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Chapter 6: Q. 6.13 (page 285)

The Bureau of Labor Statistics publishes information on average annual expenditures by consumers in the Consumer Expenditure Survey. In 2010, the mean amount spent by consumers on nonalcoholic beverages was $333. A random sample of 12consumers yielded the following data, in dollars, on last year's expenditures on nonalcoholic beverages.

Part (a): Use Table III in Appendix A to construct a normal probability plot of the given data.

Part (b): Use part (a) to identify any outliers.

Part (c): Use part (a) to access the normality of the variable under consideration.

Short Answer

Expert verified

Part (a): The required normal probability plot is given below,

Part (b): On identifying the outliers, we get 472.

Part (c): The expenditures on nonalcoholic beverages does not follow normal distribution.

Step by step solution

01

Part (a) Step 1. Given information.

Consider the given question,

02

Part (a) Step 2. Construct a normal probability plot.

Use Table III in Appendix A to construct the normal probability plot for the data on expenditures on nonalcoholic beverages.

A sample of 12 consumers on expenditures on nonalcoholic beverages is considered. From Appendix A, Table III, Normal scores, the z-scores corresponding to the column n=12for the data is tabulated below,


03

Part (a) Step 3. Plot the points.

Plot the points using horizontal axis for the final exam scores and the vertical axis for the normal scores,

04

Part (b) Step 1. Identify any outliers.

Consider the normal probability plot from part (a),

From the probability plot, it is clear one point is far away from other points. That is the expenditure 472represents an outlier.

05

Part (c) Step 1. Access the normality of the variable under consideration.

Consider the normal probability plot from part (a),

From the probability of expenditures, it is clear tht all the points are not close to the line. In other words, the points are non linear.

Hence, it can be concluded that the expenditures on nonalcoholic beverages does not follow normal distribution.

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