Northeast is the same as the population standard deviation in the South.

Step-by-step solution

Part a. Step 1. Given information

The sample data is given

Part a. Step 2. Calculation

Calculate the mean and standard deviation using relation

\(\bar{x}=\frac{\sum_{i-1}^{6}x_{i}}{6}\) and \(\sigma=\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{N}}\)

After solving we will get

\(\bar{x}=743\)

\(\sigma=92.0695\)

Then calculate the marginal error

\(t_{\alpha/2}=2.571\)

\(E=t_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}=2.571\times \frac{92.0695}{\sqrt{6}}\)

After solving we will get

\(\approx 96.6367\)

Then calculate the upper and lower boundary of the confidence interval

\(\bar{x}-E=743-92.0695=646.36\)

\(\bar{x}+E=743+92.0695=839.64\)

So, the confidence interval will be

confidence interval \(=[646.36, 839.64]\)

Program:

Query:

• First, we have defined the random samples.
• Then calculate the 95% confidence interval.
• Sketch a plot of the confidence interval.

Part b. Step 1. Calculation

Calculate the mean and standard deviation using relation

\(\bar{x_{1}}=\frac{\sum_{i-1}^{5}x_{i}}{5}\) and \(\bar{x_{2}}=\frac{\sum_{i-1}^{4}x_{i}}{4}\)

After solving we will get

\(\bar{}x_{1}=1055.2\)

\(\bar{}x_{2}=854.5\)

Calculate the standard deviation for both the apartments using relation

\(\sigma_{1}=\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{5-1}}, \sigma_{2}=\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{4-1}}\)

Then calculate the pooled standard deviation

\(\sigma = \sqrt{\frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2}}\approx 250.7839\)

Calculate the marginal error

\(t_{\alpha/2}=2.571\)

\(E=t_{\alpha/2}\times \sigma \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}=2.571\times 250.7839 \sqrt{\frac{1}{5}+\frac{1}{4}}\)

After solving we will get

\(\approx 397.895\)

Then calculate the upper and lower boundary of the confidence interval

\((\bar{x_{1}-\bar{x_{2}}}-E=(1055.2-854.5)-397.8646=-197.1646\)

\((\bar{x_{1}-\bar{x_{2}}}+E=(1055.2-854.5)+397.8646=598.5646\)

So, the confidence interval will be

confidence interval \(=[-197.1646, 598.5646]\)

Program:

Query:

• First, we have defined the random samples.
• Then calculate the \(95%\) confidence interval.

• Sketch a plot of the confidence interval.

Part c. Step 1. Calculation

Since the examples are given to be autonomous arbitrary examples, we at that point need to expect that the information start from typical populaces and the populace standard deviation in the upper east is equivalent to the populace standard deviation in the south.