The US Census Bureau collect data on monthly rents of newly completed apartments and publishes the results, in Current Housing Reports. Independent random samples of newly completed apartments in the four US regions yielded the data on monthly rents, in dollars given on WeissStats site. At the \(5%\) significance level, do the data provide sufficient evidence to conclude that a difference exists in mean monthly rents among newly completed apartments in the four US regions?

a. conduct a one-way ANOVA test on the data

b. Interpret your results from part (a)

c. decide whether presuming that the assumptions of normal populations and equal population standard are met is reasonable.

The data on monthly rents in dollars for independent random samples of newly completed apartments in the four U.S regions are given.

The four independent random samples of apartments have the rents as shown below:

The aim is to find \(95%\) confidence interval for the apartment Midwest

Calculate the mean and Standard deviation t0 the Midwest:

Mean\(=\frac{\sum_{i-1}^{n}X_{i}}{n}\)

\(=\frac{(870+748+...+606)}{6}\)

Mean\(=743.00\)

Standard deviation \(=\frac{\sum_{i-1}^{n}(X_{i}-\bar{X})^{2}}{n-1}\)

\(=\frac{(870-743)^{2}+(748-743)^{2}+..+(606-743)^{2}}{6-1}\)

Standard deviation \(=92.07\)

The \(95%\) confidence interval is defined as follows:

\(=\left ( \bar{x}-t_{(\alpha/2,n-1)}\times\left ( \frac{s}{\sqrt{n}} \right ),\bar{x}+t_{(\alpha/2,n-1)}\times\left ( \frac{s}{\sqrt{n}} \right ) \right )\)

\(=\left ( 743-t_{(0.05/2.5)}\times\left ( \frac{92.07}{\sqrt{6}} \right )743+t_{(0.05/2.5)}\times\left ( \frac{92.07}{\sqrt{6}} \right ) \right )\)

\(=\left ( 743-2.5705\times\left ( \frac{92.07}{\sqrt{6}} \right )743+2.5705\times\left ( \frac{92.07}{\sqrt{6}} \right ) \right )\)

\(=\left ( 743-96.6215,743+96.6215 \right )\)

\(=\left ( 646.3784,839.6215 \right )\)

Hence, the 95% confidence interval to the Midwest is \(\left ( 646.3784,839.6215 \right )\)

The aim is to find \(95%\) confidence interval for difference between Northeast and South apartments.

Let 'x' denote the apartment ‘north east’.

Let 'y' denote the apartment ‘south'.

The \(95%\) confidence interval for the difference of means is defined as follows:

\(\left ( {\bar{x}}- \bar{y}\right )\pm t_{(0.05/2,n_{1}+n_{2}-2)^{s}}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}\)

where, \(s=\sqrt{\frac{(n_{1}-1)s_{x}^{2}+(n_{2}-1)s_{y}^{2}}{n_{1}+n_{2}-2}}\)

\(\bar{x},\bar{y}s_{x}^{2}\) and \(s_{y}^{2}\)

\(\bar{x}=\frac{(1005+898+..+1244)}{5}\)

\(\bar{x}=1055.20\)

\(\bar{y}=\frac{(891+630+861+..+1036)}{4}\)

\(\bar{y}=854.50\)

\(s_{x}^{2}=\frac{\sum_{i-1}^{n}(X_{i}-\bar{X})^{2}}{n-1}\)

\(=\frac{(1005-1055.2)^{2}+(898-1055.2)^{2}+..+(1244-1055.2)^{2}}{n-1}\)

\(s_{x}^{2}= 22548.0256\)

\(s_{y}^{2}=\frac{\sum_{i-1}^{n}(Y_{i}-\bar{Y})^{2}}{n-1}\)

\(=\frac{(891-854.5)^{2}+(630-854.5)^{2}+..+(1036-854.5)^{2}}{4-1}\)

\(s_{y}^{2}= 28239\)

Substitute the values of \(n_{1}, n_{2},s_{1}^{2}\) and \(s_{2}^{2}\) in \(S\)

Therefore,

\(s=\sqrt{\frac{(n_{1}-1)s_{x}^{2}+(n_{2}-1)s_{y}^{2}}{n_{1}+n_{2}-2}}\)

\(= \sqrt{\frac{(5-1)22548.0256+(4-1)28239}{5+4-2}}\)

\(=\sqrt{24987.4}\)

\(=158.074\)

Now, find the confidence interval:

\(\left ( {\bar{x}}- \bar{y}\right )\pm t_{(0.05/2,n_{1}+n_{2}-2)^{s}}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}\)

\(=\left ( 1055.20-854\right )\pm t_{(0.05/2.5+4-2)}\cdot 158.074 \sqrt{\frac{1}{5}+\frac{1}{4}}\)

\(=200.700\pm 2.36462\times 158.074 \sqrt{\frac{1}{5}+\frac{1}{4}}\)

\(=200.700\pm 250.743\)

\((-50.043, 451.443)\)

Hence, the \(95%\) confidence interval for difference between Northeast and south apartments is \((-50.043, 451.443)\).

The aim is to write the assumptions to solve part [a] and [b].

The assumptions to construct \(95%\) confidence interval are as follows:

1. The populations are normally distributed.

2. Samples are drawn independently.

3. The two populations have same variance.