A real estate company is analyzing the selling prices of residential homes in a given community.140 homes that have been sold in the past month are randomly selected and their selling prices are recorded.The statistician working on the project has stated that in order to perform various statistical tests,the data must be distributed according to a normal distribution.In order to determine whether the selling prices of homes included in the random sample are normally distributed,the statistician divides the data into 6 classes of equal size and records the number of observations in each class.She then performs a chi-square goodness of fit test for normal distribution.The results are summarized in the following table.
Goodness of Fit Test $$\begin{array} { r | r | r | r | r }
\text { Observed } & \text { Expected } & \mathrm { O } - \mathrm { E } & ( \mathrm { O } - \mathrm { E } ) ^ { 2 } \mathrm { E } & \% \text { of chi-square } \
\hline 10 & 3.192 & 6.808 & 14.520 & 64.81 \
\hline 23 & 19.026 & 3.974 & 0.830 & 3.70 \
\hline 37 & 47.782 & - 10.782 & 2.433 & 10.86 \
\hline 40 & 47.782 & - 7.782 & 1.267 & 5.66 \
\hline 27 & 19.026 & 7.974 & 3.342 & 14.92 \
\hline 3 & 3.192 & - 0.192 & 0.012 & 0.05 \
\hline 140 & 140.000 & 0.000 & 22.404 & 100.00 \
\hline & & & & \
\hline 22.40 & \text { chi-square } & & & \
\hline .0001 & \text { p-value } & & &
\end{array}$$
At a significance level of .05,we:

Question

A real estate company is analyzing the selling prices of residential homes in a given community.140 homes that have been sold in the past month are randomly selected and their selling prices are recorded.The statistician working on the project has stated that in order to perform various statistical tests,the data must be distributed according to a normal distribution.In order to determine whether the selling prices of homes included in the random sample are normally distributed,the statistician divides the data into 6 classes of equal size and records the number of observations in each class.She then performs a chi-square goodness of fit test for normal distribution.The results are summarized in the following table.
Goodness of Fit Test $$\begin{array} { r | r | r | r | r } 
\text { Observed } & \text { Expected } & \mathrm { O } - \mathrm { E } & ( \mathrm { O } - \mathrm { E } ) ^ { 2 } \mathrm { E } & \% \text { of chi-square } \
\hline 10 & 3.192 & 6.808 & 14.520 & 64.81 \
\hline 23 & 19.026 & 3.974 & 0.830 & 3.70 \
\hline 37 & 47.782 & - 10.782 & 2.433 & 10.86 \
\hline 40 & 47.782 & - 7.782 & 1.267 & 5.66 \
\hline 27 & 19.026 & 7.974 & 3.342 & 14.92 \
\hline 3 & 3.192 & - 0.192 & 0.012 & 0.05 \
\hline 140 & 140.000 & 0.000 & 22.404 & 100.00 \
\hline & & & & \
\hline 22.40 & \text { chi-square } & & & \
\hline .0001 & \text { p-value } & & &
\end{array}$$
At a significance level of .05,we:

Quizplus · Accepted Answer

The null hypothesis (H₀) is that the selling prices of residential homes are normally distributed. The alternative hypothesis is that they are not normally distributed. At a .05 significance level, we would reject the null hypothesis and conclude that the data is not normally distributed if the p-value is less than .05. In this case, the p-value is .0001, which is less than .05, so we reject the null hypothesis and conclude that the selling prices of residential homes in this community are not normally distributed. Therefore, the best choice is A.

A Real Estate Company Is Analyzing the Selling Prices of Residential