A Fruit and Vegetable Market Evaluation in Sydney and Melbourne – Report Example

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The paper "A Fruit and Vegetable Market Evaluation in Sydney and Melbourne" is a perfect example of a report on statistics. From the pivot table above, the total production in winter exceeded the total produce in summer by 106. Also, for all types of fruits, the produce in winter was generally more than the summer produce. The only similarity is that 26 fruits were produced in summer both for the oranges and tomatoes, 26 was also the highest number of fruits produced by type in summer. In winter there was no equal produce, however, oranges were more produced at 60 fruits whereas tomatoes were the least produced at 42 fruits. Pivot Table B   Column Labels         1   2   Row Labels Average of Price $/kg Average of Quantity sold (ton/day) Average of Price $/kg Average of Quantity sold (ton/day) 1 3.565714286 286.177619 5.6128 101.9612 2 5.46 83.54692308 2.968666667 324.5673333 3 3.1056 360.3368 1.534615385 443.7503846 4 2.717692308 399.9669231 4.514285714 65.89380952 Grand Total 3.725918367 281.5256122 3.569411765 247.1306863 Summer had the highest average of price in dollars per kilogram for oranges (5.46) whereas winter had the highest average of price in dollars per kilogram for apples (5.6128). Winter had the highest average of quantity sold in tonnes per day for oranges (324.567) whereas summer had the highest average of quantity sold in tonnes per day for apples (268.178). T-test Summary statistics for oranges in winter in price in dollars per kilogram Oranges Price $/kg in Winter     Mean 2.968667 Standard Error 0.108383 Standard Deviation 0.593637 Hypotheses: H0: The average price of $/kg in winter is less than the summer price of $5.50 H1: The average price of $/kg in winter is not less than the summer price of $5.50 Assumption: The level of significance is 0.05 Test statistic: Computation: = -23.487.

Absolute answer = 23.487 Critical value: = 1.701 Conclusion: Since 23.487 > 1.701, we fail to reject the null hypothesis and conclude that the average price of $/kg in winter is less than the summer price of $5.50. Scatter Plot   Nature and Relationship between Price $/kg and Quantity ton/day The scatter plot shows a downhill pattern moving from left to right indicative of a negative relationship between the price in dollars per kilogram and the quantity sold in tonnes per day.

As the price in dollars per kilogram increases, the quantity sold in tonnes per day decreases. Regression Analysis Excel Output SUMMARY OUTPUT                                 Regression Statistics               Multiple R 0.861065               R Square 0.741433               Adjusted R Square 0.740127               Standard Error 76.88356               Observations 200                                 ANOVA                   df SS MS F Significance F       Regression 1 3356064 3356064 567.758 4.59E-60       Residual 198 1170394 5911.082           Total 199 4526459                                 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 580.7437 14.36244 40.4349 1.24E-97 552.4207 609.0666 552.4207 609.0666 Price $/kg -86.8762 3.646023 -23.8277 4.59E-60 -94.0663 -79.6862 -94.0663 -79.6862                                                       Regression equation Quantity sold in tonnes per day = 580.744 – 86.876 Price in dollars per kilogram R-squared and Standard error R-squared is 0.741.

This represents the Quantity sold in tonnes per day that is explained by the model. The standard error is 76.884. This represents the average distance that the quantity sold in tonnes per day falls from the regression line. The gradient of the regression line The gradient of the regression line is – 86.876. This represents the variance in quantity in tonnes per day depending on the price in dollars per kilogram. Significance of Gradient and Intercept To test the significance of the gradient and the intercept, I pick on the last output from the sorted random data pictured in this paper (200th variable).

The actual figures for the quantity sold in tonnes per day and the price in dollars per kilogram are 60.80 and 3.74 respectively. Using the regression equation, the following are the workings: Quantity sold in tonnes per day = 580.744 – 86.876 (3.74) = 255.828 From the above result, I can conclude that the value of the intercept and the gradient are not significant. Good model? I think the regression model is a good model. This is because the value of R-squared is quite high, also the regression sum of squares is high and the error the sum of squares is low.

In addition, the p-value is low, the F-statistic is high and the adjusted R-squared is high. All these parameters are indicative of a good model.

References

Freedman, D.H et al, 2007, Statistics, 4th edn, New York, W.W Norton & Company.
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